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O, CONTENTS, , v, vii, , FOREWORD, PREFACE, CHAPTER, NINE, CH T, RAY OPTICS AND, AN OPTICAL, AL INSTRUMENTS, T, T, 9, .1, 9.1, 9 .2, 9 .3, 9 .4, 9 .5, 9 .6, 9 .7, 9 .8, 9 .9, , Introduction, Reflection of Light by Spherical Mirrors, Refraction, Total Internal Reflection, Refraction at Spherical Surfaces and by Lenses, Refraction through a Prism, Dispersion by a Prism, Some Natural Phenomena due to Sunlight, Optical Instruments, , 309, 310, 316, 319, 323, 330, 332, 333, 335, , C, A, R TEN, CHAPTER, WAV, P I, AVE OPTICS, 1 0.1, 1 0.2, 1 0.3, 0.4, 10.4, 1, 0.5, 10.5, 1, 1 0.6, 1 0.7, , Introduction, Huygens Principle, Refraction and reflection of plane waves using Huygens Principle, Coherent and Incoherent Addition of Waves, Interference of Light Waves and Young’s Experiment, Diffraction, Polarisation, , C, A, EV, CHAPTER, ELEVEN, DUAL NATURE, AT O, R OF RADIATION, 1 1.1, 1 1.2, 1 1.3, 1 1.4, 1 1.5, 1 1.6, 1 1.7, 1 1.8, 1 1.9, , AN, AND, , 351, 353, 355, 360, 362, 367, 376, , T, MATTER, , Introduction, Electron Emission, Photoelectric Effect, Experimental Study of Photoelectric Effect, Photoelectric Effect and Wave Theory of Light, Einstein’s Photoelectric Equation: Energy Quantum of Radiation, Particle Nature of Light: The Photon, Wave Nature of Matter, Davisson and Germer Experiment, , 386, 387, 388, 389, 393, 393, 395, 398, 403

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CHAPTER, C A, R TWELVE, E, A TOMS, OM, 12, 12.1, 12, 12.2, 12, 12.3, 12, 12.4, 12, 12.5, 12, 12.6, , Introduction, Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom, Atomic Spectra, Bohr Model of the Hydrogen Atom, The Line Spectra of the Hydrogen Atom, DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation, , 414, 415, 420, 422, 428, 430, , C, A, CHAPTER, THIRTEEN, EI, NUCLEI, 13, 13.1, 13.2, 13, 13.3, 13, 13, 13.5, 13, 13.6, 13, 13.7, , Introduction, Atomic Masses and Composition of Nucleus, Size of the Nucleus, Nuclear Force, Radioactivity, Nuclear Energy, , 438, 438, 441, 445, 446, 451, , C, A, R FOURTEEN, R, CHAPTER, ELECTRONICS, SEMICONDUCTOR, I, N S: MATERIALS, DEVICES AND SIMPLE, P, RCU, CIRCUITS, 14, 14.1, 14, 14.2, 14, 14.3, 14, 14.4, 14.5, 14, 14, 14.6, 14, 14.7, 14, 14.8, 14, 14.9, 1 4.10, 1 4.11, , Introduction, Classification of Metals, Conductors and Semiconductors, Intrinsic Semiconductor, Extrinsic Semiconductor, p-n Junction, Semiconductor diode, Application of Junction Diode as a Rectifier, Special Purpose p-n Junction Diodes, Junction Transistor, Digital Electronics and Logic Gates, Integrated Circuits, , C, A, IFTEE N, CHAPTER, FIFTEEN, SYSTEMS, I, Y E, COMMUNICATION, 15, 15.1 Introduction, 15, 15.2 Elements of a Communication System, 15, 15.3 Basic Terminology Used in Electronic Communication Systems, 15.4 Bandwidth of Signals, 15, 15.5 Bandwidth of Transmission Medium, 15, 15, 15.6 Propagation of Electromagnetic Waves, , xiv, , 467, 468, 472, 474, 478, 479, 483, 485, 490, 501, 505, , 513, 513, 515, 517, 518, 519

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1 5.7, 1 5.8, 1 5.9, 5.10, 15.10, 1, , Modulation and its Necessity, Amplitude Modulation, Production of Amplitude Modulated Wave, Detection of Amplitude Modulated Wave, , 522, 524, 525, 526, , A, IT, IO, MA, N, ADDITIONAL, ITIO, INFORMATION, , 528, , A, I CE S, APPENDICES, , 532, , A, W, S, ANSWERS, , 534, , B IBLIOGRAPHY, PHY, , 552, , I NDEX, , 554, , xv

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Chapter Nine, , RAY OPTICS, AND OPTICAL, INSTRUMENTS, 9.1 INTRODUCTION, Nature has endowed the human eye (retina) with the sensitivity to detect, electromagnetic waves within a small range of the electromagnetic, spectrum. Electromagnetic radiation belonging to this region of the, spectrum (wavelength of about 400 nm to 750 nm) is called light. It is, mainly through light and the sense of vision that we know and interpret, the world around us., There are two things that we can intuitively mention about light from, common experience. First, that it travels with enormous speed and second,, that it travels in a straight line. It took some time for people to realise that, the speed of light is finite and measurable. Its presently accepted value, in vacuum is c = 2.99792458 × 108 m s–1. For many purposes, it suffices, to take c = 3 × 108 m s–1. The speed of light in vacuum is the highest, speed attainable in nature., The intuitive notion that light travels in a straight line seems to, contradict what we have learnt in Chapter 8, that light is an, electromagnetic wave of wavelength belonging to the visible part of the, spectrum. How to reconcile the two facts? The answer is that the, wavelength of light is very small compared to the size of ordinary objects, that we encounter commonly (generally of the order of a few cm or larger)., In this situation, as you will learn in Chapter 10, a light wave can be, considered to travel from one point to another, along a straight line joining

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Physics, them. The path is called a ray of light, and a bundle of such rays, constitutes a beam of light., In this chapter, we consider the phenomena of reflection, refraction, and dispersion of light, using the ray picture of light. Using the basic, laws of reflection and refraction, we shall study the image formation by, plane and spherical reflecting and refracting surfaces. We then go on to, describe the construction and working of some important optical, instruments, including the human eye., , PARTICLE, , MODEL OF LIGHT, , Newton’s fundamental contributions to mathematics, mechanics, and gravitation often blind, us to his deep experimental and theoretical study of light. He made pioneering contributions, in the field of optics. He further developed the corpuscular model of light proposed by, Descartes. It presumes that light energy is concentrated in tiny particles called corpuscles., He further assumed that corpuscles of light were massless elastic particles. With his, understanding of mechanics, he could come up with a simple model of reflection and, refraction. It is a common observation that a ball bouncing from a smooth plane surface, obeys the laws of reflection. When this is an elastic collision, the magnitude of the velocity, remains the same. As the surface is smooth, there is no force acting parallel to the surface,, so the component of momentum in this direction also remains the same. Only the component, perpendicular to the surface, i.e., the normal component of the momentum, gets reversed, in reflection. Newton argued that smooth surfaces like mirrors reflect the corpuscles in a, similar manner., In order to explain the phenomena of refraction, Newton postulated that the speed of, the corpuscles was greater in water or glass than in air. However, later on it was discovered, that the speed of light is less in water or glass than in air., In the field of optics, Newton – the experimenter, was greater than Newton – the theorist., He himself observed many phenomena, which were difficult to understand in terms of, particle nature of light. For example, the colours observed due to a thin film of oil on water., Property of partial reflection of light is yet another such example. Everyone who has looked, into the water in a pond sees image of the face in it, but also sees the bottom of the pond., Newton argued that some of the corpuscles, which fall on the water, get reflected and some, get transmitted. But what property could distinguish these two kinds of corpuscles? Newton, had to postulate some kind of unpredictable, chance phenomenon, which decided whether, an individual corpuscle would be reflected or not. In explaining other phenomena, however,, the corpuscles were presumed to behave as if they are identical. Such a dilemma does not, occur in the wave picture of light. An incoming wave can be divided into two weaker waves, at the boundary between air and water., , 9.2 REFLECTION, , 310, , OF, , LIGHT, , BY, , SPHERICAL MIRRORS, , We are familiar with the laws of reflection. The angle of reflection (i.e., the, angle between reflected ray and the normal to the reflecting surface or, the mirror) equals the angle of incidence (angle between incident ray and, the normal). Also that the incident ray, reflected ray and the normal to, the reflecting surface at the point of incidence lie in the same plane, (Fig. 9.1). These laws are valid at each point on any reflecting surface, whether plane or curved. However, we shall restrict our discussion to the, special case of curved surfaces, that is, spherical surfaces. The normal in

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Ray Optics and, Optical Instruments, this case is to be taken as normal to the tangent, to surface at the point of incidence. That is, the, normal is along the radius, the line joining the, centre of curvature of the mirror to the point of, incidence., We have already studied that the geometric, centre of a spherical mirror is called its pole while, that of a spherical lens is called its optical centre., The line joining the pole and the centre of curvature, of the spherical mirror is known as the principal, axis. In the case of spherical lenses, the principal, axis is the line joining the optical centre with its, principal focus as you will see later., , FIGURE 9.1 The incident ray, reflected ray, and the normal to the reflecting surface lie, in the same plane., , 9.2.1 Sign convention, To derive the relevant formulae for reflection by spherical mirrors and, refraction by spherical lenses, we must first adopt a sign convention for, measuring distances. In this book, we shall follow the Cartesian sign, convention. According to this, convention, all distances are measured, from the pole of the mirror or the optical, centre of the lens. The distances, measured in the same direction as the, incident light are taken as positive and, those measured in the direction, opposite to the direction of incident, light are taken as negative (Fig. 9.2)., The heights measured upwards with, respect to x-axis and normal to the, principal axis (x-axis) of the mirror/, lens are taken as positive (Fig. 9.2). The, heights measured downwards are, FIGURE 9.2 The Cartesian Sign Convention., taken as negative., With a common accepted convention, it turns out that a single formula, for spherical mirrors and a single formula for spherical lenses can handle, all different cases., , 9.2.2 Focal length of spherical mirrors, Figure 9.3 shows what happens when a parallel beam of light is incident, on (a) a concave mirror, and (b) a convex mirror. We assume that the rays, are paraxial, i.e., they are incident at points close to the pole P of the mirror, and make small angles with the principal axis. The reflected rays converge, at a point F on the principal axis of a concave mirror [Fig. 9.3(a)]., For a convex mirror, the reflected rays appear to diverge from a point F, on its principal axis [Fig. 9.3(b)]. The point F is called the principal focus, of the mirror. If the parallel paraxial beam of light were incident, making, some angle with the principal axis, the reflected rays would converge (or, appear to diverge) from a point in a plane through F normal to the principal, axis. This is called the focal plane of the mirror [Fig. 9.3(c)]., , 311

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Physics, , FIGURE 9.3 Focus of a concave and convex mirror., , The distance between the focus F and the pole P of the mirror is called, the focal length of the mirror, denoted by f. We now show that f = R/2,, where R is the radius of curvature of the mirror. The geometry, of reflection of an incident ray is shown in Fig. 9.4., Let C be the centre of curvature of the mirror. Consider a, ray parallel to the principal axis striking the mirror at M. Then, CM will be perpendicular to the mirror at M. Let T be the angle, of incidence, and MD be the perpendicular from M on the, principal axis. Then,, ‘MCP = T and ‘MFP = 2T, Now,, MD, MD, and tan 2T , (9.1), CD, FD, For small T, which is true for paraxial rays, tanT  |T, tan 2T |2TTherefore, Eq. (9.1) gives, , tanT =, , FIGURE 9.4 Geometry of, reflection of an incident ray on, (a) concave spherical mirror,, and (b) convex spherical mirror., , MD, MD,  2, FD, CD, CD, or, FD =, (9.2), 2, Now, for small T, the point D is very close to the point P., Therefore, FD = f and CD = R. Equation (9.2) then gives, f = R/2, (9.3), , 9.2.3 The mirror equation, 312, , If rays emanating from a point actually meet at another point after, reflection and/or refraction, that point is called the image of the first, point. The image is real if the rays actually converge to the point; it is

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Ray Optics and, Optical Instruments, virtual if the rays do not actually meet but appear, to diverge from the point when produced, backwards. An image is thus a point-to-point, correspondence with the object established, through reflection and/or refraction., In principle, we can take any two rays, emanating from a point on an object, trace their, paths, find their point of intersection and thus,, obtain the image of the point due to reflection at a, spherical mirror. In practice, however, it is, convenient to choose any two of the following rays:, (i) The ray from the point which is parallel to the, principal axis. The reflected ray goes through, FIGURE 9.5 Ray diagram for image, the focus of the mirror., formation by a concave mirror., (ii) The ray passing through the centre of, curvature of a concave mirror or appearing to pass through it for a, convex mirror. The reflected ray simply retraces the path., (iii) The ray passing through (or directed towards) the focus of the concave, mirror or appearing to pass through (or directed towards) the focus, of a convex mirror. The reflected ray is parallel to the principal axis., (iv) The ray incident at any angle at the pole. The reflected ray follows, laws of reflection., Figure 9.5 shows the ray diagram considering three rays. It shows, the image AcBc (in this case, real) of an object A B formed by a concave, mirror. It does not mean that only three rays emanate from the point A., An infinite number of rays emanate from any source, in all directions., Thus, point Ac is image point of A if every ray originating at point A and, falling on the concave mirror after reflection passes through the point Ac., We now derive the mirror equation or the relation between the object, distance (u), image distance (v) and the focal length ( f )., From Fig. 9.5, the two right-angled triangles AcBcF and MPF are, similar. (For paraxial rays, MP can be considered to be a straight line, perpendicular to CP.) Therefore,, B cA c, PM, , B cF, FP, , B c A c B cF, (' PM = AB), (9.4), BA, FP, Since ‘ APB = ‘ AcPBc, the right angled triangles AcBcP and ABP are, also similar. Therefore,, , or, , B cA c B c P, B A BP, Comparing Eqs. (9.4) and (9.5), we get, , (9.5), , B cF B cP – FP B cP, (9.6), FP, FP, BP, Equation (9.6) is a relation involving magnitude of distances. We now, apply the sign convention. We note that light travels from the object to, the mirror MPN. Hence this is taken as the positive direction. To reach, , 313

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Physics, the object AB, image AcBc as well as the focus F from the pole P, we have, to travel opposite to the direction of incident light. Hence, all the three, will have negative signs. Thus,, Bc P = –v, FP = –f, BP = –u, Using these in Eq. (9.6), we get, –v  f, –f, , –v, –u, , v– f, v, f, u, 1 1, 1, , (9.7), v u, f, This relation is known as the mirror equation., The size of the image relative to the size of the object is another, important quantity to consider. We define linear magnification (m) as the, ratio of the height of the image (hc) to the height of the object (h):, , or, , hc, (9.8), h, h and hc will be taken positive or negative in accordance with the accepted, sign convention. In triangles AcBcP and ABP, we have,, , m=, , B cA c, B cP, BA, BP, With the sign convention, this becomes, , – hc, h, so that, , –v, –u, , hc, v, –, (9.9), h, u, We have derived here the mirror equation, Eq. (9.7), and the, magnification formula, Eq. (9.9), for the case of real, inverted image formed, by a concave mirror. With the proper use of sign convention, these are,, in fact, valid for all the cases of reflection by a spherical mirror (concave, or convex) whether the image formed is real or virtual. Figure 9.6 shows, the ray diagrams for virtual image formed by a concave and convex mirror., You should verify that Eqs. (9.7) and (9.9) are valid for these cases, as well., , m=, , 314, , FIGURE 9.6 Image formation by (a) a concave mirror with object between, P and F, and (b) a convex mirror.

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Ray Optics and, Optical Instruments, Example 9.1 Suppose that the lower half of the concave mirror’s, reflecting surface in Fig. 9.5 is covered with an opaque (non-reflective), material. What effect will this have on the image of an object placed, in front of the mirror?, , EXAMPLE 9.1, , Solution You may think that the image will now show only half of the, object, but taking the laws of reflection to be true for all points of the, remaining part of the mirror, the image will be that of the whole object., However, as the area of the reflecting surface has been reduced, the, intensity of the image will be low (in this case, half)., Example 9.2 A mobile phone lies along the principal axis of a concave, mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation, of its image. Explain why the magnification is not uniform. Will the, distortion of image depend on the location of the phone with respect, to the mirror?, , EXAMPLE 9.2, , FIGURE 9.7, , Solution, The ray diagram for the formation of the image of the phone is shown, in Fig. 9.7. The image of the part which is on the plane perpendicular, to principal axis will be on the same plane. It will be of the same size,, i.e., BcC = BC. You can yourself realise why the image is distorted., Example 9.3 An object is placed at (i) 10 cm, (ii) 5 cm in front of a, concave mirror of radius of curvature 15 cm. Find the position, nature,, and magnification of the image in each case., Solution, The focal length f = –15/2 cm = –7.5 cm, (i) The object distance u = –10 cm. Then Eq. (9.7) gives, 1, 1, , v, 10, , or, , v=, , 1, 7. 5, , The image is 30 cm from the mirror on the same side as the object., , ( 30), v, –, –3, u, ( 10), The image is magnified, real and inverted., Also, magnification m = –, , EXAMPLE 9.3, , 10 u 7.5, = – 30 cm, 2 .5, , 315

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Physics, (ii) The object distance u = –5 cm. Then from Eq. (9.7),, 1 1, , v 5, , EXAMPLE 9.3, , or, , 1, 7.5, , v=, , 5 u 7 .5, 7.5 5, , 15 cm, , This image is formed at 15 cm behind the mirror. It is a virtual image., , v, 15, –, 3, u, ( 5), The image is magnified, virtual and erect., Magnification m = –, , Example 9.4 Suppose while sitting in a parked car, you notice a, jogger approaching towards you in the side view mirror of R = 2 m. If, the jogger is running at a speed of 5 m s–1, how fast the image of the, jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,, and (d) 9 m away., Solution, From the mirror equation, Eq. (9.7), we get, v, , fu, u f, , For convex mirror, since R = 2 m, f = 1 m. Then, ( 39) u 1 39, m, 39  1, 40, Since the jogger moves at a constant speed of 5 m s–1, after 1 s the, position of the image v (for u = –39 + 5 = –34) is (34/35 )m., The shift in the position of image in 1 s is, , for u = –39 m, v, , EXAMPLE 9.4, , 39 34 1365  1360, 5, 1, m, , 40 35, 1400, 1400 280, Therefore, the average speed of the image when the jogger is between, 39 m and 34 m from the mirror, is (1/280) m s–1, Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the, speed with which the image appears to move is, 1, 1, 1, m s –1 ,, m s –1 and, m s –1 , respectively., 150, 60, 10, Although the jogger has been moving with a constant speed, the speed, of his/her image appears to increase substantially as he/she moves, closer to the mirror. This phenomenon can be noticed by any person, sitting in a stationary car or a bus. In case of moving vehicles, a, similar phenomenon could be observed if the vehicle in the rear is, moving closer with a constant speed., , 9.3 REFRACTION, , 316, , When a beam of light encounters another transparent medium, a part of, light gets reflected back into the first medium while the rest enters the, other. A ray of light represents a beam. The direction of propagation of, an obliquely incident ray of light that enters the other medium, changes

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Ray Optics and, Optical Instruments, at the interface of the two media. This, phenomenon is called refraction of light. Snell, experimentally obtained the following laws of, refraction:, (i) The incident ray, the refracted ray and the, normal to the interface at the point of, incidence, all lie in the same plane., (ii) The ratio of the sine of the angle of incidence, to the sine of angle of refraction is constant., Remember that the angles of incidence (i ) and, refraction (r ) are the angles that the incident, and its refracted ray make with the normal,, respectively. We have, sin i, sin r, , n 21, , (9.10), , FIGURE 9.8 Refraction and reflection of light., , where n 21 is a constant, called the refractive index of the second medium, with respect to the first mediumEquation (9.10) is the well-known Snell’s, law of refraction. We note that n 21 is a characteristic of the pair of media, (and also depends on the wavelength of light), but is independent of the, angle of incidence, From Eq. (9.10), if n 21 > 1, r < i , i.e., the refracted ray bends towards, the normal. In such a case medium 2 is said to be optically denser (or, denser, in short) than medium 1On the other hand, if n 21 <1, r > i, the, refracted ray bends away from the normal. This is the case when incident, ray in a denser medium refracts into a rarer medium., Note: Optical density should not be confused with mass density,, which is mass per unit volume. It is possible that mass density of, an optically denser medium may be less than that of an optically, rarer medium (optical density is the ratio of the speed of light in, two media). For example, turpentine and water. Mass density of, turpentine is less than that of water but its optical density is higher., If n 21 is the refractive index of medium 2, with respect to medium 1 and n12 the refractive, index of medium 1 with respect to medium 2,, then it should be clear that, n12, , 1, n 21, , (9.11), , It also follows that if n 32 is the refractive, index of medium 3 with respect to medium 2, then n 32 = n 31 × n 12, where n 31 is the refractive, index of medium 3 with respect to medium 1., FIGURE 9.9 Lateral shift of a ray refracted, Some elementary results based on the laws, through a parallel-sided slab., of refraction follow immediately. For a, rectangular slab, refraction takes place at two, interfaces (air-glass and glass-air). It is easily seen from Fig. 9.9 that, 317, r2 = i1, i.e., the emergent ray is parallel to the incident ray—there is no

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Physics, , FIGURE 9.10 Apparent depth for, (a) normal, and (b) oblique viewing., , deviation, but it does suffer lateral displacement/, shift with respect to the incident ray. Another familiar, observation is that the bottom of a tank filled with, water appears to be raised (Fig. 9.10). For viewing, near the normal direction, it can be shown that the, apparent depth, (h1) is real depth (h 2) divided by, the refractive index of the medium (water)., The refraction of light through the atmosphere, is responsible for many interesting phenomena. For, example, the sun is visible a little before the actual, sunrise and until a little after the actual sunset, due to refraction of light through the atmosphere, (Fig. 9.11). By actual sunrise we mean the actual, crossing of the horizon by the sun. Figure 9.11, shows the actual and apparent positions of the sun, with respect to the horizon. The figure is highly, exaggerated to show the effect. The refractive index, of air with respect to vacuum is 1.00029. Due to, this, the apparent shift in the direction of the sun, is by about half a degree and the corresponding, time difference between actual sunset and apparent, sunset is about 2 minutes (see Example 9.5). The, apparent flattening (oval shape) of the sun at sunset, and sunrise is also due to the same phenomenon., , 318, , EXAMPLE 9.5, , FIGURE 9.11 Advance sunrise and delayed sunset due to, atmospheric refraction., , Example 9.5 The earth takes 24 h to rotate once about its axis. How, much time does the sun take to shift by 1º when viewed from, the earth?, Solution, Time taken for 360° shift = 24 h, Time taken for 1° shift = 24/360 h = 4 min.

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Ray Optics and, Optical Instruments, THE, , DROWNING CHILD, LIFEGUARD AND, , SNELL’S, , LAW, , Consider a rectangular swimming pool PQSR; see figure here. A lifeguard sitting at G, outside the pool notices a child drowning at a point C. The guard wants to reach the, child in the shortest possible time. Let SR be the, side of the pool between G and C. Should he/she, take a straight line path GAC between G and C or, GBC in which the path BC in water would be the, shortest, or some other path GXC? The guard knows, that his/her running speed v1 on ground is higher, than his/her swimming speed v2., Suppose the guard enters water at X. Let GX =l1, and XC =l 2. Then the time taken to reach from G to, C would be, t, , l1 l 2, , v1 v 2, , To make this time minimum, one has to, differentiate it (with respect to the coordinate of X ) and find the point X when t is a, minimum. On doing all this algebra (which we skip here), we find that the guard should, enter water at a point where Snell’s law is satisfied. To understand this, draw a, perpendicular LM to side SR at X. Let ‘GXM = i and ‘CXL = r. Then it can be seen that t, is minimum when, sin i, sin r, , v1, v2, , In the case of light v1/v2, the ratio of the velocity of light in vacuum to that in the, medium, is the refractive index n of the medium., In short, whether it is a wave or a particle or a human being, whenever two mediums, and two velocities are involved, one must follow Snell’s law if one wants to take the, shortest time., , 9.4 TOTAL INTERNAL REFLECTION, When light travels from an optically denser medium to a rarer medium, at the interface, it is partly reflected back into the same medium and, partly refracted to the second medium. This reflection is called the internal, reflection., When a ray of light enters from a denser medium to a rarer medium,, it bends away from the normal, for example, the ray AO1 B in Fig. 9.12., The incident ray AO1 is partially reflected (O1C) and partially transmitted, (O1B) or refracted, the angle of refraction (r ) being larger than the angle of, incidence (i ). As the angle of incidence increases, so does the angle of, refraction, till for the ray AO3, the angle of refraction is S/2The refracted, ray is bent so much away from the normal that it grazes the surface at, the interface between the two media. This is shown by the ray AO3 D in, Fig. 9.12. If the angle of incidence is increased still further (e.g., the ray, AO4), refraction is not possible, and the incident ray is totally reflected., , 319

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Physics, This is called total internal reflection. When, light gets reflected by a surface, normally, some fraction of it gets transmitted. The, reflected ray, therefore, is always less intense, than the incident ray, howsoever smooth the, reflecting surface may be. In total internal, reflection, on the other hand, no, transmission of light takes place., The angle of incidence corresponding to, an angle of refraction 90º, say ‘AO3N, is, called the critical angle (ic ) for the given pair, of media. We see from Snell’s law [Eq. (9.10)], FIGURE 9.12 Refraction and internal reflection, of rays from a point A in the denser medium, that if the relative refractive index is less, (water) incident at different angles at the interface than one then, since the maximum value, with a rarer medium (air)., of sin r is unity, there is an upper limit, to the value of sin i for which the law can be satisfied, that is, i = ic, such that, sin ic = n 21, (9.12), For values of i larger than ic, Snell’s law of refraction cannot be, satisfied, and hence no refraction is possible., The refractive index of denser medium 2 with respect to rarer medium, 1 will be n12 = 1/sin ic. Some typical critical angles are listed in Table 9.1., , TABLE 9.1 CRITICAL ANGLE OF SOME TRANSPARENT MEDIA, Substance medium, , Refractive index, , Critical angle, , Water, , 1.33, , 48.75°, , Crown glass, , 1.52, , 41.14°, , Dense flint glass, , 1.62, , 37.31°, , Diamond, , 2.42, , 24.41°, , A demonstration for total internal reflection, , 320, , All optical phenomena can be demonstrated very easily with the use of a, laser torch or pointer, which is easily available nowadays. Take a glass, beaker with clear water in it. Stir the water a few times with a piece of, soap, so that it becomes a little turbid. Take a laser pointer and shine its, beam through the turbid water. You will find that the path of the beam, inside the water shines brightly., Shine the beam from below the beaker such that it strikes at the, upper water surface at the other end. Do you find that it undergoes partial, reflection (which is seen as a spot on the table below) and partial refraction, [which comes out in the air and is seen as a spot on the roof; Fig. 9.13(a)]?, Now direct the laser beam from one side of the beaker such that it strikes, the upper surface of water more obliquely [Fig. 9.13(b)]. Adjust the, direction of laser beam until you find the angle for which the refraction

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Ray Optics and, Optical Instruments, above the water surface is totally absent and the beam is totally reflected, back to water. This is total internal reflection at its simplest., Pour this water in a long test tube and shine the laser light from top,, as shown in Fig. 9.13(c). Adjust the direction of the laser beam such that, it is totally internally reflected every time it strikes the walls of the tube., This is similar to what happens in optical fibres., Take care not to look into the laser beam directly and not to point it, at anybody’s face., , 9.4.1 Total internal reflection in nature and, its technological applications, (i) Mirage: On hot summer days, the air near the ground becomes hotter, than the air at higher levels. The refractive index of air increases with, its density. Hotter air is less dense, and has smaller refractive index, than the cooler airIf the air currents are small, that is, the air is still,, the optical density at different layers of air increases with height. As a, result, light from a tall object such as a tree, passes through a medium, whose refractive index decreases towards the ground. Thus, a ray of, light from such an object successively bends away from the normal, and undergoes total internal reflection, if the angle of incidence for, the air near the ground exceeds the critical angle. This is shown in, Fig. 9.14(b). To a distant observer, the light appears to be coming, FIGURE 9.13, from somewhere below the ground. The observer naturally assumes, Observing total, that light is being reflected from the ground, say, by a pool of water internal reflection in, near the tall object. Such inverted images of distant tall objects cause, water with a laser, an optical illusion to the observer. This phenomenon is called mirage. beam (refraction due, to glass of beaker, This type of mirage is especially common in hot deserts. Some of you, might have noticed that while moving in a bus or a car during a hot neglected being very, thin)., summer day, a distant patch of road, especially on a highway, appears, to be wet. But, you do not find any evidence of wetness when you, reach that spot. This is also due to mirage., , FIGURE 9.14 (a) A tree is seen by an observer at its place when the air above the ground is, at uniform temperature, (b) When the layers of air close to the ground have varying, temperature with hottest layers near the ground, light from a distant tree may, undergo total internal reflection, and the apparent image of the tree may create, 321, an illusion to the observer that the tree is near a pool of water.

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Physics, (ii) Diamond : Diamonds are known for their, spectacular brilliance. Their brilliance, is mainly due to the total internal, reflection of light inside them. The critical, angle for diamond-air interface (# 24.4°), is very small, therefore once light enters, a diamond, it is very likely to undergo, total internal reflection inside it., Diamonds found in nature rarely exhibit, the brilliance for which they are known., It is the technical skill of a diamond, cutter which makes diamonds to, sparkle so brilliantly. By cutting the, diamond suitably, multiple total, internal reflections can be made, to occur., (iii) Prism : Prisms designed to bend light by, FIGURE 9.15 Prisms designed to bend rays by, 90º or by 180º make use of total internal, 90º and 180º or to invert image without changing, reflection [Fig. 9.15(a) and (b)]. Such a, its size make use of total internal reflection., prism is also used to invert images, without changing their size [Fig. 9.15(c)]., In the first two cases, the critical angle ic for the material of the prism, must be less than 45º. We see from Table 9.1 that this is true for both, crown glass and dense flint glass., (iv) Optical fibres: Now-a-days optical fibres are extensively used for, transmitting audio and video signals through long distances. Optical, fibres too make use of the phenomenon of total internal reflection., Optical fibres are fabricated with high quality composite glass/quartz, fibres. Each fibre consists of a core and cladding. The refractive index, of the material of the core is higher than that of the cladding., When a signal in the form of light is, directed at one end of the fibre at a suitable, angle, it undergoes repeated total internal, reflections along the length of the fibre and, finally comes out at the other end (Fig. 9.16)., Since light undergoes total internal reflection, at each stage, there is no appreciable loss in, the intensity of the light signal. Optical fibres, FIGURE 9.16 Light undergoes successive total, are fabricated such that light reflected at one, internal reflections as it moves through an, side of inner surface strikes the other at an, optical fibre., angle larger than the critical angle. Even if the, fibre is bent, light can easily travel along its, length. Thus, an optical fibre can be used to act as an optical pipe., A bundle of optical fibres can be put to several uses. Optical fibres, are extensively used for transmitting and receiving electrical signals which, are converted to light by suitable transducers. Obviously, optical fibres, can also be used for transmission of optical signals. For example, these, are used as a ‘light pipe’ to facilitate visual examination of internal organs, 322, like esophagus, stomach and intestines. You might have seen a commonly

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Ray Optics and, Optical Instruments, available decorative lamp with fine plastic fibres with their free ends, forming a fountain like structure. The other end of the fibres is fixed over, an electric lamp. When the lamp is switched on, the light travels from the, bottom of each fibre and appears at the tip of its free end as a dot of light., The fibres in such decorative lamps are optical fibres., The main requirement in fabricating optical fibres is that there should, be very little absorption of light as it travels for long distances inside, them. This has been achieved by purification and special preparation of, materials such as quartz. In silica glass fibres, it is possible to transmit, more than 95% of the light over a fibre length of 1 km. (Compare with, what you expect for a block of ordinary window glass 1 km thick.), , 9.5 REFRACTION AT SPHERICAL SURFACES, AND BY LENSES, We have so far considered refraction at a plane interface. We shall now, consider refraction at a spherical interface between two transparent media., An infinitesimal part of a spherical surface can be regarded as planar, and the same laws of refraction can be applied at every point on the, surface. Just as for reflection by a spherical mirror, the normal at the, point of incidence is perpendicular to the tangent plane to the spherical, surface at that point and, therefore, passes through its centre of curvature., We first consider refraction by a single spherical surface and follow it by, thin lenses. A thin lens is a transparent optical medium bounded by two, surfaces; at least one of which should be spherical. Applying the formula, for image formation by a single spherical surface successively at the two, surfaces of a lens, we shall obtain the lens maker’s formula and then the, lens formula., , 9.5.1 Refraction at a spherical surface, Figure 9.17 shows the geometry of formation of image I of an object O on, the principal axis of a spherical surface with centre of curvature C, and, radius of curvature R. The rays are incident from a medium of refractive, index n1, to another of refractive index n 2. As before, we take the aperture, (or the lateral size) of the surface to be small, compared to other distances involved, so that small, angle approximation can be made. In particular,, NM will be taken to be nearly equal to the length of, the perpendicular from the point N on the principal, axis. We have, for small angles,, tan ‘NOM =, , MN, OM, , tan ‘NCM =, , MN, MC, , tan ‘NIM =, , MN, MI, , FIGURE 9.17 Refraction at a spherical, surface separating two media., , 323

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Physics, LIGHT SOURCES AND PHOTOMETRY, It is known that a body above absolute zero temperature emits electromagnetic radiation., The wavelength region in which the body emits the radiation depends on its absolute, temperature. Radiation emitted by a hot body, for example, a tungsten filament lamp, having temperature 2850 K are partly invisible and mostly in infrared (or heat) region., As the temperature of the body increases radiation emitted by it is in visible region. The, sun with temperature of about 5500 K emits radiation whose energy versus wavelength, graph peaks approximately at 550 nm corresponding to green light and is almost in the, middle of the visible region. The energy versus wavelength distribution graph for a given, body peaks at some wavelength, which is inversely proportional to the absolute, temperature of that body., The measurement of light as perceived by human eye is called photometry. Photometry, is measurement of a physiological phenomenon, being the stimulus of light as received, by the human eye, transmitted by the optic nerves and analysed by the brain. The main, physical quantities in photometry are (i) the luminous intensity of the source,, (ii) the luminous flux or flow of light from the source, and (iii) illuminance of the surface., The SI unit of luminous intensity (I ) is candela (cd). The candela is the luminous intensity,, in a given direction, of a source that emits monochromatic radiation of frequency, 540 × 1012 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian., If a light source emits one candela of luminous intensity into a solid angle of one steradian,, the total luminous flux emitted into that solid angle is one lumen (lm). A standard, 100 watt incadescent light bulb emits approximately 1700 lumens., In photometry, the only parameter, which can be measured directly is illuminance. It, is defined as luminous flux incident per unit area on a surface (lm/m2 or lux ). Most light, meters measure this quantity. The illuminance E, produced by a source of luminous, intensity I, is given by E = I/r2, where r is the normal distance of the surface from the, source. A quantity named luminance (L), is used to characterise the brightness of emitting, or reflecting flat surfaces. Its unit is cd/m2 (sometimes called ‘nit’ in industry) . A good, LCD computer monitor has a brightness of about 250 nits., , Now, for 'NOC, i is the exterior angle. Therefore, i = ‘NOM + ‘NCM, i=, , MN MN, , OM MC, , (9.13), , Similarly,, r = ‘NCM – ‘NIM, i.e., r =, , MN, MN, , MC, MI, , Now, by Snell’s law, n1 sin i = n 2 sin r, or for small angles, , 324, , n1i = n 2r, , (9.14)

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Ray Optics and, Optical Instruments, Substituting i and r from Eqs. (9.13) and (9.14), we get, n1, OM, , , , n2, , n 2  n1, , MI, , MC, , (9.15), , Here, OM, MI and MC represent magnitudes of distances. Applying the, Cartesian sign convention,, OM = –u, MI = +v, MC = +R, Substituting these in Eq. (9.15), we get, n 2 n1 n 2  n1, , (9.16), v, u, R, Equation (9.16) gives us a relation between object and image distance, in terms of refractive index of the medium and the radius of, curvature of the curved spherical surface. It holds for any curved, spherical surface., Example 9.6 Light from a point source in air falls on a spherical, glass surface (n = 1.5 and radius of curvature = 20 cm). The distance, of the light source from the glass surface is 100 cm. At what position, the image is formed?, Solution, We use the relation given by Eq. (9.16). Here, u = – 100 cm, v = ?, R = + 20 cm, n1 = 1, and n 2 = 1.5., We then have, 0.5, 20, , or v = +100 cm, The image is formed at a distance of 100 cm from the glass surface,, in the direction of incident light., , E XAMPLE 9.6, , 1.5, 1, , v 100, , 9.5.2 Refraction by a lens, Figure 9.18(a) shows the geometry of image formation by a double convex, lens. The image formation can be seen in terms of two steps:, (i) The first refracting surface forms the image I1 of the object O, [Fig. 9.18(b)]. The image I1 acts as a virtual object for the second surface, that forms the image at I [Fig. 9.18(c)]. Applying Eq. (9.15) to the first, interface ABC, we get, n1 n 2, , OB BI1, , n 2  n1, BC1, , (9.17), , A similar procedure applied to the second interface* ADC gives,, , , n 2 n1 n 2  n 1, , DI1 DI, DC2, , (9.18), , * Note that now the refractive index of the medium on the right side of ADC is n1, while on its left it is n 2. Further DI1 is negative as the distance is measured, against the direction of incident light., , 325

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Physics, For a thin lens, BI 1 = DI 1. Adding, Eqs. (9.17) and (9.18), we get, § 1, n1 n1, 1 ·, , , (n 2  n1 ) ¨, ¸, OB DI, BC, DC, ©, 1, 2¹, , (9.19), , Suppose the object is at infinity, i.e.,, OB o f and DI = f, Eq. (9.19) gives, n1, f, , § 1, 1 ·, , (n 2  n1 ) ¨, © BC1 DC2 ¸¹, , (9.20), , The point where image of an object, placed at infinity is formed is called the, focus F, of the lens and the distance f gives, its focal length. A lens has two foci, F and, Fc, on either side of it (Fig. 9.19). By the, sign convention,, BC1 = + R1,, DC2 = –R 2, So Eq. (9.20) can be written as, 1, f, , FIGURE 9.18 (a) The position of object, and the, image formed by a double convex lens,, (b) Refraction at the first spherical surface and, (c) Refraction at the second spherical surface., , § 1, 1 ·, , n 21  1 ¨, ¸, R, R, © 1, 2¹, , §, ¨©' n 21, , n2 ·, n1 ¸¹, , (9.21), , Equation (9.21) is known as the lens, maker’s formula. It is useful to design, lenses of desired focal length using surfaces, of suitable radii of curvatureNote that the, formula is true for a concave lens also. In, that case R1is negative, R 2 positive and, therefore, f is negative, From Eqs. (9.19) and (9.20), we get, n1 n1, , OB DI, , n1, f, , (9.22), , Again, in the thin lens approximation, B and D are both close to the, optical centre of the lens. Applying the sign convention,, BO = – u, DI = +v, we get, 1 1, , v u, , 326, , 1, f, , (9.23), , Equation (9.23) is the familiar thin lens formula. Though we derived, it for a real image formed by a convex lens, the formula is valid for both, convex as well as concave lenses and for both real and virtual images., It is worth mentioning that the two foci, F and Fc, of a double convex, or concave lens are equidistant from the optical centre. The focus on the, side of the (original) source of light is called the first focal point, whereas, the other is called the second focal point., To find the image of an object by a lens, we can, in principle, take any, two rays emanating from a point on an object; trace their paths using

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Ray Optics and, Optical Instruments, the laws of refraction and find the point where, the refracted rays meet (or appear to meet). In, practice, however, it is convenient to choose any, two of the following rays:, (i) A ray emanating from the object parallel to, the principal axis of the lens after refraction, passes through the second principal focus, Fc (in a convex lens) or appears to diverge (in, a concave lens) from the first principal focus F., (ii) A ray of light, passing through the optical, centre of the lens, emerges without any, deviation after refraction., (iii) A ray of light passing through the first, principal focus (for a convex lens) or, appearing to meet at it (for a concave lens), emerges parallel to the principal axis after, refraction., Figures 9.19(a) and (b) illustrate these rules, for a convex and a concave lens, respectively., You should practice drawing similar ray, diagrams for different positions of the object with, respect to the lens and also verify that the lens, FIGURE 9.19 Tracing rays through (a), formula, Eq. (9.23), holds good for all cases., convex lens (b) concave lens., Here again it must be remembered that each, point on an object gives out infinite number of, rays. All these rays will pass through the same image point after refraction, at the lens., Magnification (m) produced by a lens is defined, like that for a mirror,, as the ratio of the size of the image to that of the object. Proceeding in the, same way as for spherical mirrors, it is easily seen that for a lens, hc, v, =, (9.24), h, u, When we apply the sign convention, we see that, for erect (and virtual), image formed by a convex or concave lens, m is positive, while for an, inverted (and real) image, m is negative., , m=, , Example 9.7 A magician during a show makes a glass lens with, n = 1.47 disappear in a trough of liquid. What is the refractive index, of the liquid? Could the liquid be water?, , E XAMPLE 9.7, , Solution, The refractive index of the liquid must be equal to 1.47 in order to, make the lens disappear. This means n1 = n 2.. This gives 1/f = 0 or, f o f. The lens in the liquid will act like a plane sheet of glass. No,, the liquid is not water. It could be glycerine., , 9.5.3 Power of a lens, Power of a lens is a measure of the convergence or divergence, which a, lens introduces in the light falling on it. Clearly, a lens of shorter focal, , 327

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Physics, length bends the incident light more,while converging it, in case of a convex lens and diverging it in case of a, concave lens. The power P of a lens is defined as the, tangent of the angle by which it converges or diverges a, beam of light falling at unit distant from the optical centre, (Fig. 9.20)., tan G, , h, ; if h, f, , 1 tan G, , 1, f, , or G, , 1, f for small, , value of G. Thus,, FIGURE 9.20 Power of a lens., , P=, , 1, f, , (9.25), , The SI unit for power of a lens is dioptre (D): 1D = 1m–1. The power of, a lens of focal length of 1 metre is one dioptre. Power of a lens is positive, for a converging lens and negative for a diverging lens. Thus, when an, optician prescribes a corrective lens of power + 2.5 D, the required lens is, a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a, concave lens of focal length – 25 cm., Example 9.8 (i) If f = 0.5 m for a glass lens, what is the power of the, lens? (ii) The radii of curvature of the faces of a double convex lens, are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive, index of glass? (iii) A convex lens has 20 cm focal length in air. What, is focal length in water? (Refractive index of air-water = 1.33, refractive, index for air-glass = 1.5.), Solution, (i) Power = +2 dioptre., (ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm., Refractive index of air is taken as unity., We use the lens formula of Eq. (9.22). The sign convention has to, be applied for fR1 and R 2., Substituting the values, we have, , EXAMPLE 9.8, , 1, 1 ·, § 1, (n  1) ¨, , ¸, ©, 12, 10 15 ¹, This gives n = 1.5., (iii) For a glass lens in air, n 2 = 1.5, n1 = 1, f = +20 cm. Hence, the lens, formula gives, ª1, 1, 1 º, 0.5 « , », 20, ¬ R1 R 2 ¼, For the same glass lens in water, n 2 = 1.5, n1 = 1.33. Therefore,, ª1, 1 º, « , », R, R, 2¼, ¬ 1, Combining these two equations, we find f = + 78.2 cm., 1.33, (1.5  1.33), f, , (9.26), , 9.5.4 Combination of thin lenses in contact, 328, , Consider two lenses A and B of focal length f1 and f2 placed in contact, with each other. Let the object be placed at a point O beyond the focus of

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Ray Optics and, Optical Instruments, the first lens A (Fig. 9.21). The first lens produces, an image at I1. Since image I1 is real, it serves as a, virtual object for the second lens B, producing the, final image at I. It must, however, be borne in mind, that formation of image by the first lens is presumed, only to facilitate determination of the position of the, final image. In fact, the direction of rays emerging, FIGURE 9.21 Image formation by a, from the first lens gets modified in accordance with, combination, of two thin lenses in contact., the angle at which they strike the second lens. Since, the lenses are thin, we assume the optical centres of the lenses to be, coincident. Let this central point be denoted by P., For the image formed by the first lens A, we get, 1 1 1, , v1 u f1, For the image formed by the second lens B, we get, , (9.27), , 1 1 1, , v v1 f 2, Adding Eqs. (9.27) and (9.28), we get, , (9.28), , 1 1, , v u, , (9.29), , 1, 1, , f1, f2, , If the two lens-system is regarded as equivalent to a single lens of, focal length f, we have, 1 1, , v u, , 1, f, , so that we get, 1, f, , 1, 1, , f1, f2, , (9.30), , The derivation is valid for any number of thin lenses in contact. If, several thin lenses of focal length f1, f2, f3,... are in contact, the effective, focal length of their combination is given by, 1, f, , 1, 1, 1, , , …, f1, f2, f3, , In terms of power, Eq. (9.31) can be written as, P = P1 + P2 + P3 + …, , (9.31), , (9.32), , where P is the net power of the lens combination. Note that the sum in, Eq. (9.32) is an algebraic sum of individual powers, so some of the terms, on the right side may be positive (for convex lenses) and some negative, (for concave lenses). Combination of lenses helps to obtain diverging or, converging lenses of desired magnification. It also enhances sharpness, of the image. Since the image formed by the first lens becomes the object, for the second, Eq. (9.25) implies that the total magnification m of the, combination is a product of magnification (m1, m 2, m 3,...) of individual, lenses, m = m1 m 2 m 3 ..., , (9.33), , 329

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Physics, Such a system of combination of lenses is commonly used in designing, lenses for cameras, microscopes, telescopes and other optical instruments., Example 9.9 Find the position of the image formed by the lens, combination given in the Fig. 9.22., , FIGURE 9.22, , Solution Image formed by the first lens, 1 1, , v1 u1, , 1, f1, , 1, 1, 1, , v1 30 10, , or, v1 = 15 cm, The image formed by the first lens serves as the object for the second, This is at a distance of (15 – 5) cm = 10 cm to the right of the second, lens. Though the image is real, it serves as a virtual object for the, second lens, which means that the rays appear to come from it for, the second lens., 1 1, , v 2 10, , 1, 10, , EXAMPLE 9.9, , or, v2 = f, The virtual image is formed at an infinite distance to the left of the, second lens. This acts as an object for the third lens., 1 1, , v3 u 3, , 1, f3, , or, , 1, v3, , 1, 1, , f 30, , or, , v3 = 30 cm, , The final image is formed 30 cm to the right of the third lens., , 9.6 REFRACTION, , 330, , THROUGH A, , PRISM, , Figure 9.23 shows the passage of light through a triangular prism ABC., The angles of incidence and refraction at the first face AB are i and r1,, while the angle of incidence (from glass to air) at the second face AC is r, and the angle of refraction or emergence e. The angle between the, emergent ray RS and the direction of the incident ray PQ is called the, angle of deviation, G.

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Ray Optics and, Optical Instruments, In the quadrilateral AQNR, two of the angles, (at the vertices Q and R) are right angles., Therefore, the sum of the other angles of the, quadrilateral is 180º., ‘A + ‘QNR = 180º, From the triangle QNR,, r1 + r + ‘QNR = 180º, Comparing these two equations, we get, (9.34), r 1 + r = A, The total deviation G is the sum of deviations, at the two faces,, G= (i – r1 ) + (e – r ), , FIGURE 9.23 A ray of light passing through, a triangular glass prism., , that is,, , G = i + e– A, (9.35), Thus, the angle of deviation depends on the angle of incidence. A plot, between the angle of deviation and angle of incidence is shown in, Fig. 9.24. You can see that, in general, any given value of G except for, i = ecorresponds to two values i and hence of e. This, in fact, is expected, from the symmetry of i and e in Eq. (9.35), i.e., G remains the same if i, and e are interchanged. Physically, this is related, to the fact that the path of ray in Fig. 9.23 can be, traced back, resulting in the same angle of, deviation. At the minimum deviation Dm, the, refracted ray inside the prism becomes parallel, to its base. We have, G= Dm, i = e which implies r1 = r2., Equation (9.34) gives, A, 2, In the same way, Eq. (9.35) gives, , 2r = A or r =, , (9.36), , Dm = 2i – A, or i = (A + Dm)/2, , (9.37), , The refractive index of the prism is, n 21, , n2, n1, , sin[( A  Dm )/2], sin [ A / 2], , (9.38), , FIGURE 9.24 Plot of angle of deviation (G ), versus angle of incidence (i ) for a, triangular prism., , The angles A and D m can be measured, experimentally. Equation (9.38) thus provides a, method of determining refractive index of the material of the prism., For a small angle prism, i.e., a thin prism, Dm is also very small, and, we get, , n 21, , A  Dm /2, sin[( A  Dm )/2], , A /2, sin[ A /2], , Dm = (n 21–1)A, It implies that, thin prisms do not deviate light much., , 331

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Physics, 9.7 DISPERSION, , BY A, , PRISM, , It has been known for a long time that when a narrow beam of sunlight,, usually called white light, is incident on a glass prism, the emergent, light is seen to be consisting of several colours. There is actually a, continuous variation of colour, but broadly, the different component, colours that appear in sequence are:, violet, indigo, blue, green, yellow, orange, and red (given by the acronym, VIBGYOR). The red light bends the, least, while the violet light bends the most, (Fig. 9.25)., The phenomenon of splitting of light, into its component colours is known as, dispersion. The pattern of colour, components of light is called the spectrum, of light. The word spectrum is now used, in a much more general sense: we, discussed in Chapter 8 the electroFIGURE 9.25 Dispersion of sunlight or white light, magnetic spectrum over the large range, on passing through a glass prism. The relative, of wavelengths, from J-rays to radio, deviation of different colours shown is highly, waves, of which the spectrum of light, exaggerated., (visible spectrum) is only a small part., Though the reason for appearance of, spectrum is now common knowledge, it was a matter of much debate in, the history of physics. Does the prism itself create colour in some way or, does it only separate the colours already present in white light?, In a classic experiment known for its simplicity but great significance,, Isaac Newton settled the issue once for all. He put another similar prism,, but in an inverted position, and let the emergent beam from the first, prism fall on the second prism (Fig. 9.26). The resulting emergent beam, was found to be white light. The explanation was clear— the first prism, splits the white light into its component colours, while the inverted prism, recombines them to give white light. Thus, white, light itself consists of light of different colours,, which are separated by the prism., It must be understood here that a ray of light,, as defined mathematically, does not exist. An, actual ray is really a beam of many rays of light., Each ray splits into component colours when it, enters the glass prism. When those coloured rays, come out on the other side, they again produce a, white beam., FIGURE 9.26 Schematic diagram of, We now know that colour is associated with, Newton’s classic experiment on, wavelength of light. In the visible spectrum, red, dispersion of white light., light is at the long wavelength end (~700 nm) while, the violet light is at the short wavelength end, (~ 400 nm). Dispersion takes place because the refractive index of medium, 332, for different wavelengths (colours) is different. For example, the bending

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Ray Optics and, Optical Instruments, of red component of white light is least while it is most for the violet., Equivalently, red light travels faster than violet light in a glass prism., Table 9.2 gives the refractive indices for different wavelength for crown, glass and flint glass. Thick lenses could be assumed as made of many, prisms, therefore, thick lenses show chromatic aberration due to, dispersion of light., , TABLE 9.2 REFRACTIVE INDICES FOR DIFFERENT WAVELENGTHS, Colour, , Wavelength (nm), , Crown glass, , Flint glass, , Violet, , 396.9, , 1.533, , 1.663, , Blue, , 486.1, , 1.523, , 1.639, , Yellow, , 589.3, , 1.517, , 1.627, , Red, , 656.3, , 1.515, , 1.622, , The variation of refractive index with wavelength may be more, pronounced in some media than the other. In vacuum, of course, the, speed of light is independent of wavelength. Thus, vacuum (or air, approximately) is a non-dispersive medium in which all colours travel, with the same speed. This also follows from the fact that sunlight reaches, us in the form of white light and not as its components. On the other, hand, glass is a dispersive medium., , 9.8 SOME NATURAL PHENOMENA, , DUE TO, , SUNLIGHT, , The rainbow is an example of the dispersion of sunlight by the water, drops in the atmosphere. This is a phenomenon due to combined effect, of dispersion, refraction and reflection of sunlight by spherical water, droplets of rain. The conditions for observing a rainbow are that the sun, should be shining in one part of the sky (say near western horizon) while, it is raining in the opposite part of the sky (say eastern horizon)., An observer can therefore see a rainbow only when his back is towards, the sun., In order to understand the formation of rainbows, consider, Fig. (9.27(a). Sunlight is first refracted as it enters a raindrop, which, causes the different wavelengths (colours) of white light to separate., Longer wangelength of light (red) are bent the least while the shorter, wavelength (violet) are bent the most. Next, these component rays strike, , Formation of rainbows, , 9.8.1 The rainbow, , http://www.eo.ucar.edu/rainbows, http://www.atoptics.co.uk/bows.htm, , The interplay of light with things around us gives rise to several beautiful, phenomena. The spectacle of colour that we see around us all the time is, possible only due to sunlight. The blue of the sky, white clouds, the redhue at sunrise and sunset, the rainbow, the brilliant colours of some, pearls, shells, and wings of birds, are just a few of the natural wonders, we are used to. We describe some of them here from the point of view, of physics., , 333

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Physics, , FIGURE 9.27 Rainbow: (a) The sun rays incident on a water drop get refracted twice, and reflected internally by a drop; (b) Enlarge view of internal reflection and, refraction of a ray of light inside a drop form primary rainbow; and, (c) secondary rainbow is formed by rays, undergoing internal reflection twice, inside the drop., , 334, , the inner surface of the water drop and get internally reflected if the angle, between the refracted ray and normal to the drop surface is greater then, the critical angle (48º, in this case). The reflected light is refracted again, as it comes out of the drop as shown in the figure. It is found that the, violet light emerges at an angle of 40º related to the incoming sunlight, and red light emerges at an angle of 42º. For other colours, angles lie in, between these two values.

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Ray Optics and, Optical Instruments, Figure 9.27(b) explains the formation of primary rainbow. We see, that red light from drop 1 and violet light from drop 2 reach the observers, eye. The violet from drop 1 and red light from drop 2 are directed at level, above or below the observer. Thus the observer sees a rainbow with, red colour on the top and violet on the bottom. Thus, the primary, rainbow is a result of three-step process, that is, refraction, reflection, and refraction., When light rays undergoes two internal reflections inside a raindrop,, instead of one as in the primary rainbow, a secondary rainbow is formed, as shown in Fig. 9.27(c). It is due to four-step process. The intensity of, light is reduced at the second reflection and hence the secondary rainbow, is fainter than the primary rainbow. Further, the order of the colours is, reversed in it as is clear from Fig. 9.27(c)., , 9.8.2 Scattering of light, As sunlight travels through the earth’s atmosphere, it gets scattered, (changes its direction) by the atmospheric particles. Light of shorter, wavelengths is scattered much more than light of longer wavelengths., (The amount of scattering is inversely proportional to the fourth power, of the wavelength. This is known as Rayleigh scattering). Hence, the bluish, colour predominates in a clear sky, since blue has a shorter wavelength than red and is scattered much more strongly. In fact, violet, gets scattered even more than blue, having a shorter wavelength., But since our eyes are more sensitive to blue than violet, we see the, sky blue., Large particles like dust and water, droplets present in the atmosphere, behave differently. The relevant quantity, here is the relative size of the wavelength, of light O, and the scatterer (of typical size,, say, a). For a  O, one has Rayleigh, scattering which is proportional to (1/O)4., For a >> O, i.e., large scattering objects, (for example, raindrops, large dust or ice, particles) this is not true; all wavelengths, are scattered nearly equally. Thus, clouds, FIGURE 9.28 Sunlight travels through a longer, which have droplets of water with a >> O, distance in the atmosphere at sunset and sunrise., are generally white., At sunset or sunrise, the sun’s rays, have to pass through a larger distance in the atmosphere (Fig. 9.28)., Most of the blue and other shorter wavelengths are removed by scattering., The least scattered light reaching our eyes, therefore, the sun looks, reddish. This explains the reddish appearance of the sun and full moon, near the horizon., , 9.9 OPTICAL INSTRUMENTS, A number of optical devices and instruments have been designed utilising, reflecting and refracting properties of mirrors, lenses and prisms., Periscope, kaleidoscope, binoculars, telescopes, microscopes are some, , 335

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Physics, examples of optical devices and instruments that are in common use., Our eye is, of course, one of the most important optical device the nature, has endowed us with. Starting with the eye, we then go on to describe, the principles of working of the microscope and the telescope., , 9.9.1 The eye, , 336, , Figure 9.29 (a) shows the eye. Light enters the eye through a curved, front surface, the cornea. It passes through the pupil which is the central, hole in the iris. The size of the pupil can change under control of muscles., The light is further focussed by the eye lens on the retina. The retina is a, film of nerve fibres covering the curved back surface of the eye. The retina, contains rods and cones which sense light intensity and colour,, respectively, and transmit electrical signals via the optic nerve to the brain, which finally processes this information. The shape (curvature) and, therefore the focal length of the lens can be modified somewhat by the, ciliary muscles. For example, when the muscle is relaxed, the focal length, is about 2.5 cm and objects at infinity are in sharp focus on the retina., When the object is brought closer to the eye, in order to maintain the, same image-lens distance (# 2.5 cm), the focal length of the eye lens, becomes shorter by the action of the ciliary muscles. This property of the, eye is called accommodation. If the object is too close to the eye, the lens, cannot curve enough to focus the image on to the retina, and the image, is blurred. The closest distance for which the lens can focus light on the, retina is called the least distance of distinct vision, or the near point., The standard value for normal vision is taken as 25 cm. (Often the near, point is given the symbol D.) This distance increases with age, because, of the decreasing effectiveness of the ciliary muscle and the loss of, flexibility of the lens. The near point may be as close as about 7 to 8 cm, in a child ten years of age, and may increase to as much as 200 cm at 60, years of age. Thus, if an elderly person tries to read a book at about 25 cm, from the eye, the image appears blurred. This condition (defect of the eye), is called presbyopia. It is corrected by using a converging lens for reading., Thus, our eyes are marvellous organs that have the capability to, interpret incoming electromagnetic waves as images through a complex, process. These are our greatest assets and we must take proper care to, protect them. Imagine the world without a pair of functional eyes. Yet, many amongst us bravely face this challenge by effectively overcoming, their limitations to lead a normal life. They deserve our appreciation for, their courage and conviction., In spite of all precautions and proactive action, our eyes may develop, some defects due to various reasons. We shall restrict our discussion to, some common optical defects of the eye. For example, the light from a, distant object arriving at the eye-lens may get converged at a point in, front of the retina. This type of defect is called nearsightedness or myopia ., This means that the eye is producing too much convergence in the incident, beam. To compensate this, we interpose a concave lens between the eye, and the object, with the diverging effect desired to get the image focussed, on the retina [Fig. 9.29(b)].

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Ray Optics and, Optical Instruments, , FIGURE 9.29 (a) The structure of the eye; (b) shortsighted or myopic eye and its correction;, (c) farsighted or hypermetropic eye and its correction; and (d) astigmatic eye and its correction., , Similarly, if the eye-lens focusses the incoming light at a point behind, the retina, a convergent lens is needed to compensate for the defect in vision., This defect is called farsightedness or hypermetropia [Fig. 9.29(c)]., Another common defect of vision is called astigmatism. This occurs, when the cornea is not spherical in shape. For example, the cornea could, have a larger curvature in the vertical plane than in the horizontal plane, or vice-versa. If a person with such a defect in eye-lens looks at a wire, mesh or a grid of lines, focussing in either the vertical or the horizontal, plane may not be as sharp as in the other plane. Astigmatism results in, lines in one direction being well focussed while those in a perpendicular, direction may appear distorted [Fig. 9.29(d)]. Astigmatism can be, corrected by using a cylindrical lens of desired radius of curvature with, an appropriately directed axis. This defect can occur along with myopia, or hypermetropia., Example 9.10 What focal length should the reading spectacles have, for a person for whom the least distance of distinct vision is 50 cm?, , 1, 1, 1, 1, , or f, –50, – 25 50, orf = + 50 cm (convex lens)., , E XAMPLE 9.10, , Solution The distance of normal vision is 25 cm. So if a book is at, u = –25 cm, its image should be formed at v = –50 cm. Therefore, the, desired focal length is given by, 1, 1 1, , f, v u, , 337

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Physics, , EXAMPLE 9.11, , Example 9.11, (a) The far point of a myopic person is 80 cm in front of the eye. What, is the power of the lens required to enable him to see very distant, objects clearly?, (b) In what way does the corrective lens help the above person? Does, the lens magnify very distant objects? Explain carefully., (c) The above person prefers to remove his spectacles while reading, a book. Explain why?, Solution, (a) Solving as in the previous example, we find that the person should, use a concave lens of focal length = – 80 cm, i.e., of power = – 1.25, dioptres., (b) No. The concave lens, in fact, reduces the size of the object, but, the angle subtended by the distant object at the eye is the same, as the angle subtended by the image (at the far point) at the eye., The eye is able to see distant objects not because the corrective, lens magnifies the object, but because it brings the object (i.e., it, produces virtual image of the object) at the far point of the eye, which then can be focussed by the eye-lens on the retina., (c) The myopic person may have a normal near point, i.e., about, 25 cm (or even less). In order to read a book with the spectacles,, such a person must keep the book at a distance greater than, 25 cm so that the image of the book by the concave lens is produced, not closer than 25 cm. The angular size of the book (or its image), at the greater distance is evidently less than the angular size, when the book is placed at 25 cm and no spectacles are needed, Hence, the person prefers to remove the spectacles while reading., , 338, , EXAMPLE 9.12, , Example 9.12 (a) The near point of a hypermetropic person is 75 cm, from the eye. What is the power of the lens required to enable the, person to read clearly a book held at 25 cm from the eye? (b) In what, way does the corrective lens help the above person? Does the lens, magnify objects held near the eye? (c) The above person prefers to, remove the spectacles while looking at the sky. Explain why?, Solution, (a) u = – 25 cm, v = – 75 cm, 1/f = 1/25 – 1/75, i.e., f = 37.5 cm., The corrective lens needs to have a converging power of +2.67, dioptres., (b) The corrective lens produces a virtual image (at 75 cm) of an object, at 25 cm. The angular size of this image is the same as that of the, object. In this sense the lens does not magnify the object but merely, brings the object to the near point of the hypermetric eye, which, then gets focussed on the retina. However,  the angular size is, greater than that of the same object at the near point (75 cm), viewed without the spectacles., (c) A hypermetropic eye may have normal far point i.e., it may have, enough converging power to focus parallel rays from infinity on, the retina of the shortened eyeball. Wearing spectacles of converging, lenses (used for near vision) will amount to more converging power, than needed for parallel rays. Hence the person prefers not to use, the spectacles for far objects.

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Ray Optics and, Optical Instruments, 9.9.2 The microscope, A simple magnifier or microscope is a converging lens of small focal length, (Fig. 9.30). In order to use such a lens as a microscope, the lens is held, near the object, one focal length away or less, and, the eye is positioned close to the lens on the other, side. The idea is to get an erect, magnified and, virtual image of the object at a distance so that it, can be viewed comfortably, i.e., at 25 cm or more., If the object is at a distance f, the image is at, infinity. However, if the object is at a distance, slightly less than the focal length of the lens, the, image is virtual and closer than infinity. Although, the closest comfortable distance for viewing the, image is when it is at the near point (distance, D # 25 cm), it causes some strain on the eye., Therefore, the image formed at infinity is often, considered most suitable for viewing by the relaxed, eye. We show both cases, the first in Fig. 9.30(a),, and the second in Fig. 9.30(b) and (c)., The linear magnification m , for the image, formed at the near point D, by a simple microscope, can be obtained by using the relation, m, , v, u, , §1 1· §, v·, v ¨ – ¸ ¨1– ¸, f¹, ©v f ¹ ©, , Now according to our sign convention, v is, negative, and is equal in magnitude to D. Thus,, the magnification is, m, , §, D·, ¨©1  f ¸¹, , (9.39), , Since D is about 25 cm, to have a magnification of, six, one needs a convex lens of focal length,, FIGURE 9.30 A simple microscope; (a) the, f = 5 cm., magnifying lens is located such that the, Note that m = hc/h where his the size of the, image is at the near point, (b) the angle, object and hc the size of the image. This is also the, subtanded by the object, is the same as, ratio of the angle subtended by the image, that at the near point, and (c) the object, to that subtended by the object, if placed at D for, near the focal point of the lens; the image, comfortable viewing. (Note that this is not the angle, is far off but closer than infinity., actually subtended by the object at the eye, which, is h/u.) What a single-lens simple magnifier, achieves is that it allows the object to be brought closer to the eye than D., We will now find the magnification when the image is at infinity. In, this case we will have to obtained the angular magnification. Suppose, the object has a height h. The maximum angle it can subtend, and be, clearly visible (without a lens), is when it is at the near point, i.e., a distance, D. The angle subtended is then given by, tan To, , §h·, ¨© ¸¹ |To, D, , (9.40), , 339

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Physics, We now find the angle subtended at the eye by the image when the, object is at u. From the relations, hc, v, m, h, u, we have the angle subtended by the image, hc, v, is at u = –f., , h v, ˜, v u, , tan Ti, , Ti, , h, |T . The angle subtended by the object, when it, u, , §h·, ¨© f ¸¹, , (9.41), , as is clear from Fig. 9.29(c). The angular magnification is, therefore, m, , § Ti ·, ¨© T ¸¹, o, , D, f, , (9.42), , This is one less than the magnification when the image is at the near, point, Eq. (9.39), but the viewing is more comfortable and the difference, in magnification is usually small. In subsequent discussions of optical, instruments (microscope and telescope) we shall assume the image to be, at infinity., A simple microscope has a limited maximum magnification (d 9) for, realistic focal lengths. For much larger magnifications, one uses two, lenses, one compounding the effect of the other. This is known as a, compound microscope. A schematic diagram of, a compound microscope is shown in Fig. 9.31., The lens nearest the object, called the objective,, forms a real, inverted, magnified image of the, object. This serves as the object for the second, lens, the eyepiece, which functions essentially, like a simple microscope or magnifier, produces, the final image, which is enlarged and virtual., The first inverted image is thus near (at or, within) the focal plane of the eyepiece, at a, distance appropriate for final image formation, at infinity, or a little closer for image formation, at the near point. Clearly, the final image is, inverted with respect to the original object., We now obtain the magnification due to a, FIGURE 9.31 Ray diagram for the, compound microscope. The ray diagram of, formation of image by a compound, Fig. 9.31 shows that the (linear) magnification, microscope., due to the objective, namely hc/h, equals, hc, L, h, fo, where we have used the result, mO, , 340, , tan E, , §h ·, ¨© f ¸¹, o, , § hc·, ¨© ¸¹, L, , (9.43)

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Ray Optics and, Optical Instruments, Here hc is the size of the first image, the object size being h and fo, being the focal length of the objective. The first image is formed near the, focal point of the eyepiece. The distance L, i.e., the distance between the, second focal point of the objective and the first focal point of the eyepiece, (focal length fe ) is called the tube length of the compound microscope., As the first inverted image is near the focal point of the eyepiece, we, use the result from the discussion above for the simple microscope to, obtain the (angular) magnification me due to it [Eq. (9.39)], when the, final image is formed at the near point, is, §, D·, [9.44(a)], ¨©1  f ¸¹, e, When the final image is formed at infinity, the angular magnification, due to the eyepiece [Eq. (9.42)] is, me, , me = (D/fe ), , [9.44(b)], , Thus, the total magnification [(according to Eq. (9.33)], when the, image is formed at infinity, is, , m, , m om e, , § L·, ¨© f ¸¹, o, , §D·, ¨© f ¸¹, e, , 20 25, u, 250, 1, 2, Various other factors such as illumination of the object, contribute to, the quality and visibility of the image. In modern microscopes, multicomponent lenses are used for both the objective and the eyepiece to, improve image quality by minimising various optical aberrations (defects), in lenses., , The world’s largest optical telescopes, , m om e, , http://astro.nineplanets.org/bigeyes.html, , § L· §D·, (9.45), ¨© f ¸¹ ¨© f ¸¹, o, e, Clearly, to achieve a large magnification of a small object (hence the, name microscope), the objective and eyepiece should have small focal, lengths. In practice, it is difficult to make the focal length much smaller, than 1 cm. Also large lenses are required to make L large., For example, with an objective with fo = 1.0 cm, and an eyepiece with, focal length fe = 2.0 cm, and a tube length of 20 cm, the magnification is, m, , 9.9.3 Telescope, The telescope is used to provide angular magnification of distant objects, (Fig. 9.32). It also has an objective and an eyepiece. But here, the objective, has a large focal length and a much larger aperture than the eyepiece., Light from a distant object enters the objective and a real image is formed, in the tube at its second focal point. The eyepiece magnifies this image, producing a final inverted image. The magnifying power m is the ratio of, the angle E subtended at the eye by the final image to the angle D which, the object subtends at the lens or the eye. Hence, fo, E h fo, m | |, ., (9.46), D fe h, fe, In this case, the length of the telescope tube is fo + fe ., , 341

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Physics, Terrestrial telescopes have, in, addition, a pair of inverting lenses to, make the final image erect. Refracting, telescopes can be used both for, terrestrial and astronomical, observations. For example, consider, a telescope whose objective has a focal, length of 100 cm and the eyepiece a, focal length of 1 cm. The magnifying, power of this telescope is, m = 100/1 = 100., Let us consider a pair of stars of, actual separation 1c (one minute of, arc). The stars appear as though they, FIGURE 9.32 A refracting telescope., are separated by an angle of 100 × 1c, = 100c=1.67º., The main considerations with an astronomical telescope are its light, gathering power and its resolution or resolving power. The former clearly, depends on the area of the objective. With larger diameters, fainter objects, can be observed. The resolving power, or the ability to observe two objects, distinctly, which are in very nearly the same direction, also depends on, the diameter of the objective. So, the desirable aim in optical telescopes, is to make them with objective of large diameter. The largest lens objective, in use has a diameter of 40 inch (~1.02 m). It is at the Yerkes Observatory, in Wisconsin, USA. Such big lenses tend to be very heavy and therefore,, difficult to make and support by their edges. Further, it is rather difficult, and expensive to make such large sized lenses which form images that, are free from any kind of chromatic aberration and distortions., For these reasons, modern telescopes use a concave mirror rather, than a lens for the objective. Telescopes with mirror objectives are called, reflecting telescopes. They have several advantages. First, there is no, chromatic aberration in a mirror. Second, if a parabolic reflecting surface, is chosen, spherical aberration is also removed. Mechanical support is, much less of a problem since a mirror weighs much less than a lens of, equivalent optical quality, and can be, supported over its entire back surface, not, just over its rim. One obvious problem with a, reflecting telescope is that the objective mirror, focusses light inside the telescope tube. One, must have an eyepiece and the observer right, there, obstructing some light (depending on, the size of the observer cage). This is what is, done in the very large 200 inch (~5.08 m), diameters, Mt. Palomar telescope, California., The viewer sits near the focal point of the, FIGURE 9.33 Schematic diagram of a reflecting mirror, in a small cage. Another solution to, telescope (Cassegrain)., the problem is to deflect the light being, focussed by another mirror. One such, arrangement using a convex secondary mirror to focus the incident light,, 342, which now passes through a hole in the objective primary mirror, is shown

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Ray Optics and, Optical Instruments, in Fig. 9.33. This is known as a Cassegrain telescope, after its inventor., It has the advantages of a large focal length in a short telescope. The, largest telescope in India is in Kavalur, Tamil Nadu. It is a 2.34 m diameter, reflecting telescope (Cassegrain). It was ground, polished, set up, and is, being used by the Indian Institute of Astrophysics, Bangalore. The largest, reflecting telescopes in the world are the pair of Keck telescopes in Hawaii,, USA, with a reflector of 10 metre in diameter., , SUMMARY, 1., , 2., , 3., , 4., , Reflection is governed by the equation ‘i = ‘rc and refraction by the, Snell’s law, sini/sinr = n, where the incident ray, reflected ray, refracted, ray and normal lie in the same plane. Angles of incidence, reflection, and refraction are i, r c and r, respectively., The critical angle of incidence ic for a ray incident from a denser to rarer, medium, is that angle for which the angle of refraction is 90°. For, i > ic, total internal reflection occurs. Multiple internal reflections in, diamond (ic # 24.4°), totally reflecting prisms and mirage, are some, examples of total internal reflection. Optical fibres consist of glass, fibres coated with a thin layer of material of lower refractive index., Light incident at an angle at one end comes out at the other, after, multiple internal reflections, even if the fibre is bent., Cartesian sign convention: Distances measured in the same direction, as the incident light are positive; those measured in the opposite, direction are negative. All distances are measured from the pole/optic, centre of the mirror/lens on the principal axis. The heights measured, upwards above x-axis and normal to the principal axis of the mirror/, lens are taken as positive. The heights measured downwards are taken, as negative., Mirror equation:, , 1 1, , v u, , 5., , where u and v are object and image distances, respectively and f is the, focal length of the mirror. f is (approximately) half the radius of, curvature R. f is negative for concave mirror; f is positive for a convex, mirror., For a prism of the angle A, of refractive index n 2 placed in a medium, of refractive index n1,, , n 21, , 6., , 1, f, , sin ª¬ A  D m /2º¼, sin A /2, , n2, n1, , where Dm is the angle of minimum deviation., For refraction through a spherical interface (from medium 1 to 2 of, refractive index n1 and n 2, respectively), , n 2 n1 n 2  n 1, , v, u, R, Thin lens formula, , 1 1, , v u, , 1, f, , 343

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Physics, Lens maker’s formula, , 1, f, , n 2  n1 § 1, 1·, ¨© R  R ¸¹, n1, 1, 2, , R1 and R2 are the radii of curvature of the lens surfaces. f is positive, for a converging lens; f is negative for a diverging lens. The power of a, lens P = 1/f., The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1., If several thin lenses of focal length f1, f2, f3,.. are in contact, the, effective focal length of their combination, is given by, , 1, f, , 7., 8., , 9., , 1, 1, 1, , , …, f1, f2, f3, , The total power of a combination of several lenses is, P = P1 + P2 + P3 + …, Dispersion is the splitting of light into its constituent colours., The Eye: The eye has a convex lens of focal length about 2.5 cm. This, focal length can be varied somewhat so that the image is always formed, on the retina. This ability of the eye is called accommodation. In a, defective eye, if the image is focussed before the retina (myopia), a, diverging corrective lens is needed; if the image is focussed beyond the, retina (hypermetropia), a converging corrective lens is needed., Astigmatism is corrected by using cylindrical lenses., Magnifying power m of a simple microscope is given by m = 1 + (D/f ),, where D = 25 cm is the least distance of distinct vision and f is the, focal length of the convex lens. If the image is at infinity, m = D/f. For, a compound microscope, the magnifying power is given by, m = me × m0 where me = 1 + (D/fe ), is the magnification due to the, eyepiece and mo is the magnification produced by the objective., Approximately,, , m, , L, D, u, fo, fe, , where fo and fe are the focal lengths of the objective and eyepiece,, respectively, and L is the distance between their focal points., 10. Magnifying power m of a telescope is the ratio of the angle E subtended, at the eye by the image to the angle D subtended at the eye by the, object., , m, , E, D, , fo, fe, , where f0 and fe are the focal lengths of the objective and eyepiece,, respectively., , POINTS TO PONDER, 1., 2., , 344, , The laws of reflection and refraction are true for all surfaces and, pairs of media at the point of the incidence., The real image of an object placed between f and 2f from a convex lens, can be seen on a screen placed at the image location. If the screen is, removed, is the image still there? This question puzzles many, because, it is difficult to reconcile ourselves with an image suspended in air

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Ray Optics and, Optical Instruments, , 3., , 4., , 5., , without a screen. But the image does exist. Rays from a given point, on the object are converging to an image point in space and diverging, away. The screen simply diffuses these rays, some of which reach our, eye and we see the image. This can be seen by the images formed in air, during a laser show., Image formation needs regular reflection/refraction. In principle, all, rays from a given point should reach the same image point. This is, why you do not see your image by an irregular reflecting object, say, the page of a book., Thick lenses give coloured images due to dispersion. The variety in, colour of objects we see around us is due to the constituent colours of, the light incident on them. A monochromatic light may produce an, entirely different perception about the colours on an object as seen in, white light., For a simple microscope, the angular size of the object equals the, angular size of the image. Yet it offers magnification because we can, keep the small object much closer to the eye than 25 cm and hence, have it subtend a large angle. The image is at 25 cm which we can see., Without the microscope, you would need to keep the small object at, 25 cm which would subtend a very small angle., , EXERCISES, 9.1, , 9.2, 9.3, , 9.4, , A small candle, 2.5 cm in size is placed at 27 cm in front of a concave, mirror of radius of curvature 36 cm. At what distance from the mirror, should a screen be placed in order to obtain a sharp image? Describe, the nature and size of the image. If the candle is moved closer to the, mirror, how would the screen have to be moved?, A 4.5 cm needle is placed 12 cm away from a convex mirror of focal, length 15 cm. Give the location of the image and the magnification., Describe what happens as the needle is moved farther from the mirror., A tank is filled with water to a height of 12.5 cm. The apparent, depth of a needle lying at the bottom of the tank is measured by a, microscope to be 9.4 cm. What is the refractive index of water? If, water is replaced by a liquid of refractive index 1.63 up to the same, height, by what distance would the microscope have to be moved to, focus on the needle again?, Figures 9.34(a) and (b) show refraction of a ray in air incident at 60°, with the normal to a glass-air and water-air interface, respectively., Predict the angle of refraction in glass when the angle of incidence, in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)]., , FIGURE 9.34, , 345

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Physics, 9.5, , 9.6, , 9.7, , 9.8, , 9.9, , 9.10, , 9.11, , 9.12, , 9.13, , 9.14, , 9.15, , 346, , A small bulb is placed at the bottom of a tank containing water to a, depth of 80cm. What is the area of the surface of water through, which light from the bulb can emerge out? Refractive index of water, is 1.33. (Consider the bulb to be a point source.), A prism is made of glass of unknown refractive index. A parallel, beam of light is incident on a face of the prism. The angle of minimum, deviation is measured to be 40°. What is the refractive index of the, material of the prism? The refracting angle of the prism is 60°. If, the prism is placed in water (refractive index 1.33), predict the new, angle of minimum deviation of a parallel beam of light., Double-convex lenses are to be manufactured from a glass of, refractive index 1.55, with both faces of the same radius of, curvature. What is the radius of curvature required if the focal length, is to be 20 cm?, A beam of light converges at a point P. Now a lens is placed in the, path of the convergent beam 12 cm from P. At what point does the, beam converge if the lens is (a) a convex lens of focal length 20 cm,, and (b) a concave lens of focal length 16 cm?, An object of size 3.0 cm is placed 14cm in front of a concave lens of, focal length 21 cm. Describe the image produced by the lens. What, happens if the object is moved further away from the lens?, What is the focal length of a convex lens of focal length 30 cm in, contact with a concave lens of focal length 20 cm? Is the system a, converging or a diverging lens? Ignore thickness of the lenses., A compound microscope consists of an objective lens of focal length, 2.0 cm and an eyepiece of focal length 6.25 cm separated by a, distance of 15 cm. How far from the objective should an object be, placed in order to obtain the final image at (a) the least distance of, distinct vision (25 cm), and (b) at infinity? What is the magnifying, power of the microscope in each case?, A person with a normal near point (25 cm) using a compound, microscope with objective of focal length 8.0 mm and an eyepiece of, focal length 2.5 cm can bring an object placed at 9.0 mm from the, objective in sharp focus. What is the separation between the two, lenses? Calculate the magnifying power of the microscope,, A small telescope has an objective lens of focal length 144 cm and, an eyepiece of focal length 6.0 cm. What is the magnifying power of, the telescope? What is the separation between the objective and, the eyepiece?, (a) A giant refracting telescope at an observatory has an objective, lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is, used, what is the angular magnification of the telescope?, (b) If this telescope is used to view the moon, what is the diameter, of the image of the moon formed by the objective lens? The, diameter of the moon is 3.48 × 106 m, and the radius of lunar, orbit is 3.8 × 108 m., Use the mirror equation to deduce that:, (a) an object placed between f and 2 f of a concave mirror produces, a real image beyond 2 f., (b) a convex mirror always produces a virtual image independent, of the location of the object., (c) the virtual image produced by a convex mirror is always, diminished in size and is located between the focus and, the pole.

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Ray Optics and, Optical Instruments, , 9.16, , 9.17, , (d) an object placed between the pole and focus of a concave mirror, produces a virtual and enlarged image., [Note: This exercise helps you deduce algebraically properties of, images that one obtains from explicit ray diagrams.], A small pin fixed on a table top is viewed from above from a distance, of 50 cm. By what distance would the pin appear to be raised if it is, viewed from the same point through a 15 cm thick glass slab held, parallel to the table? Refractive index of glass = 1.5. Does the answer, depend on the location of the slab?, (a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a, glass fibre of refractive index 1.68. The outer covering of the, pipe is made of a material of refractive index 1.44. What is the, range of the angles of the incident rays with the axis of the pipe, for which total reflections inside the pipe take place, as shown, in the figure., (b) What is the answer if there is no outer covering of the pipe?, , FIGURE 9.35, , 9.18, , 9.19, , 9.20, , 9.21, , Answer the following questions:, (a) You have learnt that plane and convex mirrors produce virtual, images of objects. Can they produce real images under some, circumstances? Explain., (b) A virtual image, we always say, cannot be caught on a screen., Yet when we ‘see’ a virtual image, we are obviously bringing it, on to the ‘screen’ (i.e., the retina) of our eye. Is there a, contradiction?, (c) A diver under water, looks obliquely at a fisherman standing on, the bank of a lake. Would the fisherman look taller or shorter to, the diver than what he actually is?, (d) Does the apparent depth of a tank of water change if viewed, obliquely? If so, does the apparent depth increase or decrease?, (e) The refractive index of diamond is much greater than that of, ordinary glass. Is this fact of some use to a diamond cutter?, The image of a small electric bulb fixed on the wall of a room is to be, obtained on the opposite wall 3 m away by means of a large convex, lens. What is the maximum possible focal length of the lens required, for the purpose?, A screen is placed 90 cm from an object. The image of the object on, the screen is formed by a convex lens at two different locations, separated by 20 cm. Determine the focal length of the lens., (a) Determine the ‘effective focal length’ of the combination of the, two lenses in Exercise 9.10, if they are placed 8.0 cm apart with, their principal axes coincident. Does the answer depend on, which side of the combination a beam of parallel light is incident?, Is the notion of effective focal length of this system useful at all?, (b) An object 1.5 cm in size is placed on the side of the convex lens, in the arrangement (a) above The distance between the object, , 347

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Physics, 9.22, , 9.23, , 9.24, , 9.25, , 9.26, , 9.27, , 9.28, , 9.29, , 9.30, , 9.31, , 348, , and the convex lens is 40 cm Determine the magnification, produced by the two-lens system, and the size of the image., At what angle should a ray of light be incident on the face of a prism, of refracting angle 60° so that it just suffers total internal reflection, at the other face? The refractive index of the material of the prism is, 1.524., You are given prisms made of crown glass and flint glass with a, wide variety of angles. Suggest a combination of prisms which will, (a) deviate a pencil of white light without much dispersion,, (b) disperse (and displace) a pencil of white light without much, deviation., For a normal eye, the far point is at infinity and the near point of, distinct vision is about 25 cm in front of the eye. The cornea of the, eye provides a converging power of about 40 dioptres, and the least, converging power of the eye-lens behind the cornea is about 20, dioptres. From this rough data estimate the range of accommodation, (i.e., the range of converging power of the eye-lens) of a normal eye., Does short-sightedness (myopia) or long-sightedness (hyper metropia) imply necessarily that the eye has partially lost its ability, of accommodation? If not, what might cause these defects of vision?, A myopic person has been using spectacles of power –1.0 dioptre, for distant vision. During old age he also needs to use separate, reading glass of power + 2.0 dioptres. Explain what may have, happened., A person looking at a person wearing a shirt with a pattern, comprising vertical and horizontal lines is able to see the vertical, lines more distinctly than the horizontal ones. What is this defect, due to? How is such a defect of vision corrected?, A man with normal near point (25 cm) reads a book with small print, using a magnifying glass: a thin convex lens of focal length 5 cm., (a) What is the closest and the farthest distance at which he should, keep the lens from the page so that he can read the book when, viewing through the magnifying glass?, (b) What is the maximum and the minimum angular magnification, (magnifying power) possible using the above simple microscope?, A card sheet divided into squares each of size 1 mm2 is being viewed, at a distance of 9 cm through a magnifying glass (a converging lens, of focal length 9 cm) held close to the eye., (a) What is the magnification produced by the lens? How much is, the area of each square in the virtual image?, (b) What is the angular magnification (magnifying power) of the, lens ?, (c) Is the magnification in (a) equal to the magnifying power in (b)?, Explain., (a) At what distance should the lens be held from the figure in, Exercise 9.29 in order to view the squares distinctly with the, maximum possible magnifying power?, (b) What is the magnification in this case?, (c) Is the magnification equal to the magnifying power in this case?, Explain., What should be the distance between the object in Exercise 9.30, and the magnifying glass if the virtual image of each square in the, figure is to have an area of 6.25 mm2. Would you be able to see the, squares distinctly with your eyes very close to the magnifier ?

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Ray Optics and, Optical Instruments, , 9.32, , 9.33, , 9.34, , 9.35, , 9.36, , 9.37, , [Note: Exercises 9.29 to 9.31 will help you clearly understand the, difference between magnification in absolute size and the angular, magnification (or magnifying power) of an instrument.], Answer the following questions:, (a) The angle subtended at the eye by an object is equal to the, angle subtended at the eye by the virtual image produced by a, magnifying glass. In what sense then does a magnifying glass, provide angular magnification?, (b) In viewing through a magnifying glass, one usually positions, one’s eyes very close to the lens. Does angular magnification, change if the eye is moved back?, (c) Magnifying power of a simple microscope is inversely proportional, to the focal length of the lens. What then stops us from using a, convex lens of smaller and smaller focal length and achieving, greater and greater magnifying power?, (d) Why must both the objective and the eyepiece of a compound, microscope have short focal lengths?, (e) When viewing through a compound microscope, our eyes should, be positioned not on the eyepiece but a short distance away, from it for best viewing. Why? How much should be that short, distance between the eye and eyepiece?, An angular magnification (magnifying power) of 30X is desired using, an objective of focal length 1.25 cm and an eyepiece of focal length, 5 cm. How will you set up the compound microscope?, A small telescope has an objective lens of focal length 140 cm and, an eyepiece of focal length 5.0 cm. What is the magnifying power of, the telescope for viewing distant objects when, (a) the telescope is in normal adjustment (i.e., when the final image, is at infinity)?, (b) the final image is formed at the least distance of distinct vision, (25 cm)?, (a) For the telescope described in Exercise 9.34 (a), what is the, separation between the objective lens and the eyepiece?, (b) If this telescope is used to view a 100 m tall tower 3 km away,, what is the height of the image of the tower formed by the objective, lens?, (c) What is the height of the final image of the tower if it is formed at, 25 cm?, A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such, a telescope is built with the mirrors 20 mm apart. If the radius of, curvature of the large mirror is 220 mm and the small mirror is, 140 mm, where will the final image of an object at infinity be?, Light incident normally on a plane mirror attached to a galvanometer, coil retraces backwards as shown in Fig. 9.36. A current in the coil, produces a deflection of 3.5o of the mirror. What is the displacement, of the reflected spot of light on a screen placed 1.5 m away?, , FIGURE 9.36, , 349

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Physics, 9.38, , Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in, contact with a liquid layer on top of a plane mirror. A small needle, with its tip on the principal axis is moved along the axis until its, inverted image is found at the position of the needle. The distance of, the needle from the lens is measured to be 45.0 cm. The liquid is, removed and the experiment is repeated. The new distance is, measured to be 30.0 cm. What is the refractive index of the liquid?, , FIGURE 9.37, , 350

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Chapter Ten, , WAVE OPTICS, , 10.1 INTRODUCTION, In 1637 Descartes gave the corpuscular model of light and derived Snell’s, law. It explained the laws of reflection and refraction of light at an interface., The corpuscular model predicted that if the ray of light (on refraction), bends towards the normal then the speed of light would be greater in the, second medium. This corpuscular model of light was further developed, by Isaac Newton in his famous book entitled OPTICKS and because of, the tremendous popularity of this book, the corpuscular model is very, often attributed to Newton., In 1678, the Dutch physicist Christiaan Huygens put forward the, wave theory of light – it is this wave model of light that we will discuss in, this chapter. As we will see, the wave model could satisfactorily explain, the phenomena of reflection and refraction; however, it predicted that on, refraction if the wave bends towards the normal then the speed of light, would be less in the second medium. This is in contradiction to the, prediction made by using the corpuscular model of light. It was much, later confirmed by experiments where it was shown that the speed of, light in water is less than the speed in air confirming the prediction of the, wave model; Foucault carried out this experiment in 1850., The wave theory was not readily accepted primarily because of, Newton’s authority and also because light could travel through vacuum

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Physics, and it was felt that a wave would always require a medium to propagate, from one point to the other. However, when Thomas Young performed, his famous interference experiment in 1801, it was firmly established, that light is indeed a wave phenomenon. The wavelength of visible, light was measured and found to be extremely small; for example, the, wavelength of yellow light is about 0.5 Pm. Because of the smallness, of the wavelength of visible light (in comparison to the dimensions of, typical mirrors and lenses), light can be assumed to approximately, travel in straight lines. This is the field of geometrical optics, which we, had discussed in the previous chapter. Indeed, the branch of optics in, which one completely neglects the finiteness of the wavelength is called, geometrical optics and a ray is defined as the path of energy, propagation in the limit of wavelength tending to zero., After the interference experiment of Young in 1801, for the next 40, years or so, many experiments were carried out involving the, interference and diffraction of lightwaves; these experiments could only, be satisfactorily explained by assuming a wave model of light. Thus,, around the middle of the nineteenth century, the wave theory seemed, to be very well established. The only major difficulty was that since it, was thought that a wave required a medium for its propagation, how, could light waves propagate through vacuum. This was explained, when Maxwell put forward his famous electromagnetic theory of light., Maxwell had developed a set of equations describing the laws of, electricity and magnetism and using these equations he derived what, is known as the wave equation from which he predicted the existence, of electromagnetic waves*. From the wave equation, Maxwell could, calculate the speed of electromagnetic waves in free space and he found, that the theoretical value was very close to the measured value of speed, o f l i g h t . F r o m t h i s , h e p r o p o u n d e d t h a t light must be an, electromagnetic wave. Thus, according to Maxwell, light waves are, associated with changing electric and magnetic fields; changing electric, field produces a time and space varying magnetic field and a changing, magnetic field produces a time and space varying electric field. The, changing electric and magnetic fields result in the propagation of, electromagnetic waves (or light waves) even in vacuum., In this chapter we will first discuss the original formulation of the, Huygens principle and derive the laws of reflection and refraction. In, Sections 10.4 and 10.5, we will discuss the phenomenon of interference, which is based on the principle of superposition. In Section 10.6 we, will discuss the phenomenon of diffraction which is based on HuygensFresnel principle. Finally in Section 10.7 we will discuss the, phenomenon of polarisation which is based on the fact that the light, waves are transverse electromagnetic waves., , 352, , * Maxwell had predicted the existence of electromagnetic waves around 1855; it, was much later (around 1890) that Heinrich Hertz produced radiowaves in the, laboratory. J.C. Bose and G. Marconi made practical applications of the Hertzian, waves

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Wave Optics, DOES LIGHT TRAVEL IN A STRAIGHT LINE?, Light travels in a straight line in Class VI; it does not do so in Class XII and beyond! Surprised,, aren’t you?, In school, you are shown an experiment in which you take three cardboards with, pinholes in them, place a candle on one side and look from the other side. If the flame of the, candle and the three pinholes are in a straight line, you can see the candle. Even if one of, them is displaced a little, you cannot see the candle. This proves, so your teacher says,, that light travels in a straight line., In the present book, there are two consecutive chapters, one on ray optics and the other, on wave optics. Ray optics is based on rectilinear propagation of light, and deals with, mirrors, lenses, reflection, refraction, etc. Then you come to the chapter on wave optics,, and you are told that light travels as a wave, that it can bend around objects, it can diffract, and interfere, etc., In optical region, light has a wavelength of about half a micrometre. If it encounters an, obstacle of about this size, it can bend around it and can be seen on the other side. Thus a, micrometre size obstacle will not be able to stop a light ray. If the obstacle is much larger,, however, light will not be able to bend to that extent, and will not be seen on the other side., This is a property of a wave in general, and can be seen in sound waves too. The sound, wave of our speech has a wavelength of about 50 cm to 1 m. If it meets an obstacle of the, size of a few metres, it bends around it and reaches points behind the obstacle. But when it, comes across a larger obstacle of a few hundred metres, such as a hillock, most of it is, reflected and is heard as an echo., Then what about the primary school experiment? What happens there is that when we, move any cardboard, the displacement is of the order of a few millimetres, which is much, larger than the wavelength of light. Hence the candle cannot be seen. If we are able to move, one of the cardboards by a micrometer or less, light will be able to diffract, and the candle, will still be seen., One could add to the first sentence in this box : It learns how to bend as it grows up!, , 10.2 HUYGENS PRINCIPLE, We would first define a wavefront: when we drop a small stone on a calm, pool of water, waves spread out from the point of impact. Every point on, the surface starts oscillating with time. At any instant, a photograph of, the surface would show circular rings on which the disturbance is, maximum. Clearly, all points on such a circle are oscillating in phase, because they are at the same distance from the source. Such a locus of, points, which oscillate in phase is called a wavefront ; thus a wavefront, is defined as a surface of constant phase. The speed with which the, wavefront moves outwards from the source is called the speed of the FIGURE 10.1 (a) A, wave. The energy of the wave travels in a direction perpendicular to the diverging spherical, wave emanating from, wavefront., If we have a point source emitting waves uniformly in all directions, a point source. The, wavefronts are, then the locus of points which have the same amplitude and vibrate in, spherical., the same phase are spheres and we have what is known as a spherical, 353, wave as shown in Fig. 10.1(a). At a large distance from the source, a

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Physics, small portion of the sphere can be considered as a plane and we have, what is known as a plane wave [Fig. 10.1(b)]., Now, if we know the shape of the wavefront at t = 0, then Huygens, principle allows us to determine the shape of the wavefront at a later, time W. Thus, Huygens principle is essentially a geometrical construction,, which given the shape of the wafefront at any time allows us to determine, FIGURE 10.1 (b) At a, large distance from the shape of the wavefront at a later time. Let us consider a diverging, the source, a small wave and let F1F2 represent a portion of the spherical wavefront at t = 0, (Fig. 10.2). Now, according to Huygens principle, each point of the, portion of the, spherical wave can wavefront is the source of a secondary disturbance and the wavelets, be approximated by a emanating from these points spread out in all directions with the speed, plane wave., of the wave. These wavelets emanating from the wavefront are usually, referred to as secondary wavelets and if we draw a common tangent, to all these spheres, we obtain the new position of the wavefront at a, later time., , FIGURE 10.2 F1F2 represents the spherical wavefront (with O as, centre) at t = 0. The envelope of the secondary wavelets, emanating from F1F2 produces the forward moving wavefront G1G2., The backwave D1D2 does not exist., , Thus, if we wish to determine the shape of the wavefront at t = W, we, draw, spheres of radius vW from each point on the spherical wavefront, FIGURE 10.3, where, v represents the speed of the waves in the medium. If we now draw, Huygens geometrical, a, common, tangent to all these spheres, we obtain the new position of the, construction for a, wavefront, at, t = W. The new wavefront shown as G1G2 in Fig. 10.2 is again, plane wave, propagating to the, spherical with point O as the centre., right. F1 F2 is the, The above model has one shortcoming: we also have a backwave which, plane wavefront at is shown as D D in Fig. 10.2. Huygens argued that the amplitude of the, 1 2, t = 0 and G1G2 is the secondary wavelets is maximum in the forward direction and zero in the, wavefront at a later, backward direction; by making this adhoc assumption, Huygens could, time W. The lines A1A2,, explain the absence of the backwave. However, this adhoc assumption is, B1B2 … etc, are, normal to both F1F2 not satisfactory and the absence of the backwave is really justified from, more rigorous wave theory., and G1G2 and, In a similar manner, we can use Huygens principle to determine the, represent rays., shape of the wavefront for a plane wave propagating through a medium, 354, (Fig. 10.3).

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Wave Optics, , 10.3 REFRACTION AND REFLECTION OF, P LANE W AVES USING H UYGENS, PRINCIPLE, 10.3.1 Refraction of a plane wave, , BC = v1 W, , FIGURE 10.4 A plane wave AB is incident at an angle i, on the surface PPc separating medium 1 and medium 2., The plane wave undergoes refraction and CE represents, the refracted wavefront. The figure corresponds to v2 < v1, so that the refracted waves bends towards the normal., , Christiaan Huygens, (1629 – 1695) Dutch, physicist, astronomer,, mathematician and the, founder of the wave, theory of light. His book,, Treatise on light, makes, fascinating reading even, today. He brilliantly, explained the double, refraction shown by the, mineral calcite in this, work in addition to, reflection and refraction., He was the first to, analyse circular and, simple harmonic motion, and designed and built, improved clocks and, telescopes. He discovered, the true geometry of, Saturn’s rings., , CHRISTIAAN HUYGENS (1629 – 1695), , We will now use Huygens principle to derive the laws of, refraction. Let PPc represent the surface separating medium, 1 and medium 2, as shown in Fig. 10.4. Let v1 and v2, represent the speed of light in medium 1 and medium 2,, respectively. We assume a plane wavefront AB propagating, in the direction AcA incident on the interface at an angle i, as shown in the figure. Let W be the time taken by the, wavefront to travel the distance BC. Thus,, , In order to determine the shape of the refracted wavefront, we draw a, sphere of radius v2W from the point A in the second medium (the speed of, the wave in the second medium is v2). Let CE represent a tangent plane, drawn from the point C on to the sphere. Then, AE = v2 Wand CE would, represent the refracted wavefront. If we now consider the triangles ABC, and AEC, we readily obtain, sin i =, , BC, AC, , v1W, AC, , (10.1), , and, AE v 2W, (10.2), AC AC, where i and r are the angles of incidence and refraction, respectively., , sin r =, , 355

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Physics, Thus we obtain, v1, v2, , (10.3), , From the above equation, we get the important result that if r < i (i.e.,, if the ray bends toward the normal), the speed of the light wave in the, second medium (v2) will be less then the speed of the light wave in the, first medium (v1). This prediction is opposite to the prediction from the, corpuscular model of light and as later experiments showed, the prediction, of the wave theory is correct. Now, if c represents the speed of light in, vacuum, then,, c, v1, , (10.4), , c, n2 = v, 2, , (10.5), , n1, and, , are known as the refractive indices of medium 1 and medium 2,, respectively. In terms of the refractive indices, Eq. (10.3) can be, written as, n1 sin i = n2 sin r, , (10.6), , This is the Snell’s law of refraction. Further, if O1 and O 2 denote the, wavelengths of light in medium 1 and medium 2, respectively and if the, distance BC is equal to O 1 then the distance AE will be equal to O 2 ( because, if the crest from B has reached C in time W, then the crest from A should, have also reached E in time W ); thus,, http://www.falstad.com/ripple/, , Demonstration of interference, diffraction, refraction, resonance and Doppler effect, , sin i, sin r, , O1, O2, , BC, AE, , v1, , v2, , O1, , O2, , v1, v2, , or, (10.7), , The above equation implies that when a wave gets refracted into a, denser medium (v1 > v2) the wavelength and the speed of propagation, decrease but the frequency Q(= v/O) remains the same., , 10.3.2 Refraction at a rarer medium, We now consider refraction of a plane wave at a rarer medium, i.e.,, v2 > v1. Proceeding in an exactly similar manner we can construct a, refracted wavefront as shown in Fig. 10.5. The angle of refraction, will now be greater than angle of incidence; however, we will still have, n1 sin i = n2 sin r . We define an angle ic by the following equation, n2, (10.8), n1, Thus, if i = ic then sin r = 1 and r = 90°. Obviously, for i > ic, there can, not be any refracted wave. The angle ic is known as the critical angle and, for all angles of incidence greater than the critical angle, we will not have, sin i c, , 356

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Wave Optics, any refracted wave and the wave will undergo what is known as total, internal reflection. The phenomenon of total internal reflection and its, applications was discussed in Section 9.4., , FIGURE 10.5 Refraction of a plane wave incident on a rarer medium for, which v2 > v1. The plane wave bends away from the normal., , 10.3.3 Reflection of a plane wave by a plane surface, We next consider a plane wave AB incident at an angle i on a reflecting, surface MN. If v represents the speed of the wave in the medium and if W, represents the time taken by the wavefront to advance from the point B, to C then the distance, BC = vW, In order the construct the reflected wavefront we draw a sphere of, radius vW from the point A as shown in Fig. 10.6. Let CE represent the, tangent plane drawn from the point C to this sphere. Obviously, AE = BC = vW, , FIGURE 10.6 Reflection of a plane wave AB by the reflecting surface MN., AB and CE represent incident and reflected wavefronts., , If we now consider the triangles EAC and BAC we will find that they, are congruent and therefore, the angles i and r (as shown in Fig. 10.6), would be equal. This is the law of reflection., Once we have the laws of reflection and refraction, the behaviour of, prisms, lenses, and mirrors can be understood. These phenomena were, , 357

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Physics, discussed in detail in Chapter 9 on the basis of rectilinear propagation of, light. Here we just describe the behaviour of the wavefronts as they, undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave, passing through a thin prism. Clearly, since the speed of light waves is, less in glass, the lower portion of the incoming wavefront (which travels, through the greatest thickness of glass) will get delayed resulting in a tilt, in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we, consider a plane wave incident on a thin convex lens; the central part of, the incident plane wave traverses the thickest portion of the lens and is, delayed the most. The emerging wavefront has a depression at the centre, and therefore the wavefront becomes spherical and converges to the point, F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on, a concave mirror and on reflection we have a spherical wave converging, to the focal point F. In a similar manner, we can understand refraction, and reflection by concave lenses and convex mirrors., , FIGURE 10.7 Refraction of a plane wave by (a) a thin prism, (b) a convex lens. (c) Reflection of a, plane wave by a concave mirror., , From the above discussion it follows that the total time taken from a, point on the object to the corresponding point on the image is the same, measured along any ray. For example, when a convex lens focusses light, to form a real image, although the ray going through the centre traverses, a shorter path, but because of the slower speed in glass, the time taken, is the same as for rays travelling near the edge of the lens., , 10.3.4 The doppler effect, , 358, , We should mention here that one should be careful in constructing the, wavefronts if the source (or the observer) is moving. For example, if there, is no medium and the source moves away from the observer, then later, wavefronts have to travel a greater distance to reach the observer and, hence take a longer time. The time taken between the arrival of two, successive wavefronts is hence longer at the observer than it is at the, source. Thus, when the source moves away from the observer the, frequency as measured by the source will be smaller. This is known as, the Doppler effect. Astronomers call the increase in wavelength due to, doppler effect as red shift since a wavelength in the middle of the visible, region of the spectrum moves towards the red end of the spectrum. When, waves are received from a source moving towards the observer, there is, an apparent decrease in wavelength, this is referred to as blue shift.

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Wave Optics, You have already encountered Doppler effect for sound waves in, Chapter 15 of Class XI textbook. For velocities small compared to the, speed of light, we can use the same formulae which we use for sound, waves. The fractional change in frequency 'Q/Q is given by –vradial/c, where, vradial is the component of the source velocity along the line joining the, observer to the source relative to the observer; vradial is considered positive, when the source moves away from the observer. Thus, the Doppler shift, can be expressed as:, 'Q, , –, , Q, , vradial, c, , (10.9), , The formula given above is valid only when the speed of the source is, small compared to that of light. A more accurate formula for the Doppler, effect which is valid even when the speeds are close to that of light, requires, the use of Einstein’s special theory of relativity. The Doppler effect for, light is very important in astronomy. It is the basis for the measurements, of the radial velocities of distant galaxies., Example 10.1 What speed should a galaxy move with respect, to us so that the sodium line at 589.0 nm is observed, at 589.6 nm?, Solution Since QO = c,, , 'Q, , Q, , –, , 'O, , O, , (for small changes in Q and O). For, , 'O = 589.6 – 589.0 = + 0.6 nm, we get [using Eq. (10.9)], 'Q, , Q, , –, , 'O, , O, , –, , vradial, c, , = 306 km/s, Therefore, the galaxy is moving away from us., , Solution, (a) Reflection and refraction arise through interaction of incident light, with the atomic constituents of matter. Atoms may be viewed as, , EXAMPLE 10.2, , Example 10.2, (a) When monochromatic light is incident on a surface separating, two media, the reflected and refracted light both have the same, frequency as the incident frequency. Explain why?, (b) When light travels from a rarer to a denser medium, the speed, decreases. Does the reduction in speed imply a reduction in the, energy carried by the light wave?, (c) In the wave picture of light, intensity of light is determined by the, square of the amplitude of the wave. What determines the intensity, of light in the photon picture of light., , EXAMPLE 10.1, , § 0.6 ·, , 5, –1, or, vradial #  c ¨©, ¸  3.06 u 10 m s, 589.0 ¹, , 359

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EXAMPLE 10.2, , Physics, oscillators, which take up the frequency of the external agency (light), causing forced oscillations. The frequency of light emitted by a charged, oscillator equals its frequency of oscillation. Thus, the frequency of, scattered light equals the frequency of incident light., (b) No. Energy carried by a wave depends on the amplitude of the, wave, not on the speed of wave propagation., (c) For a given frequency, intensity of light in the photon picture is, determined by the number of photons crossing an unit area per, unit time., , 10.4 COHERENT AND INCOHERENT ADDITION OF WAVES, , (a), , In this section we will discuss the interference pattern produced by, the superposition of two waves. You may recall that we had discussed, the superposition principle in Chapter 15 of your Class XI textbook., Indeed the entire field of interference is based on the superposition, principle according to which at a particular point in the medium, the, resultant displacement produced by a number of waves is the vector, sum of the displacements produced by each of the waves., Consider two needles S1 and S2 moving periodically up and down, in an identical fashion in a trough of water [Fig. 10.8(a)]. They produce, two water waves, and at a particular point, the phase difference between, the displacements produced by each of the waves does not change, with time; when this happens the two sources are said to be coherent., Figure 10.8(b) shows the position of crests (solid circles) and troughs, (dashed circles) at a given instant of time. Consider a point P for which, S1 P = S 2 P, , (b), , FIGURE 10.8 (a) Two, needles oscillating in, phase in water, represent two coherent, sources., (b) The pattern of, displacement of water, molecules at an, instant on the surface, of water showing nodal, N (no displacement), and antinodal A, (maximum, displacement) lines., , 360, , Since the distances S1 P and S2 P are equal, waves from S1 and S2, will take the same time to travel to the point P and waves that emanate, from S1 and S2 in phase will also arrive, at the point P, in phase., Thus, if the displacement produced by the source S1 at the point P, is given by, y1 = a cos Zt, then, the displacement produced by the source S2 (at the point P) will, also be given by, y2 = a cos Zt, Thus, the resultant of displacement at P would be given by, y = y1 + y2 = 2 a cos Zt, Since the intensity is the proportional to the square of the, amplitude, the resultant intensity will be given by, I = 4 I0, where I0 represents the intensity produced by each one of the individual, sources; I0 is proportional to a2. In fact at any point on the perpendicular, bisector of S1S2, the intensity will be 4I0. The two sources are said to

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Wave Optics, interfere constructively and we have what is referred to as constructive, interference. We next consider a point Q [Fig. 10.9(a)], for which, S2Q –S1Q = 2O, The waves emanating from S1 will arrive exactly two cycles earlier, than the waves from S2 and will again be in phase [Fig. 10.9(a)]. Thus, if, the displacement produced by S1 is given by, y1 = a cos Zt, then the displacement produced by S2 will be given by, y2 = a cos (Zt – 4S) = a cos Zt, where we have used the fact that a path difference of 2O corresponds to a, phase difference of 4S. The two displacements are once again in phase, and the intensity will again be 4 I0 giving rise to constructive interference., In the above analysis we have assumed that the distances S1Q and S2Q, are much greater than d (which represents the distance between S1 and, S2) so that although S1Q and S2Q are not equal, the amplitudes of the, displacement produced by each wave are very nearly the same., We next consider a point R [Fig. 10.9(b)] for which, FIGURE 10.9, (a) Constructive, The waves emanating from S1 will arrive exactly two and a half cycles, interference at a, point, Q for which the, later than the waves from S2 [Fig. 10.10(b)]. Thus if the displacement, path, difference, is 2O., produced by S1 is given by, (b) Destructive, y1 = a cos Zt, interference at a, point R for which the, then the displacement produced by S2 will be given by, path difference is, y2 = a cos (Zt + 5S) = – a cos Zt, 2.5 O ., , S2R – S1R = –2.5O, , where we have used the fact that a path difference of 2.5O corresponds to, a phase difference of 5S. The two displacements are now out of phase, and the two displacements will cancel out to give zero intensity. This is, referred to as destructive interference., To summarise: If we have two coherent sources S1 and S2 vibrating, in phase, then for an arbitrary point P whenever the path difference,, S1P ~S2P = nO, , (n = 0, 1, 2, 3,...), , (10.10), , we will have constructive interference and the resultant intensity will be, 4I0; the sign ~between S1P and S2 P represents the difference between, S1P and S2 P. On the other hand, if the point P is such that the path, difference,, 1, ) O (n = 0, 1, 2, 3, ...), (10.11), 2, FIGURE 10.10 Locus, we will have destructive interference and the resultant intensity will be of points for which, zero. Now, for any other arbitrary point G (Fig. 10.10) let the phase S P – S P is equal to, 1, 2, difference between the two displacements be I. Thus, if the displacement, zero, rO, r 2O, r 3O ., produced by S1 is given by, , S1P ~S2P = (n+, , y1 = a cos Zt, , 361

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Physics, then, the displacement produced by S2 would be, y2 = a cos (Zt + I ), and the resultant displacement will be given by, y = y1 + y2, = a [cos Zt + cos (Zt +I ], = 2 a cos (I/2) cos (Zt + I/2), The amplitude of the resultant displacement is 2a cos (I/2) and, therefore the intensity at that point will be, , http://www.colorado.edu/physics/2000/applets/fourier.html, , Ripple Tank experiments on wave interference, , I = 4 I0 cos2 I/2), , (10.12), , If I = 0, r 2 S, r 4S,… which corresponds to the condition given by, Eq. (10.10) we will have constructive interference leading to maximum, intensity. On the other hand, if I = r S, r 3S, r 5S … [which corresponds to, the condition given by Eq. (10.11)] we will have destructive interference, leading to zero intensity., Now if the two sources are coherent (i.e., if the two needles are going, up and down regularly) then the phase difference I at any point will not, change with time and we will have a stable interference pattern; i.e., the, positions of maxima and minima will not change with time. However, if, the two needles do not maintain a constant phase difference, then the, interference pattern will also change with time and, if the phase difference, changes very rapidly with time, the positions of maxima and minima will, also vary rapidly with time and we will see a “time-averaged” intensity, distribution. When this happens, we will observe an average intensity, that will be given by, ,  I ! 4I 0  cos2 I/2 !, , (10.13), , where angular brackets represent time averaging. Indeed it is shown in, Section 7.2 that if I (t ) varies randomly with time, the time-averaged, quantity < cos2 (I/2) > will be 1/2. This is also intuitively obvious because, the function cos2 (I/2) will randomly vary between 0 and 1 and the, average value will be 1/2. The resultant intensity will be given by, I = 2 I0, , (10.14), , at all points., When the phase difference between the two vibrating sources changes, rapidly with time, we say that the two sources are incoherent and when, this happens the intensities just add up. This is indeed what happens, when two separate light sources illuminate a wall., , 10.5 INTERFERENCE, EXPERIMENT, 362, , OF, , LIGHT WAVES, , AND, , YOUNG’S, , We will now discuss interference using light waves. If we use two sodium, lamps illuminating two pinholes (Fig. 10.11) we will not observe any, interference fringes. This is because of the fact that the light wave emitted, from an ordinary source (like a sodium lamp) undergoes abrupt phase

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Wave Optics, changes in times of the order of 10–10 seconds. Thus, the light waves coming out from two independent, sources of light will not have any fixed phase, relationship and would be incoherent, when this, happens, as discussed in the previous section, the, intensities on the screen will add up., The British physicist Thomas Young used an, ingenious technique to “lock” the phases of the waves, emanating from S1 and S2. He made two pinholes S1, FIGURE 10.11 If two sodium, and S2 (very close to each other) on an opaque screen, lamps illuminate two pinholes, [Fig. 10.12(a)]. These were illuminated by another, S1 and S2, the intensities will add, pinholes that was in turn, lit by a bright source. Light, up and no interference fringes will, be observed on the screen., waves spread out from S and fall on both S1 and S2., S1 and S2 then behave like two coherent sources, because light waves coming out from S1 and S2 are derived from the, same original source and any abrupt phase change in S will manifest in, exactly similar phase changes in the light coming out from S1 and S2., Thus, the two sources S1 and S2 will be locked in phase; i.e., they will be, coherent like the two vibrating needle in our water wave example, [Fig. 10.8(a)]., , (a), , (b), , FIGURE 10.12 Young’s arrangement to produce interference pattern., , Thus spherical waves emanating from S1 and S2 will produce, interference fringes on the screen GGc, as shown in Fig. 10.12(b). The, positions of maximum and minimum intensities can be calculated by, using the analysis given in Section 10.4 where we had shown that for an, arbitrary point P on the line GGc [Fig. 10.12(b)] to correspond to a, maximum, we must have, S2P – S1P = n O;, , n = 0, 1, 2 ..., , (10.15), , Now,, 2, ª 2 §, d· º, (S2P ) – (S1P ) = «D  ¨© x  2 ¸¹ » –, «¬, »¼, , 2, , 2, , 2, ª 2 §, d· º, «D  ¨ x – ¸ » = 2x d, ©, 2 ¹ »¼, «¬, , 363

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Physics, where S1S2 = d and OP = x . Thus, , THOMAS YOUNG (1773 – 1829), , 2 xd, S2P – S1P = S P+S P, 2, 1, , (10.16), , If x, d<<D then negligible error will be introduced if, S2P + S1P (in the denominator) is replaced by 2D. For, example, for d = 0.1 cm, D = 100 cm, OP = 1 cm (which, correspond to typical values for an interference, experiment using light waves), we have, Thomas, Young, (1773 – 1829) English, physicist, physician and, Egyptologist. Young worked, on a wide variety of, scientific problems, ranging, from the structure of the eye, and the mechanism of, vision to the decipherment, of the Rosetta stone. He, revived the wave theory of, light and recognised that, interference phenomena, provide proof of the wave, properties of light., , S2P + S1P = [(100)2 + (1.05)2]½ + [(100)2 + (0.95)2]½, | 200.01 cm, Thus if we replace S2P + S1P by 2 D, the error involved is, about 0.005%. In this approximation, Eq. (10.16), becomes, S2P – S1P |, , (10.17), , Hence we will have constructive interference resulting in, a bright region when, n OD, ; n = 0, r 1, r 2, ..., (10.18), d, On the other hand, we will have a dark region near, , x = xn =, , OD, , ; n 0, r 1, r 2, (10.19), d, Thus dark and bright bands appear on the screen, as shown in, Fig. 10.13. Such bands are called fringes. Equations (10.18) and (10.19), show that dark and bright fringes are equally spaced and the distance, between two consecutive bright and dark fringes is given by, , x = xn = (n+, , ), , E = xn+1 –xn, or E =, , 364, , OD, , (10.20), d, which is the expression for the fringe width. Obviously, the central point, O (in Fig. 10.12) will be bright because S1O = S2O and it will correspond, to n = 0. If we consider the line perpendicular to the plane of the paper, and passing through O [i.e., along the y-axis] then all points on this line, will be equidistant from S1 and S2 and we will have a bright central fringe, which is a straight line as shown in Fig. 10.13. In order to determine the, shape of the interference pattern on the screen we note that a particular, fringe would correspond to the locus of points with a constant value of, S2P – S1P. Whenever this constant is an integral multiple of O, the fringe, will be bright and whenever it is an odd integral multiple of O/2 it will be, a dark fringe. Now, the locus of the point P lying in the x-y plane such, that S2P – S1P (= ' is a constant, is a hyperbola. Thus the fringe pattern, will strictly be a hyperbola; however, if the distance D is very large compared, to the fringe width, the fringes will be very nearly straight lines as shown, in Fig. 10.13.

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Wave Optics, , FIGURE 10.13 Computer generated fringe pattern produced by two point source S1 and S2 on the, screen GGc (Fig. 10.12); (a) and (b) correspond to d = 0.005 mm and 0.025 mm, respectively (both, figures correspond to D = 5 cm and O = 5 × 10–5 cm.) (Adopted from OPTICS by A. Ghatak, Tata, McGraw Hill Publishing Co. Ltd., New Delhi, 2000.), , In the double-slit experiment shown in Fig. 10.12, we have taken the, source hole S on the perpendicular bisector of the two slits, which is, shown as the line SO. What happens if the source S is slightly away from, the perpendicular bisector. Consider that the source is moved to some, new point Sc and suppose that Q is the mid-point of S1 and S2. If the, angle ScQS is I, then the central bright fringe occurs at an angle –I, on, the other side. Thus, if the source S is on the perpendicular bisector,, then the central fringe occurs at O, also on the perpendicular bisector. If, S is shifted by an angle I to point Sc, then the central fringe appears at a, point Oc at an angle –I, which means that it is shifted by the same angle, on the other side of the bisector. This also means that the source Sc, the, mid-point Q and the point Oc of the central fringe are in a straight line., We end this section by quoting from the Nobel lecture of Dennis Gabor*, The wave nature of light was demonstrated convincingly for the, first time in 1801 by Thomas Young by a wonderfully simple, experiment. He let a ray of sunlight into a dark room, placed a, dark screen in front of it, pierced with two small pinholes, and, beyond this, at some distance, a white screen. He then saw two, darkish lines at both sides of a bright line, which gave him, sufficient encouragement to repeat the experiment, this time with, spirit flame as light source, with a little salt in it to produce the, bright yellow sodium light. This time he saw a number of dark, lines, regularly spaced; the first clear proof that light added to, light can produce darkness. This phenomenon is called, * Dennis Gabor received the 1971 Nobel Prize in Physics for discovering the, principles of holography., , 365

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Physics, interference. Thomas Young had expected it because he believed, in the wave theory of light., , FIGURE 10.14 Photograph and the graph of the intensity, distribution in Young’s double-slit experiment., , EXAMPLE 10.3, , http://vsg.quasihome.com/interfer.html, , Interactive animation of Young’s experiment, , We should mention here that the fringes are straight lines although, S1 and S2 are point sources. If we had slits instead of the point sources, (Fig. 10.14), each pair of points would have produced straight line fringes, resulting in straight line fringes with increased intensities., , Example 10.3 Two slits are made one millimetre apart and the screen, is placed one metre away. What is the fringe separation when bluegreen light of wavelength 500 nm is used?, Solution Fringe spacing =, , DO, d, , 1 u 5 u 10 –7, m, 1 u 10 –3, , = 5 × 10–4 m = 0.5 mm, , 366, , EXAMPLE 10.4, , Example 10.4 What is the effect on the interference fringes in a, Young’s double-slit experiment due to each of the following operations:, (a) the screen is moved away from the plane of the slits;, (b) the (monochromatic) source is replaced by another, (monochromatic) source of shorter wavelength;, (c) the separation between the two slits is increased;, (d) the source slit is moved closer to the double-slit plane;, (e) the width of the source slit is increased;, (f ) the monochromatic source is replaced by a source of white, light?

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Wave Optics, ( In each operation, take all parameters, other than the one specified,, to remain unchanged.), Solution, (a) Angular separation of the fringes remains constant, (= O /d). The actual, separation of the fringes increases in, proportion to the distance of the screen from the plane of the, two slits., (b) The separation of the fringes (and also angular separation), decreases. See, however, the condition mentioned in (d) below., (c) The separation of the fringes (and also angular separation), decreases. See, however, the condition mentioned in (d) below., (d) Let s be the size of the source and S its distance from the plane of, the two slits. For interference fringes to be seen, the condition, s/S < O/d should be satisfied; otherwise, interference patterns, produced by different parts of the source overlap and no fringes, are seen. Thus, as S decreases (i.e., the source slit is brought, closer), the interference pattern gets less and less sharp, and, when the source is brought too close for this condition to be valid,, the fringes disappear. Till this happens, the fringe separation, remains fixed., (e) Same as in (d). As the source slit width increases, fringe pattern, gets less and less sharp. When the source slit is so wide that the, condition s/S d O/d is not satisfied, the interference pattern, disappears., , Thus, the fringe closest on either side of the central white fringe, is red and the farthest will appear blue. After a few fringes, no, clear fringe pattern is seen., , EXAMPLE 10.4, , (f ) The interference patterns due to different component colours of, white light overlap (incoherently). The central bright fringes for, different colours are at the same position. Therefore, the central, fringe is white. For a point P for which S2P –S1P = Ob/2, where Ob, (| 4000 Å) represents the wavelength for the blue colour, the blue, component will be absent and the fringe will appear red in colour., Slightly farther away where S2Q–S1Q = Ob = Or/2 where Or (| 8000 Å), is the wavelength for the red colour, the fringe will be predominantly, blue., , 10.6 DIFFRACTION, If we look clearly at the shadow cast by an opaque object, close to the, region of geometrical shadow, there are alternate dark and bright regions, just like in interference. This happens due to the phenomenon of, diffraction. Diffraction is a general characteristic exhibited by all types of, waves, be it sound waves, light waves, water waves or matter waves. Since, the wavelength of light is much smaller than the dimensions of most, obstacles; we do not encounter diffraction effects of light in everyday, observations. However, the finite resolution of our eye or of optical, , 367

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Physics, instruments such as telescopes or microscopes is limited due to the, phenomenon of diffraction. Indeed the colours that you see when a CD is, viewed is due to diffraction effects. We will now discuss the phenomenon, of diffraction., , 10.6.1 The single slit, In the discussion of Young’s experiment, we stated that a single narrow, slit acts as a new source from which light spreads out. Even before Young,, early experimenters – including Newton – had noticed that light spreads, out from narrow holes and slits. It seems to turn around corners and, enter regions where we would expect a shadow. These effects, known as, diffraction, can only be properly understood using wave ideas. After all,, you are hardly surprised to hear sound waves from someone talking, around a corner !, When the double slit in Young’s experiment is replaced by a single, narrow slit (illuminated by a monochromatic source), a broad pattern, with a central bright region is seen. On both sides, there are alternate, dark and bright regions, the intensity becoming weaker away from the, centre (Fig. 10.16). To understand this, go to Fig. 10.15, which shows a, parallel beam of light falling normally on a single slit LN of width a. The, diffracted light goes on to meet a screen. The midpoint of the slit is M., A straight line through M perpendicular to the slit plane meets the, screen at C. We want the intensity at any point P on the screen. As before,, straight lines joining P to the different points L,M,N, etc., can be treated as, parallel, making an angle T with the normal MC., The basic idea is to divide the slit into much smaller parts, and add, their contributions at P with the proper phase differences. We are treating, different parts of the wavefront at the slit as secondary sources. Because, the incoming wavefront is parallel to the plane of the slit, these sources, are in phase., The path difference NP – LP between the two edges of the slit can be, calculated exactly as for Young’s experiment. From Fig. 10.15,, NP – LP = NQ, = a sin T, , | aT, , 368, , (10.21), , Similarly, if two points M1 and M2 in the slit plane are separated by y, the, path difference M2 P – M1P | yT. We now have to sum up equal, coherent, contributions from a large number of sources, each with a different phase., This calculation was made by Fresnel using integral calculus, so we omit, it here. The main features of the diffraction pattern can be understood by, simple arguments., At the central point C on the screen, the angle T is zero. All path, differences are zero and hence all the parts of the slit contribute in phase., This gives maximum intensity at C. Experimental observation shown in

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Wave Optics, Fig. 10.15 indicates that the intensity has a, central maximum at T = 0 and other, secondary maxima at T O(n+1/2) O/a, and, has minima (zero intensity) at T O nO/a,, n = ±1, ±2, ±3, .... It is easy to see why it has, minima at these values of angle. Consider, first the angle T where the path difference aT, is O. Then,, , T | O /a ., , (10.22), , Now, divide the slit into two equal halves, FIGURE 10.15 The geometry of path, differences, for diffraction by a single slit., LM and MN each of size a/2. For every point, M1 in LM, there is a point M2 in MN such that, M1M2 = a/2. The path difference between M1 and M2 at P = M2P – M1P, =T a/2 =O/2 for the angle chosen. This means that the contributions, from M1 and M2 are 180º out of phase and cancel in the direction, T = O/a. Contributions from the two halves of the slit LM and MN,, therefore, cancel each other. Equation (10.22) gives the angle at which, the intensity falls to zero. One can similarly show that the intensity is, zero for T = n O/a, with n being any integer (except zero!). Notice that the, angular size of the central maximum increases when the slit width a, decreases., It is also easy to see why there are maxima at T  (n + 1/2) O/a and, why they go on becoming weaker and weaker with increasing n. Consider, an angle T = 3O/2a which is midway between two of the dark fringes., Divide the slit into three equal parts. If we take the first two thirds of the, slit, the path difference between the two ends would be, 2, a uT, 3, , 2a 3O, u, 3, 2a, , O, , (10.23), , The first two-thirds of the slit can therefore be divided, into two halves which have a O/2 path difference. The, contributions of these two halves cancel in the same manner, as described earlier. Only the remaining one-third of the, slit contributes to the intensity at a point between the two, minima. Clearly, this will be much weaker than the central, maximum (where the entire slit contributes in phase). One, can similarly show that there are maxima at (n + 1/2)T/a, with n = 2, 3, etc. These become weaker with increasing n,, since only one-fifth, one-seventh, etc., of the slit contributes, in these cases. The photograph and intensity pattern, corresponding to it is shown in Fig. 10.16., There has been prolonged discussion about difference, between intereference and diffraction among scientists since, the discovery of these phenomena. In this context, it is, , FIGURE 10.16 Intensity, distribution and photograph of, fringes due to diffraction, at single slit., , 369

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Physics, interesting to note what Richard Feynman* has said in his famous, Feynman Lectures on Physics:, , In the double-slit experiment, we must note that the pattern on the, screen is actually a superposition of single-slit diffraction from each slit, or hole, and the double-slit interference pattern. This is shown in, Fig. 10.17. It shows a broader diffraction peak in which there appear, several fringes of smaller width due to double-slit interference. The, number of interference fringes occuring in the broad diffraction peak, depends on the ratio d/a, that is the ratio of the distance between the, two slits to the width of a slit. In the limit of a becoming very small, the, diffraction pattern will become very flat and we will obsrve the two-slit, interference pattern [see Fig. 10.13(b)]., , http://www.phys.hawaii.edu/~teb/optics/java/slitdiffr/, , FIGURE 10.17 The actual double-slit interference pattern., The envelope shows the single slit diffraction., Example 10.5 In Example 10.3, what should the width of each slit be, to obtain 10 maxima of the double slit pattern within the central, maximum of the single slit pattern?, , EXAMPLE 10.5, , Interactive animation on single slit diffraction pattern, , No one has ever been able to define the difference between, interference and diffraction satisfactorily. It is just a question, of usage, and there is no specific, important physical difference, between them. The best we can do is, roughly speaking, is to, say that when there are only a few sources, say two interfering, sources, then the result is usually called interference, but if there, is a large number of them, it seems that the word diffraction is, more often used., , Solution We want aT, 10, , O, d, , 2, , O, a, a, , d, 5, , O ,T, , O, a, , 0.2 mm, , Notice that the wavelength of light and distance of the screen do not, enter in the calculation of a., , In the double-slit interference experiment of Fig. 10.12, what happens, if we close one slit? You will see that it now amounts to a single slit. But, you will have to take care of some shift in the pattern. We now have a, source at S, and only one hole (or slit) S1 or S2. This will produce a single-, , 370, , * Richand Feynman was one of the recipients of the 1965 Nobel Prize in Physics, for his fundamental work in quantum electrodynamics.

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Wave Optics, slit diffraction pattern on the screen. The centre of the central bright fringe, will appear at a point which lies on the straight line SS1 or SS2, as the, case may be., We now compare and contrast the interference pattern with that seen, for a coherently illuminated single slit (usually called the single slit, diffraction pattern)., (i) The interference pattern has a number of equally spaced bright and, dark bands. The diffraction pattern has a central bright maximum, which is twice as wide as the other maxima. The intensity falls as we, go to successive maxima away from the centre, on either side., (ii) We calculate the interference pattern by superposing two waves, originating from the two narrow slits. The diffraction pattern is a, superposition of a continuous family of waves originating from each, point on a single slit., (iii) For a single slit of width a, the first null of the interference pattern, occurs at an angle of O/a. At the same angle of O/a, we get a maximum, (not a null) for two narrow slits separated by a distance a., One must understand that both d and a have to be quite small, to be, able to observe good interference and diffraction patterns. For example,, the separation d between the two slits must be of the order of a milimetre, or so. The width a of each slit must be even smaller, of the order of 0.1 or, 0.2 mm., In our discussion of Young’s experiment and the single-slit diffraction,, we have assumed that the screen on which the fringes are formed is at a, large distance. The two or more paths from the slits to the screen were, treated as parallel. This situation also occurs when we place a converging, lens after the slits and place the screen at the focus. Parallel paths from, the slit are combined at a single point on the screen. Note that the lens, does not introduce any extra path differences in a parallel beam. This, arrangement is often used since it gives more intensity than placing the, screen far away. If f is the focal length of the lens, then we can easily work, out the size of the central bright maximum. In terms of angles, the, separation of the central maximum from the first null of the diffraction, pattern is O/a . Hence, the size on the screen will be f O/a., , 10.6.2 Seeing the single slit diffraction pattern, It is surprisingly easy to see the single-slit diffraction pattern for oneself., The equipment needed can be found in most homes –– two razor blades, and one clear glass electric bulb preferably with a straight filament. One, has to hold the two blades so that the edges are parallel and have a, narrow slit in between. This is easily done with the thumb and forefingers, FIGURE 10.18, (Fig. 10.18)., Holding, two blades to, Keep the slit parallel to the filament, right in front of the eye. Use, form a single slit. A, spectacles if you normally do. With slight adjustment of the width of the, bulb filament viewed, slit and the parallelism of the edges, the pattern should be seen with its through this shows, bright and dark bands. Since the position of all the bands (except the, clear diffraction, central one) depends on wavelength, they will show some colours. Using, bands., a filter for red or blue will make the fringes clearer. With both filters, 371, available, the wider fringes for red compared to blue can be seen.

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Physics, In this experiment, the filament plays the role of the first slit S in, Fig. 10.16. The lens of the eye focuses the pattern on the screen (the, retina of the eye)., With some effort, one can cut a double slit in an aluminium foil with, a blade. The bulb filament can be viewed as before to repeat Young’s, experiment. In daytime, there is another suitable bright source subtending, a small angle at the eye. This is the reflection of the Sun in any shiny, convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage, the eye and will not give fringes anyway as the Sun subtends an angle, of (1/2)º., In interference and diffraction, light energy is redistributed. If it, reduces in one region, producing a dark fringe, it increases in another, region, producing a bright fringe. There is no gain or loss of energy,, which is consistent with the principle of conservation of energy., , 10.6.3 Resolving power of optical instruments, In Chapter 9 we had discussed about telescopes. The angular resolution, of the telescope is determined by the objective of the telescope. The stars, which are not resolved in the image produced by the objective cannot be, resolved by any further magnification produced by the eyepiece. The, primary purpose of the eyepiece is to provide magnification of the image, produced by the objective., Consider a parallel beam of light falling on a convex lens. If the lens is, well corrected for aberrations, then geometrical optics tells us that the, beam will get focused to a point. However, because of diffraction, the, beam instead of getting focused to a point gets focused to a spot of finite, area. In this case the effects due to diffraction can be taken into account, by considering a plane wave incident on a circular aperture followed by, a convex lens (Fig. 10.19). The analysis of the corresponding diffraction, pattern is quite involved; however, in principle, it is similar to the analysis, carried out to obtain the single-slit diffraction pattern. Taking into account, the effects due to diffraction, the pattern on the focal plane would consist, of a central bright region surrounded by concentric dark and bright rings, (Fig. 10.19). A detailed analysis shows that the radius of the central bright, region is approximately given by, r0 |, , 372, , 1.22 O f, 2a, , 0.61O f, a, , (10.24), , FIGURE 10.19 A parallel beam of light is incident on a convex lens., Because of diffraction effects, the beam gets focused to a, spot of radius | 0.61 Of/a .

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Wave Optics, where f is the focal length of the lens and 2a is the diameter of the circular, aperture or the diameter of the lens, whichever is smaller. Typically if, O | 0.5 Pm, f | 20 cm and a | 5 cm, we have, r0 | 1.2 Pm, Although the size of the spot is very small, it plays an important role, in determining the limit of resolution of optical instruments like a telescope, or a microscope. For the two stars to be just resolved, f 'T | r0 |, , 0.61O f, a, , implying, 0.61O, (10.25), a, Thus 'T will be small if the diameter of the objective is large. This, implies that the telescope will have better resolving power if a is large. It, is for this reason that for better resolution, a telescope must have a large, diameter objective., 'T |, , Example 10.6 Assume that light of wavelength 6000Å is coming from, a star. What is the limit of resolution of a telescope whose objective, has a diameter of 100 inch?, , 'T |, , 0.61 u 6 u 10 –5, 127, , | 2.9 u 10, , –7, , radians, , EXAMPLE 10.6, , Solution A 100 inch telescope implies that 2a = 100 inch, = 254 cm. Thus if,, O|6000Å = 6×10–5 cm, then, , We can apply a similar argument to the objective lens of a microscope., In this case, the object is placed slightly beyond f, so that a real image is, formed at a distance v [Fig. 10.20]. The magnification – ratio of, image size to object size – is given by m O v/f. It can be seen from, Fig. 10.20 that, D/f O 2 tan E, (10.26), where 2E is the angle subtended by the diameter of the objective lens at, the focus of the microscope., , FIGURE 10.20 Real image formed by the objective lens of the microscope., , 373

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Physics, DETERMINE, , THE RESOLVING POWER OF YOUR EYE, , You can estimate the resolving power of your eye with a simple experiment. Make, black stripes of equal width separated by white stripes; see figure here. All the black, stripes should be of equal width, while the width of the intermediate white stripes should, increase as you go from the left to the right. For example, let all black stripes have a width, of 5 mm. Let the width of the first two white stripes be 0.5 mm each, the next two white, stripes be 1 mm each, the next two 1.5 mm each, etc. Paste this pattern on a wall in a, room or laboratory, at the height of your eye., , Now watch the pattern, preferably with one eye. By moving away or closer to the wall,, find the position where you can just see some two black stripes as separate stripes. All, the black stripes to the left of this stripe would merge into one another and would not be, distinguishable. On the other hand, the black stripes to the right of this would be more, and more clearly visible. Note the width d of the white stripe which separates the two, regions, and measure the distance D of the wall from your eye. Then d/D is the resolution, of your eye., You have watched specks of dust floating in air in a sunbeam entering through your, window. Find the distance (of a speck) which you can clearly see and distinguish from a, neighbouring speck. Knowing the resolution of your eye and the distance of the speck,, estimate the size of the speck of dust., , When the separation between two points in a microscopic specimen, is comparable to the wavelength O of the light, the diffraction effects, become important. The image of a point object will again be a diffraction, pattern whose size in the image plane will be, vT, , § 1.22 O ·, v¨, © D ¸¹, , (10.27), , Two objects whose images are closer than this distance will not be, resolved, they will be seen as one. The corresponding minimum, separation, dmin, in the object plane is given by, , ª § 122O · º, ¸ m, dmin = «v ¨©, D ¹ »¼, ¬, =, , 122 O v, ., D, m, , 122 f O, D, Now, combining Eqs. (10.26) and (10.28), we get, , =, , 374, , d min, , 1.22 O, 2tanE, , (10.28)

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Wave Optics, , , 1.22 O, 2 sin E, , (10.29), , If the medium between the object and the objective lens is not air but, a medium of refractive index n, Eq. (10.29) gets modified to, 1.22 O, 2 n sin E, , d min, , (10.30), , The product n sinEis called the numerical aperture and is sometimes, marked on the objective., The resolving power of the microscope is given by the reciprocal of, the minimum separation of two points seen as distinct. It can be seen, from Eq. (10.30) that the resolving power can be increased by choosing a, medium of higher refractive index. Usually an oil having a refractive index, close to that of the objective glass is used. Such an arrangement is called, an ‘oil immersion objective’. Notice that it is not possible to make sinE, larger than unity. Thus, we see that the resolving power of a microscope, is basically determined by the wavelength of the light used., There is a likelihood of confusion between resolution and, magnification, and similarly between the role of a telescope and a, microscope to deal with these parameters. A telescope produces images, of far objects nearer to our eye. Therefore objects which are not resolved, at far distance, can be resolved by looking at them through a telescope., A microscope, on the other hand, magnifies objects (which are near to, us) and produces their larger image. We may be looking at two stars or, two satellites of a far-away planet, or we may be looking at different, regions of a living cell. In this context, it is good to remember that a, telescope resolves whereas a microscope magnifies., , 10.6.4 The validity of ray optics, An aperture (i.e., slit or hole) of size a illuminated by a parallel beam, sends diffracted light into an angle of approximately |O/a . This is the, angular size of the bright central maximum. In travelling a distance z,, the diffracted beam therefore acquires a width zO/a due to diffraction. It, is interesting to ask at what value of z the spreading due to diffraction, becomes comparable to the size a of the aperture. We thus approximately, equate zO/a with a. This gives the distance beyond which divergence of, the beam of width a becomes significant. Therefore,, z, , a2, , O, , (10.31), , We define a quantity zF called the Fresnel distance by the following, equation, , zF  a 2 / O, Equation (10.31) shows that for distances much smaller than zF , the, spreading due to diffraction is smaller compared to the size of the beam., It becomes comparable when the distance is approximately zF . For, distances much greater than zF , the spreading due to diffraction, , 375

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Physics, dominates over that due to ray optics (i.e., the size a of the aperture)., Equation (10.31) also shows that ray optics is valid in the limit of, wavelength tending to zero., , EXAMPLE 10.7, , Example 10.7 For what distance is ray optics a good approximation, when the aperture is 3 mm wide and the wavelength is 500 nm?, , Solution z F, , a, , 2, , O, , 3 u 10 –3, 5 u 10, , –7, , 2, , 18 m, , This example shows that even with a small aperture, diffraction, spreading can be neglected for rays many metres in length. Thus, ray, optics is valid in many common situations., , 10.7 POLARISATION, Consider holding a long string that is held horizontally, the other end of, which is assumed to be fixed. If we move the end of the string up and, down in a periodic manner, we will generate a wave propagating in the, +x direction (Fig. 10.22). Such a wave could be described by the following, equation, , 376, , FIGURE 10.21 (a) The curves represent the displacement of a string at, t = 0 and at t = 't, respectively when a sinusoidal wave is propagating, in the +x-direction. (b) The curve represents the time variation, of the displacement at x = 0 when a sinusoidal wave is propagating, in the +x-direction. At x = 'x, the time variation of the, displacement will be slightly displaced to the right.

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Wave Optics, y (x,t ) = a sin (kx – Zt), , (10.32), , where a and Z = 2SQ ) represent the amplitude and the angular frequency, of the wave, respectively; further,, 2S, (10.33), k, represents the wavelength associated with the wave. We had discussed, propagation of such waves in Chapter 15 of Class XI textbook. Since the, displacement (which is along the y direction) is at right angles to the, direction of propagation of the wave, we have what is known as a, transverse wave. Also, since the displacement is in the y direction, it is, often referred to as a y-polarised wave. Since each point on the string, moves on a straight line, the wave is also referred to as a linearly polarised, wave. Further, the string always remains confined to the x-y plane and, therefore it is also referred to as a plane polarised wave., In a similar manner we can consider the vibration of the string in the, x-z plane generating a z-polarised wave whose displacement will be given, by, , O, , z (x,t ) = a sin (kx – Zt ), , (10.34), , It should be mentioned that the linearly polarised waves [described, by Eqs. (10.33) and (10.34)] are all transverse waves; i.e., the, displacement of each point of the string is always at right angles to the, direction of propagation of the wave. Finally, if the plane of vibration of, the string is changed randomly in very short intervals of time, then we, have what is known as an unpolarised wave. Thus, for an unpolarised, wave the displacement will be randomly changing with time though it, will always be perpendicular to the direction of propagation., Light waves are transverse in nature; i.e., the electric field associated, with a propagating light wave is always at right angles to the direction of, propagation of the wave. This can be easily demonstrated using a simple, polaroid. You must have seen thin plastic like sheets, which are called, polaroids. A polaroid consists of long chain molecules aligned in a, particular direction. The electric vectors (associated with the propagating, light wave) along the direction of the aligned molecules get absorbed., Thus, if an unpolarised light wave is incident on such a polaroid then, the light wave will get linearly polarised with the electric vector oscillating, along a direction perpendicular to the aligned molecules; this direction, is known as the pass-axis of the polaroid., Thus, if the light from an ordinary source (like a sodium lamp) passes, through a polaroid sheet P1, it is observed that its intensity is reduced by, half. Rotating P1 has no effect on the transmitted beam and transmitted, intensity remains constant. Now, let an identical piece of polaroid P2 be, placed before P1. As expected, the light from the lamp is reduced in, intensity on passing through P2 alone. But now rotating P1 has a dramatic, effect on the light coming from P2. In one position, the intensity transmitted, , 377

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Physics, by P2 followed by P1 is nearly zero. When turned by 90º from this position,, P1 transmits nearly the full intensity emerging from P2 (Fig. 10.22)., , FIGURE 10.22 (a) Passage of light through two polaroids P2 and P1. The, transmitted fraction falls from 1 to 0 as the angle between them varies, from 0º to 90º. Notice that the light seen through a single polaroid, P1 does not vary with angle. (b) Behaviour of the electric vector, when light passes through two polaroids. The transmitted, polarisation is the component parallel to the polaroid axis., The double arrows show the oscillations of the electric vector., , The above experiment can be easily understood by assuming that, light passing through the polaroid P2 gets polarised along the pass-axis, of P2. If the pass-axis of P2 makes an angle T with the pass-axis of P1,, then when the polarised beam passes through the polaroid P2, the, component E cos T (along the pass-axis of P2) will pass through P2., Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as:, , 378, , EXAMPLE 10.8, , (10.35), I = I0 cos2T, where I0 is the intensity of the polarized light after passing through, P1. This is known as Malus’ law. The above discussion shows that the, intensity coming out of a single polaroid is half of the incident intensity., By putting a second polaroid, the intensity can be further controlled, from 50% to zero of the incident intensity by adjusting the angle between, the pass-axes of two polaroids., Polaroids can be used to control the intensity, in sunglasses,, windowpanes, etc. Polaroids are also used in photographic cameras and, 3D movie cameras., Example 10.8 Discuss the intensity of transmitted light when a, polaroid sheet is rotated between two crossed polaroids?, Solution Let I0 be the intensity of polarised light after passing through, the first polariser P1. Then the intensity of light after passing through, second polariser P2 will be, , I, , I 0 cos 2T ,

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Wave Optics, , I, , §S, ·, I 0 cos 2T cos 2 ¨ – T ¸, ©2, ¹, , = I0 cos2T sin2T =(I0/4) sin22T, Therefore, the transmitted intensity will be maximum when T = S/4., , EXAMPLE 10.8, , where T is the angle between pass axes of P1 and P2. Since P1 and P3, are crossed the angle between the pass axes of P2 and P3 will be, (S/2 – T ). Hence the intensity of light emerging from P3 will be, , 10.7.1 Polarisation by scattering, The light from a clear blue portion of the sky shows a rise and fall of, intensity when viewed through a polaroid which is rotated. This is nothing, but sunlight, which has changed its direction (having been scattered) on, encountering the molecules of the earth’s atmosphere. As Fig. 10.24(a), shows, the incident sunlight is unpolarised. The dots stand for polarisation, perpendicular to the plane of the figure. The double arrows show, polarisation in the plane of the figure. (There is no phase relation between, these two in unpolarised light). Under the influence of the electric field of, the incident wave the electrons in the molecules acquire components of, motion in both these directions. We have drawn an observer looking at, 90° to the direction of the sun. Clearly, charges accelerating parallel to, the double arrows do not radiate energy towards this observer since their, acceleration has no transverse component. The radiation scattered by, the molecule is therefore represented by dots. It is polarised, perpendicular to the plane of the figure. This explains the polarisation of, scattered light from the sky., , FIGURE 10.23 (a) Polarisation of the blue scattered light from the sky., The incident sunlight is unpolarised (dots and arrows). A typical, molecule is shown. It scatters light by 90º polarised normal to, the plane of the paper (dots only). (b) Polarisation of light, reflected from a transparent medium at the Brewster angle, (reflected ray perpendicular to refracted ray)., , The scattering of light by molecules was intensively investigated by, C.V. Raman and his collaborators in Kolkata in the 1920s. Raman was, awarded the Nobel Prize for Physics in 1930 for this work., , 379

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Physics, A, , SPECIAL CASE OF TOTAL TRANSMISSION, , When light is incident on an interface of two media, it is observed that some part of it, gets reflected and some part gets transmitted. Consider a related question: Is it possible, that under some conditions a monochromatic beam of light incident on a surface, (which is normally reflective) gets completely transmitted with no reflection? To your, surprise, the answer is yes., , Let us try a simple experiment and check what happens. Arrange a laser, a good, polariser, a prism and screen as shown in the figure here., Let the light emitted by the laser source pass through the polariser and be incident, on the surface of the prism at the Brewster’s angle of incidence iB. Now rotate the, polariser carefully and you will observe that for a specific alignment of the polariser, the, light incident on the prism is completely transmitted and no light is reflected from the, surface of the prism. The reflected spot will completely vanish., , 10.7.2 Polarisation by reflection, Figure 10.23(b) shows light reflected from a transparent medium, say,, water. As before, the dots and arrows indicate that both polarisations are, present in the incident and refracted waves. We have drawn a situation, in which the reflected wave travels at right angles to the refracted wave., The oscillating electrons in the water produce the reflected wave. These, move in the two directions transverse to the radiation from wave in the, medium, i.e., the refracted wave. The arrows are parallel to the direction, of the reflected wave. Motion in this direction does not contribute to the, reflected wave. As the figure shows, the reflected light is therefore linearly, polarised perpendicular to the plane of the figure (represented by dots)., This can be checked by looking at the reflected light through an analyser., The transmitted intensity will be zero when the axis of the analyser is in, the plane of the figure, i.e., the plane of incidence., When unpolarised light is incident on the boundary between two, transparent media, the reflected light is polarised with its electric vector, perpendicular to the plane of incidence when the refracted and reflected, rays make a right angle with each other. Thus we have seen that when, reflected wave is perpendicular to the refracted wave, the reflected wave, is a totally polarised wave. The angle of incidence in this case is called, Brewster’s angle and is denoted by i B. We can see that i B is related to the, refractive index of the denser medium. Since we have i B+r = S/2, we get, from Snell’s law, , 380, , P, , sin i B, sin r, , sin i B, sin S / 2 – i B

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Wave Optics, sin i B, cos i B, , tan i B, , (10.36), , This is known as Brewster’s law., , Solution For i + r to be equal to S/2, we should have tan iB = P = 1.5., This gives i B = 57°. This is the Brewster’s angle for air to glass, interface., , EXAMPLE 10.9, , Example 10.9 Unpolarised light is incident on a plane glass surface., What should be the angle of incidence so that the reflected and, refracted rays are perpendicular to each other?, , For simplicity, we have discussed scattering of light by 90º, and, reflection at the Brewster angle. In this special situation, one of the two, perpendicular components of the electric field is zero. At other angles,, both components are present but one is stronger than the other. There is, no stable phase relationship between the two perpendicular components, since these are derived from two perpendicular components of an, unpolarised beam. When such light is viewed through a rotating analyser,, one sees a maximum and a minimum of intensity but not complete, darkness. This kind of light is called partially polarised., Let us try to understand the situation. When an unpolarised beam of, light is incident at the Brewster’s angle on an interface of two media, only, part of light with electric field vector perpendicular to the plane of, incidence will be reflected. Now by using a good polariser, if we completely, remove all the light with its electric vector perpendicular to the plane of, incidence and let this light be incident on the surface of the prism at, Brewster’s angle, you will then observe no reflection and there will be, total transmission of light., We began this chapter by pointing out that there are some phenomena, which can be explained only by the wave theory. In order to develop a, proper understanding, we first described how some phenomena like, reflection and refraction, which were studied on this basis of Ray Optics, in Chapter 9, can also be understood on the basis of Wave Optics. Then, we described Young’s double slit experiment which was a turning point, in the study of optics. Finally, we described some associated points such, as diffraction, resolution, polarisation, and validity of ray optics. In the, next chapter, you will see how new experiments led to new theories at, the turn of the century around 1900 A.D., , SUMMARY, 1., 2., , Huygens’ principle tells us that each point on a wavefront is a source, of secondary waves, which add up to give the wavefront at a later time., Huygens’ construction tells us that the new wavefront is the forward, envelope of the secondary waves. When the speed of light is, independent of direction, the secondary waves are spherical. The rays, are then perpendicular to both the wavefronts and the time of travel, , 381

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Physics, 3., , 4., , 5., , 6., , is the same measured along any ray. This principle leads to the well, known laws of reflection and refraction., The principle of superposition of waves applies whenever two or more, sources of light illuminate the same point. When we consider the, intensity of light due to these sources at the given point, there is an, interference term in addition to the sum of the individual intensities., But this term is important only if it has a non-zero average, which, occurs only if the sources have the same frequency and a stable, phase difference., Young’s double slit of separation d gives equally spaced fringes of, angular separation O/d. The source, mid-point of the slits, and central, bright fringe lie in a straight line. An extended source will destroy, the fringes if it subtends angle more than O/d at the slits., A single slit of width a gives a diffraction pattern with a central, O 2O, , etc.,, maximum. The intensity falls to zero at angles of r , r, a, a, with successively weaker secondary maxima in between. Diffraction, limits the angular resolution of a telescope to O/D where D is the, diameter. Two stars closer than this give strongly overlapping images., Similarly, a microscope objective subtending angle 2E at the focus,, in a medium of refractive index n, will just separate two objects spaced, at a distance O /(2n sin E , which is the resolution limit of a, microscope. Diffraction determines the limitations of the concept of, light rays. A beam of width a travels a distance a2/O, called the Fresnel, distance, before it starts to spread out due to diffraction., Natural light, e.g., from the sun is unpolarised. This means the electric, vector takes all possible directions in the transverse plane, rapidly, and randomly, during a measurement. A polaroid transmits only one, component (parallel to a special axis). The resulting light is called, linearly polarised or plane polarised. When this kind of light is viewed, through a second polaroid whose axis turns through 2S, two maxima, and minima of intensity are seen. Polarised light can also be produced, by reflection at a special angle (called the Brewster angle) and by, scattering through S/2 in the earth’s atmosphere., , POINTS TO PONDER, 1., , 2., , 3., , 4., , 5., , 382, , Waves from a point source spread out in all directions, while light was, seen to travel along narrow rays. It required the insight and experiment, of Huygens, Young and Fresnel to understand how a wave theory could, explain all aspects of the behaviour of light., The crucial new feature of waves is interference of amplitudes from different, sources which can be both constructive and destructive, as shown in, Young’s experiment., Even a wave falling on single slit should be regarded as a large number of, sources which interefere constructively in the forward direction (T = 0),, and destructively in other directions., Diffraction phenomena define the limits of ray optics. The limit of the, ability of microscopes and telescopes to distinguish very close objects is, set by the wavelength of light., Most interference and diffraction effects exist even for longitudinal waves, like sound in air. But polarisation phenomena are special to transverse, waves like light waves.

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Wave Optics, , EXERCISES, 10.1, , Monochromatic light of wavelength 589 nm is incident from air on a, water surface. What are the wavelength, frequency and speed of, (a) reflected, and (b) refracted light? Refractive index of water is, 1.33., , 10.2, , What is the shape of the wavefront in each of the following cases:, (a) Light diverging from a point source., (b) Light emerging out of a convex lens when a point source is placed, at its focus., (c) The portion of the wavefront of light from a distant star intercepted, by the Earth., , 10.3, , (a) The refractive index of glass is 1.5. What is the speed of light in, glass? (Speed of light in vacuum is 3.0 × 108 m s–1), (b) Is the speed of light in glass independent of the colour of light? If, not, which of the two colours red and violet travels slower in a, glass prism?, , 10.4, , In a Young’s double-slit experiment, the slits are separated by, 0.28 mm and the screen is placed 1.4 m away. The distance between, the central bright fringe and the fourth bright fringe is measured, to be 1.2 cm. Determine the wavelength of light used in the, experiment., , 10.5, , In Young’s double-slit experiment using monochromatic light of, wavelength O, the intensity of light at a point on the screen where, path difference is O, is K units. What is the intensity of light at a, point where path difference is O/3 ?, , 10.6, , A beam of light consisting of two wavelengths, 650 nm and 520 nm,, is used to obtain interference fringes in a Young’s double-slit, experiment., (a) Find the distance of the third bright fringe on the screen from, the central maximum for wavelength 650 nm., (b) What is the least distance from the central maximum where the, bright fringes due to both the wavelengths coincide?, , 10.7, , In a double-slit experiment the angular width of a fringe is found to, be 0.2° on a screen placed 1 m away. The wavelength of light used is, 600 nm. What will be the angular width of the fringe if the entire, experimental apparatus is immersed in water? Take refractive index, of water to be 4/3., , 10.8, , What is the Brewster angle for air to glass transition? (Refractive, index of glass = 1.5.), , 10.9, , Light of wavelength 5000 Å falls on a plane reflecting surface. What, are the wavelength and frequency of the reflected light? For what, angle of incidence is the reflected ray normal to the incident ray?, , 10.10 Estimate the distance for which ray optics is good approximation, for an aperture of 4 mm and wavelength 400 nm., , 383

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Physics, ADDITIONAL EXERCISES, 10.11 The 6563 Å HD line emitted by hydrogen in a star is found to be redshifted by 15 Å. Estimate the speed with which the star is receding, from the Earth., 10.12 Explain how Corpuscular theory predicts the speed of light in a, medium, say, water, to be greater than the speed of light in vacuum., Is the prediction confirmed by experimental determination of the, speed of light in water ? If not, which alternative picture of light is, consistent with experiment?, 10.13 You have learnt in the text how Huygens’ principle leads to the laws, of reflection and refraction. Use the same principle to deduce directly, that a point object placed in front of a plane mirror produces a, virtual image whose distance from the mirror is equal to the object, distance from the mirror., 10.14 Let us list some of the factors, which could possibly influence the, speed of wave propagation:, (i) nature of the source., (ii) direction of propagation., (iii) motion of the source and/or observer., (iv) wavelength., (v) intensity of the wave., On which of these factors, if any, does, (a) the speed of light in vacuum,, (b) the speed of light in a medium (say, glass or water),, depend ?, 10.15 For sound waves, the Doppler formula for frequency shift differs, slightly between the two situations: (i) source at rest; observer, moving, and (ii) source moving; observer at rest. The exact Doppler, formulas for the case of light waves in vacuum are, however, strictly, identical for these situations. Explain why this should be so. Would, you expect the formulas to be strictly identical for the two situations, in case of light travelling in a medium?, 10.16 In double-slit experiment using light of wavelength 600 nm, the, angular width of a fringe formed on a distant screen is 0.1º. What is, the spacing between the two slits?, 10.17 Answer the following questions:, (a) In a single slit diffraction experiment, the width of the slit is, made double the original width. How does this affect the size, and intensity of the central diffraction band?, (b) In what way is diffraction from each slit related to the, interference pattern in a double-slit experiment?, (c) When a tiny circular obstacle is placed in the path of light from, a distant source, a bright spot is seen at the centre of the shadow, of the obstacle. Explain why?, , 384, , (d) Two students are separated by a 7 m partition wall in a room, 10 m high. If both light and sound waves can bend around

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Wave Optics, , 10.18, , 10.19, , 10.20, , 10.21, , obstacles, how is it that the students are unable to see each, other even though they can converse easily., (e) Ray optics is based on the assumption that light travels in a, straight line. Diffraction effects (observed when light propagates, through small apertures/slits or around small obstacles), disprove this assumption. Yet the ray optics assumption is so, commonly used in understanding location and several other, properties of images in optical instruments. What is the, justification?, Two towers on top of two hills are 40 km apart. The line joining, them passes 50 m above a hill halfway between the towers. What is, the longest wavelength of radio waves, which can be sent between, the towers without appreciable diffraction effects?, A parallel beam of light of wavelength 500 nm falls on a narrow slit, and the resulting diffraction pattern is observed on a screen 1 m, away. It is observed that the first minimum is at a distance of 2.5, mm from the centre of the screen. Find the width of the slit., Answer the following questions:, (a) When a low flying aircraft passes overhead, we sometimes notice, a slight shaking of the picture on our TV screen. Suggest a, possible explanation., (b) As you have learnt in the text, the principle of linear, superposition of wave displacement is basic to understanding, intensity distributions in diffraction and interference patterns., What is the justification of this principle?, In deriving the single slit diffraction pattern, it was stated that the, intensity is zero at angles of n O/a. Justify this by suitably dividing, the slit to bring out the cancellation., , 385

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Chapter Eleven, , DUAL NATURE OF, RADIATION AND, MATTER, 11.1 INTRODUCTION, The Maxwell’s equations of electromagnetism and Hertz experiments on, the generation and detection of electromagnetic waves in 1887 strongly, established the wave nature of light. Towards the same period at the end, of 19th century, experimental investigations on conduction of electricity, (electric discharge) through gases at low pressure in a discharge tube led, to many historic discoveries. The discovery of X-rays by Roentgen in 1895,, and of electron by J. J. Thomson in 1897, were important milestones in, the understanding of atomic structure. It was found that at sufficiently, low pressure of about 0.001 mm of mercury column, a discharge took, place between the two electrodes on applying the electric field to the gas, in the discharge tube. A fluorescent glow appeared on the glass opposite, to cathode. The colour of glow of the glass depended on the type of glass,, it being yellowish-green for soda glass. The cause of this fluorescence, was attributed to the radiation which appeared to be coming from the, cathode. These cathode rays were discovered, in 1870, by William, Crookes who later, in 1879, suggested that these rays consisted of streams, of fast moving negatively charged particles. The British physicist, J. J. Thomson (1856 -1940) confirmed this hypothesis. By applying, mutually perpendicular electric and magnetic fields across the discharge, tube, J. J. Thomson was the first to determine experimentally the speed, and the specific charge [charge to mass ratio (e/m )] of the cathode ray

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Dual Nature of Radiation, and Matter, particles. They were found to travel with speeds ranging from about 0.1, to 0.2 times the speed of light (3 ×108 m/s). The presently accepted value, of e/m is 1.76 × 1011 C/kg. Further, the value of e/m was found to be, independent of the nature of the material/metal used as the cathode, (emitter), or the gas introduced in the discharge tube. This observation, suggested the universality of the cathode ray particles., Around the same time, in 1887, it was found that certain metals, when, irradiated by ultraviolet light, emitted negatively charged particles having, small speeds. Also, certain metals when heated to a high temperature were, found to emit negatively charged particles. The value of e/m of these particles, was found to be the same as that for cathode ray particles. These, observations thus established that all these particles, although produced, under different conditions, were identical in nature. J. J. Thomson, in 1897,, named these particles as electrons, and suggested that they were, fundamental, universal constituents of matter. For his epoch-making, discovery of electron, through his theoretical and experimental, investigations on conduction of electricity by gasses, he was awarded the, Nobel Prize in Physics in 1906. In 1913, the American physicist R. A., Millikan (1868 -1953) performed the pioneering oil-drop experiment for, the precise measurement of the charge on an electron. He found that the, charge on an oil-droplet was always an integral multiple of an elementary, charge, 1.602 × 10 –19 C. Millikan’s experiment established that electric, charge is quantised. From the values of charge (e ) and specific charge, (e/m ), the mass (m) of the electron could be determined., , 11.2 ELECTRON EMISSION, We know that metals have free electrons (negatively charged particles) that, are responsible for their conductivity. However, the free electrons cannot, normally escape out of the metal surface. If an electron attempts to come, out of the metal, the metal surface acquires a positive charge and pulls the, electron back to the metal. The free electron is thus held inside the metal, surface by the attractive forces of the ions. Consequently, the electron can, come out of the metal surface only if it has got sufficient energy to overcome, the attractive pull. A certain minimum amount of energy is required to be, given to an electron to pull it out from the surface of the metal. This, minimum energy required by an electron to escape from the metal surface, is called the work function of the metal. It is generally denoted by I0and, measured in eV (electron volt). One electron volt is the energy gained by an, electron when it has been accelerated by a potential difference of 1 volt, so, that 1 eV = 1.602 ×10 –19 J., This unit of energy is commonly used in atomic and nuclear physics., The work function (I0 ) depends on the properties of the metal and the, nature of its surface. The values of work function of some metals are, given in Table 11.1. These values are approximate as they are very, sensitive to surface impurities., Note from Table 11.1 that the work function of platinum is the highest, (I0= 5.65 eV ) while it is the lowest (I0 = 2.14 eV ) for caesium., The minimum energy required for the electron emission from the metal, surface can be supplied to the free electrons by any one of the following, physical processes:, , 387

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Physics, TABLE 11.1 WORK FUNCTIONS OF SOME METALS, Metal, , Work function, IR (eV), , Metal, , Work function, IR (eV), , Cs, , 2.14, , Al, , 4.28, , K, , 2.30, , Hg, , 4.49, , Na, , 2.75, , Cu, , 4.65, , Ca, , 3.20, , Ag, , 4.70, , Mo, , 4.17, , Ni, , 5.15, , Pb, , 4.25, , Pt, , 5.65, , (i) Thermionic emission: By suitably heating, sufficient thermal energy, can be imparted to the free electrons to enable them to come out of the, metal., (ii) Field emission : By applying a very strong electric field (of the order of, 108 V m–1) to a metal, electrons can be pulled out of the metal, as in a, spark plug., (iii) Photo-electric emission : When light of suitable frequency illuminates, a metal surface, electrons are emitted from the metal surface. These, photo(light)-generated electrons are called photoelectrons., , 11.3 PHOTOELECTRIC EFFECT, 11.3.1 Hertz’s observations, The phenomenon of photoelectric emission was discovered in 1887 by, Heinrich Hertz (1857-1894), during his electromagnetic wave experiments., In his experimental investigation on the production of electromagnetic, waves by means of a spark discharge, Hertz observed that high voltage, sparks across the detector loop were enhanced when the emitter plate, was illuminated by ultraviolet light from an arc lamp., Light shining on the metal surface somehow facilitated the escape of, free, charged particles which we now know as electrons. When light falls, on a metal surface, some electrons near the surface absorb enough energy, from the incident radiation to overcome the attraction of the positive ions, in the material of the surface. After gaining sufficient energy from the, incident light, the electrons escape from the surface of the metal into the, surrounding space., , 11.3.2 Hallwachs’ and Lenard’s observations, , 388, , Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of, photoelectric emission in detail during 1886-1902., Lenard (1862-1947) observed that when ultraviolet radiations were, allowed to fall on the emitter plate of an evacuated glass tube enclosing, two electrodes (metal plates), current flows in the circuit (Fig. 11.1). As, soon as the ultraviolet radiations were stopped, the current flow also

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Dual Nature of Radiation, and Matter, , OF, , P HOTOELECTRIC, , Figure 11.1 depicts a schematic view of the arrangement used for the, experimental study of the photoelectric effect. It consists of an evacuated, glass/quartz tube having a photosensitive plate C and another metal, plate A. Monochromatic light from the source S of sufficiently short, wavelength passes through the window W and falls on the photosensitive, plate C (emitter). A transparent quartz window is sealed on to the glass, tube, which permits ultraviolet radiation to pass through it and irradiate, the photosensitive plate C. The electrons are emitted by the plate C and, are collected by the plate A (collector), by the electric field created by the, battery. The battery maintains the potential difference between the plates, C and A, that can be varied. The polarity of the plates C and A can be, reversed by a commutator. Thus, the plate A can be maintained at a desired, positive or negative potential with respect to emitter C. When the collector, plate A is positive with respect to the emitter plate C, the electrons are, , Simulate experiments on photoelectric effect, , 11.4 E XPERIMENTAL S TUDY, EFFECT, , http://physics.berea.edu/~king/Teaching/ModPhys/QM/Photoelectric/Photoelectric.html, , stopped. These observations indicate that when ultraviolet radiations fall, on the emitter plate C, electrons are ejected from it which are attracted, towards the positive, collector plate A by the electric field. The electrons, flow through the evacuated glass tube, resulting in the current flow. Thus,, light falling on the surface of the emitter causes current in the external, circuit. Hallwachs and Lenard studied how this photo current varied with, collector plate potential, and with frequency and intensity of incident light., Hallwachs, in 1888, undertook the study further and connected a, negatively charged zinc plate to an electroscope. He observed that the, zinc plate lost its charge when it was illuminated by ultraviolet light., Further, the uncharged zinc plate became positively charged when it was, irradiated by ultraviolet light. Positive charge on a positively charged, zinc plate was found to be further enhanced when it was illuminated by, ultraviolet light. From these observations he concluded that negatively, charged particles were emitted from the zinc plate under the action of, ultraviolet light., After the discovery of the electron in 1897, it became evident that the, incident light causes electrons to be emitted from the emitter plate. Due, to negative charge, the emitted electrons are pushed towards the collector, plate by the electric field. Hallwachs and Lenard also observed that when, ultraviolet light fell on the emitter plate, no electrons were emitted at all, when the frequency of the incident light was smaller than a certain, minimum value, called the threshold frequency. This minimum frequency, depends on the nature of the material of the emitter plate., It was found that certain metals like zinc, cadmium, magnesium, etc.,, responded only to ultraviolet light, having short wavelength, to cause, electron emission from the surface. However, some alkali metals such as, lithium, sodium, potassium, caesium and rubidium were sensitive, even to visible light. All these photosensitive substances emit electrons, when they are illuminated by light. After the discovery of electrons, these, electrons were termed as photoelectrons. The phenomenon is called, photoelectric effect., , 389

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Physics, , FIGURE 11.1 Experimental, arrangement for study of, photoelectric effect., , attracted to it. The emission of electrons causes flow of, electric current in the circuit. The potential difference, between the emitter and collector plates is measured by, a voltmeter (V) whereas the resulting photo current, flowing in the circuit is measured by a microammeter, (PA). The photoelectric current can be increased or, decreased by varying the potential of collector plate A, with respect to the emitter plate C. The intensity and, frequency of the incident light can be varied, as can the, potential difference V between the emitter C and the, collector A., We can use the experimental arrangement of, Fig. 11.1 to study the variation of photocurrent with, (a) intensity of radiation, (b) frequency of incident, radiation, (c) the potential difference between the, plates A and C, and (d) the nature of the material, of plate C. Light of different frequencies can be used, by putting appropriate coloured filter or coloured, glass in the path of light falling on the emitter C. The intensity, of light is varied by changing the distance of the light source, from the emitter., , 11.4.1 Effect of intensity of light on photocurrent, , FIGURE 11.2 Variation of, Photoelectric current with, intensity of light., , The collector A is maintained at a positive potential with, respect to emitter C so that electrons ejected from C are, attracted towards collector A. Keeping the frequency of the, incident radiation and the accelerating potential fixed, the, intensity of light is varied and the resulting photoelectric, current is measured each time. It is found that the, photocurrent increases linearly with intensity of incident light, as shown graphically in Fig. 11.2. The photocurrent is directly, proportional to the number of photoelectrons emitted per, second. This implies that the number of photoelectrons, emitted per second is directly proportional to the intensity, of incident radiation., , 11.4.2 Effect of potential on photoelectric current, , 390, , We first keep the plate A at some positive accelerating potential with respect, to the plate C and illuminate the plate C with light of fixed frequency Q, and fixed intensity I1. We next vary the positive potential of plate A gradually, and measure the resulting photocurrent each time. It is found that the, photoelectric current increases with increase in accelerating (positive), potential. At some stage, for a certain positive potential of plate A, all the, emitted electrons are collected by the plate A and the photoelectric current, becomes maximum or saturates. If we increase the accelerating potential, of plate A further, the photocurrent does not increase. This maximum, value of the photoelectric current is called saturation current. Saturation, current corresponds to the case when all the photoelectrons emitted by, the emitter plate C reach the collector plate A., We now apply a negative (retarding) potential to the plate A with respect, to the plate C and make it increasingly negative gradually. When the

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Dual Nature of Radiation, and Matter, polarity is reversed, the electrons are, repelled and only the most energetic, electrons are able to reach the collector A., The photocurrent is found to decrease, rapidly until it drops to zero at a certain, sharply defined, critical value of the negative, potential V0 on the plate A. For a particular, frequency of incident radiation, the, minimum negative (retarding) potential V0, given to the plate A for which the, photocurrent stops or becomes zero is, called the cut-off or stopping potential., The interpretation of the observation in, terms of photoelectrons is straightforward., All the photoelectrons emitted from the, metal do not have the same energy., FIGURE 11.3 Variation of photocurrent with, Photoelectric current is zero when the, collector plate potential for different, stopping potential is sufficient to repel even, intensity of incident radiation., the most energetic photoelectrons, with the, maximum kinetic energy (Kmax), so that, (11.1), Kmax = e V0, We can now repeat this experiment with incident radiation of the same, frequency but of higher intensity I2 and I3 (I3 ! I2 ! I1). We note that the, saturation currents are now found to be at higher values. This shows, that more electrons are being emitted per second, proportional to the, intensity of incident radiation. But the stopping potential remains the, same as that for the incident radiation of intensity I1, as shown graphically, in Fig. 11.3. Thus, for a given frequency of the incident radiation, the, stopping potential is independent of its intensity. In other words, the, maximum kinetic energy of photoelectrons depends on the light source, and the emitter plate material, but is independent of intensity of incident, radiation., , 11.4.3 Effect of frequency of incident radiation on stopping, potential, We now study the relation between the, frequency Q of the incident radiation and the, stopping potential V0. We suitably adjust the, same intensity of light radiation at various, frequencies and study the variation of, photocurrent with collector plate potential. The, resulting variation is shown in Fig. 11.4. We, obtain different values of stopping potential but, the same value of the saturation current for, incident radiation of different frequencies. The, energy of the emitted electrons depends on the, frequency of the incident radiations. The, stopping potential is more negative for higher, frequencies of incident radiation. Note from, , FIGURE 11.4 Variation of photoelectric current, with collector plate potential for different, frequencies of incident radiation., , 391

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Physics, Fig. 11.4 that the stopping potentials are in the, order V03 > V02 > V01 if the frequencies are in the, order Q3 > Q2 > Q1. This implies that greater the, frequency of incident light, greater is the, maximum kinetic energy of the photoelectrons., Consequently, we need greater retarding, potential to stop them completely. If we plot a, graph between the frequency of incident radiation, and the corresponding stopping potential for, different metals we get a straight line, as shown, in Fig. 11.5., The graph shows that, FIGURE 11.5 Variation of stopping potential V0, with frequency Q of incident radiation for a, (i) the stopping potential V0 varies linearly with, given photosensitive material., the frequency of incident radiation for a given, photosensitive material., (ii) there exists a certain minimum cut-off frequency Q0 for which the, stopping potential is zero., These observations have two implications:, (i) The maximum kinetic energy of the photoelectrons varies linearly, with the frequency of incident radiation, but is independent of its, intensity., (ii) For a frequency Q of incident radiation, lower than the cut-off, frequency Q0, no photoelectric emission is possible even if the, intensity is large., This minimum, cut-off frequency Q0, is called the threshold frequency., It is different for different metals., Different photosensitive materials respond differently to light. Selenium, is more sensitive than zinc or copper. The same photosensitive substance, gives different response to light of different wavelengths. For example,, ultraviolet light gives rise to photoelectric effect in copper while green or, red light does not., Note that in all the above experiments, it is found that, if frequency of, the incident radiation exceeds the threshold frequency, the photoelectric, emission starts instantaneously without any apparent time lag, even if, the incident radiation is very dim. It is now known that emission starts in, a time of the order of 10 – 9 s or less., We now summarise the experimental features and observations, described in this section., (i) For a given photosensitive material and frequency of incident radiation, (above the threshold frequency), the photoelectric current is directly, proportional to the intensity of incident light (Fig. 11.2)., (ii) For a given photosensitive material and frequency of incident radiation,, saturation current is found to be proportional to the intensity of, incident radiation whereas the stopping potential is independent of, its intensity (Fig. 11.3)., (iii) For a given photosensitive material, there exists a certain minimum, cut-off frequency of the incident radiation, called the threshold, frequency, below which no emission of photoelectrons takes place,, no matter how intense the incident light is. Above the threshold, 392, frequency, the stopping potential or equivalently the maximum kinetic

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Dual Nature of Radiation, and Matter, energy of the emitted photoelectrons increases linearly with the, frequency of the incident radiation, but is independent of its intensity, (Fig. 11.5)., (iv) The photoelectric emission is an instantaneous process without any, apparent time lag (a10 – 9s or less), even when the incident radiation is, made exceedingly dim., , 11.5 PHOTOELECTRIC EFFECT, OF LIGHT, , AND, , WAVE THEORY, , The wave nature of light was well established by the end of the nineteenth, century. The phenomena of interference, diffraction and polarisation were, explained in a natural and satisfactory way by the wave picture of light., According to this picture, light is an electromagnetic wave consisting of, electric and magnetic fields with continuous distribution of energy over, the region of space over which the wave is extended. Let us now see if this, wave picture of light can explain the observations on photoelectric, emission given in the previous section., According to the wave picture of light, the free electrons at the surface, of the metal (over which the beam of radiation falls) absorb the radiant, energy continuously. The greater the intensity of radiation, the greater are, the amplitude of electric and magnetic fields. Consequently, the greater, the intensity, the greater should be the energy absorbed by each electron., In this picture, the maximum kinetic energy of the photoelectrons on the, surface is then expected to increase with increase in intensity. Also, no, matter what the frequency of radiation is, a sufficiently intense beam of, radiation (over sufficient time) should be able to impart enough energy to, the electrons, so that they exceed the minimum energy needed to escape, from the metal surface . A threshold frequency, therefore, should not exist., These expectations of the wave theory directly contradict observations (i),, (ii) and (iii) given at the end of sub-section 11.4.3., Further, we should note that in the wave picture, the absorption of, energy by electron takes place continuously over the entire, wavefront of the radiation. Since a large number of electrons absorb energy,, the energy absorbed per electron per unit time turns out to be small., Explicit calculations estimate that it can take hours or more for a single, electron to pick up sufficient energy to overcome the work function and, come out of the metal. This conclusion is again in striking contrast to, observation (iv) that the photoelectric emission is instantaneous. In short,, the wave picture is unable to explain the most basic features of, photoelectric emission., , 11.6 EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY, QUANTUM OF RADIATION, In 1905, Albert Einstein (1879 -1955) proposed a radically new picture, of electromagnetic radiation to explain photoelectric effect. In this picture,, photoelectric emission does not take place by continuous absorption of, energy from radiation. Radiation energy is built up of discrete units – the, so called quanta of energy of radiation. Each quantum of radiant energy, , 393

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ALBERT EINSTEIN (1879 – 1955), , Physics, , 394, , Albert Einstein (1879 –, 1955) Einstein, one of the, greatest physicists of all, time, was born in Ulm,, Germany. In 1905, he, published three pathbreaking papers. In the, first paper, he introduced, the notion of light quanta, (now called photons) and, used it to explain the, features of photoelectric, effect. In the second paper,, he developed a theory of, Brownian, motion,, confirmed experimentally a, few years later and provided, a convincing evidence of, the atomic picture of matter., The third paper gave birth, to the special theory of, relativity. In 1916, he, published the general, theory of relativity. Some of, Einstein’s most significant, later contributions are: the, notion, of, stimulated, emission introduced in an, alternative derivation of, Planck’s, blackbody, radiation law, static model, of the universe which, started modern cosmology,, quantum statistics of a gas, of massive bosons, and a, critical analysis of the, foundations of quantum, mechanics. In 1921, he was, awarded the Nobel Prize in, physics for his contribution, to theoretical physics and, the photoelectric effect., , has energy h Q, where h is Planck’s constant and Q the, frequency of light. In photoelectric effect, an electron, absorbs a quantum of energy (hQ ) of radiation. If this, quantum of energy absorbed exceeds the minimum, energy needed for the electron to escape from the metal, surface (work function I0), the electron is emitted with, maximum kinetic energy, (11.2), Kmax = hQ – I0, More tightly bound electrons will emerge with kinetic, energies less than the maximum value. Note that the, intensity of light of a given frequency is determined by, the number of photons incident per second. Increasing, the intensity will increase the number of emitted electrons, per second. However, the maximum kinetic energy of the, emitted photoelectrons is determined by the energy of each, photon., Equation (11.2) is known as Einstein’s photoelectric, equation. We now see how this equation accounts in a, simple and elegant manner all the observations on, photoelectric effect given at the end of sub-section 11.4.3., x According to Eq. (11.2), Kmax depends linearly on Q,, and is independent of intensity of radiation, in, agreement with observation. This has happened, because in Einstein’s picture, photoelectric effect arises, from the absorption of a single quantum of radiation, by a single electron. The intensity of radiation (that is, proportional to the number of energy quanta per unit, area per unit time) is irrelevant to this basic process., x Since Kmax must be non-negative, Eq. (11.2 ) implies, that photoelectric emission is possible only if, h Q > I0, or Q > Q0 , where, , Q0 =, , x, , I0, , (11.3), h, Equation (11.3) shows that the greater the work, function I0, the higher the minimum or threshold, frequency Q0 needed to emit photoelectrons. Thus,, there exists a threshold frequencyQ0 (= I0/h) for the, metal surface, below which no photoelectric emission, is possible, no matter how intense the incident, radiation may be or how long it falls on the surface., In this picture, intensity of radiation as noted above,, is proportional to the number of energy quanta per, unit area per unit time. The greater the number of, energy quanta available, the greater is the number of, electrons absorbing the energy quanta and greater,, therefore, is the number of electrons coming out of, the metal (for Q > Q0 ). This explains why, for Q > Q0 ,, photoelectric current is proportional to intensity.

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Dual Nature of Radiation, and Matter, x, , In Einstein’s picture, the basic elementary process involved in, photoelectric effect is the absorption of a light quantum by an electron., This process is instantaneous. Thus, whatever may be the intensity, i.e., the number of quanta of radiation per unit area per unit time,, photoelectric emission is instantaneous. Low intensity does not mean, delay in emission, since the basic elementary process is the same., Intensity only determines how many electrons are able to participate, in the elementary process (absorption of a light quantum by a single, electron) and, therefore, the photoelectric current., Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be, written as, e V0 = h Q– I 0; for Q t Q0, , I, h, or V0 = §¨ ·¸ Q  0, (11.4), ©e¹, e, This is an important result. It predicts that the V0 versus Q curve is a, straight line with slope = (h/e), independent of the nature of the material., During 1906-1916, Millikan performed a series of experiments on, photoelectric effect, aimed at disproving Einstein’s photoelectric equation., He measured the slope of the straight line obtained for sodium, similar to, that shown in Fig. 11.5. Using the known value of e, he determined the, value of Planck’s constant h. This value was close to the value of Planck’s, contant (= 6.626 × 10–34J s) determined in an entirely different context., In this way, in 1916, Millikan proved the validity of Einstein’s photoelectric, equation, instead of disproving it., The successful explanation of photoelectric effect using the hypothesis, of light quanta and the experimental determination of values of h and I0,, in agreement with values obtained from other experiments, led to the, acceptance of Einstein’s picture of photoelectric effect. Millikan verified, photoelectric equation with great precision, for a number of alkali metals, over a wide range of radiation frequencies., , 11.7 PARTICLE NATURE, , OF, , LIGHT: THE PHOTON, , Photoelectric effect thus gave evidence to the strange fact that light in, interaction with matter behaved as if it was made of quanta or packets of, energy, each of energy h Q., Is the light quantum of energy to be associated with a particle? Einstein, arrived at the important result, that the light quantum can also be, associated with momentum (h Q/c ). A definite value of energy as well as, momentum is a strong sign that the light quantum can be associated, with a particle. This particle was later named photon. The particle-like, behaviour of light was further confirmed, in 1924, by the experiment of, A.H. Compton (1892-1962) on scattering of X-rays from electrons. In, 1921, Einstein was awarded the Nobel Prize in Physics for his contribution, to theoretical physics and the photoelectric effect. In 1923, Millikan was, awarded the Nobel Prize in physics for his work on the elementary, charge of electricity and on the photoelectric effect., We can summarise the photon picture of electromagnetic radiation, as follows:, , 395

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Physics, (i) In interaction of radiation with matter, radiation behaves as if it is, made up of particles called photons., (ii) Each photon has energy E (=hQ ) and momentum p (= h Q/c), and, speed c, the speed of light., (iii) All photons of light of a particular frequency Q, or wavelength O, have, the same energy E (=hQ  = hc/O) and momentum p (= hQ/c = h/O),, whatever the intensity of radiation may be. By increasing the intensity, of light of given wavelength, there is only an increase in the number of, photons per second crossing a given area, with each photon having, the same energy. Thus, photon energy is independent of intensity of, radiation., (iv) Photons are electrically neutral and are not deflected by electric and, magnetic fields., (v) In a photon-particle collision (such as photon-electron collision), the, total energy and total momentum are conserved. However, the number, of photons may not be conserved in a collision. The photon may be, absorbed or a new photon may be created., , EXAMPLE 11.1, , Example 11.1 Monochromatic light of frequency 6.0 u10 14 Hz is, produced by a laser. The power emitted is 2.0 u10–3 W. (a) What is the, energy of a photon in the light beam? (b) How many photons per second,, on an average, are emitted by the source?, Solution, (a) Each photon has an energy, E = h Q = ( 6.63 u10–34 J s) (6.0 u1014 Hz), = 3.98 u 10–19 J, (b) If N is the number of photons emitted by the source per second,, the power P transmitted in the beam equals N times the energy, per photon E, so that P = N E. Then, N=, , P, E, , 2.0 u 103 W, 3.98 u 10 19 J, , = 5.0 u1015 photons per second., Example 11.2 The work function of caesium is 2.14 eV. Find (a) the, threshold frequency for caesium, and (b) the wavelength of the incident, light if the photocurrent is brought to zero by a stopping potential of, 0.60 V., Solution, (a) For the cut-off or threshold frequency, the energy h Q0 of the incident, radiation must be equal to work function I 0, so that, , Q0 = I0, , 396, , EXAMPLE 11.2, , h, , , , 2.14 eV, 6.63 u 1034 J s, , 2.14 u 1.6 u 1019 J, 6.63 u 1034 J s, , 5.16 u 1014 Hz, , Thus, for frequencies less than this threshold frequency, no, photoelectrons are ejected., (b) Photocurrent reduces to zero, when maximum kinetic energy of, the emitted photoelectrons equals the potential energy e V0 by the, retarding potential V0. Einstein’s Photoelectric equation is

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Dual Nature of Radiation, and Matter, eV0 = hQ – I 0=, , hc, , O, or, O = hc/(eV0 +I0 ), , – I0, , 19.89 u 10 26 J m, (2.74 eV), , O, , 19.89 u 10 26 J m, 2.74 u 1.6 u 10 19 J, , 454 nm, , EXAMPLE 11.2, , (6.63 u 1034 J s) u (3 u 108 m/s), (0.60 eV  2.14 eV), , Example 11.3 The wavelength of light in the visible region is about, 390 nm for violet colour, about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour., (a) What are the energies of photons in (eV) at the (i) violet end, (ii), average wavelength, yellow-green colour, and (iii) red end of the, visible spectrum? (Take h = 6.63×10–34 J s and 1 eV = 1.6×10 –19J.), (b) From which of the photosensitive materials with work functions, listed in Table 11.1 and using the results of (i), (ii) and (iii) of (a),, can you build a photoelectric device that operates with visible, light?, Solution, (a) Energy of the incident photon, E = hQ = hc/O, E = (6.63×10–34J s) (3×108 m/s)/O, 1.989 u 10 –25 J m, , O, , EXAMPLE 11.3, , (i) For violet light, O1 = 390 nm (lower wavelength end), 1.989 u 10 –25 J m, Incident photon energy, E1 =, 390×10 –9 m, = 5.10 × 10–19J, 5.10 u 10 –19 J, 1.6×10 –19 J/eV, = 3.19 eV, (ii) For yellow-green light, O2 = 550 nm (average wavelength), 1.989 u 10 –25 J m, Incident photon energy, E2 =, 550×10 –9 m, = 3.62×10–19 J = 2.26 eV, (iii) For red light, O3 = 760 nm (higher wavelength end), 1.989 u 10 –25 J m, Incident photon energy, E3 =, 760×10 –9 m, = 2.62×10–19 J = 1.64 eV, (b) For a photoelectric device to operate, we require incident light energy, E to be equal to or greater than the work function I0 of the material., Thus, the photoelectric device will operate with violet light (with, E = 3.19 eV) photosensitive material Na (with I0 = 2.75 eV), K (with, I0 = 2.30 eV) and Cs (with I0 = 2.14 eV). It will also operate with, yellow-green light (with E = 2.26 eV) for Cs (with I0 = 2.14 eV) only., However, it will not operate with red light (with E = 1.64 eV) for any, of these photosensitive materials., , 397

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Physics, 11.8 WAVE NATURE, , OF, , MATTER, , The dual (wave-particle) nature of light (electromagnetic radiation, in, general) comes out clearly from what we have learnt in this and the, preceding chapters. The wave nature of light shows up in the phenomena, of interference, diffraction and polarisation. On the other hand, in, photoelectric effect and Compton effect which involve energy and, momentum transfer, radiation behaves as if it is made up of a bunch of, particles – the photons. Whether a particle or wave description is best, suited for understanding an experiment depends on the nature of the, experiment. For example, in the familiar phenomenon of seeing an object, by our eye, both descriptions are important. The gathering and focussing, mechanism of light by the eye-lens is well described in the wave picture., But its absorption by the rods and cones (of the retina) requires the photon, picture of light., A natural question arises: If radiation has a dual (wave-particle) nature,, might not the particles of nature (the electrons, protons, etc.) also exhibit, wave-like character? In 1924, the French physicist Louis Victor de Broglie, (pronounced as de Broy) (1892-1987) put forward the bold hypothesis, that moving particles of matter should display wave-like properties under, suitable conditions. He reasoned that nature was symmetrical and that, the two basic physical entities – matter and energy, must have symmetrical, character. If radiation shows dual aspects, so should matter. De Broglie, proposed that the wave length O associated with a particle of momentum, p is given as, , O , , h, p, , h, mv, , (11.5), , where m is the mass of the particle and v its speed. Equation (11.5) is, known as the de Broglie relation and the wavelength O of the matter, wave is called de Broglie wavelength. The dual aspect of matter is evident, in the de Broglie relation. On the left hand side of Eq. (11.5), O is the, attribute of a wave while on the right hand side the momentum p is a, typical attribute of a particle. Planck’s constant  h relates the two, attributes., Equation (11.5) for a material particle is basically a hypothesis whose, validity can be tested only by experiment. However, it is interesting to see, that it is satisfied also by a photon. For a photon, as we have seen,, p = hQ /c, , (11.6), , Therefore,, , 398, , h c, O, (11.7), p Q, That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals, the wavelength of electromagnetic radiation of which the photon is a, quantum of energy and momentum., Clearly, from Eq. (11.5 ), O is smaller for a heavier particle ( large m ) or, more energetic particle (large v). For example, the de Broglie wavelength, of a ball of mass 0.12 kg moving with a speed of 20 m s–1 is easily, calculated:

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Dual Nature of Radiation, and Matter, PHOTOCELL, A photocell is a technological application of the photoelectric effect. It is a device whose, electrical properties are affected by light. It is also sometimes called an electric eye. A photocell, consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A, (collector) supported in an evacuated glass or quartz bulb. It is connected to the external, circuit having a high-tension battery B and microammeter (PA) as shown in the Figure., Sometimes, instead of the plate C, a thin layer of photosensitive material is pasted on the, inside of the bulb. A part of the bulb is left clean for the light to enter it., When light of suitable wavelength falls on the, emitter C, photoelectrons are emitted. These, photoelectrons are drawn to the collector A., Photocurrent of the order of a few microampere, can be normally obtained from a photo cell., A photocell converts a change in intensity of, illumination into a change in photocurrent. This, current can be used to operate control systems, and in light measuring devices. A photocell of lead, sulphide sensitive to infrared radiation is used, in electronic ignition circuits., In scientific work, photo cells are used, whenever it is necessary to measure the intensity, of light. Light meters in photographic cameras, make use of photo cells to measure the intensity, of incident light. The photocells, inserted in the, door light electric circuit, are used as automatic, door opener. A person approaching a doorway, may interrupt a light beam which is incident on, a photocell. The abrupt change in photocurrent, A photo cell, may be used to start a motor which opens the, door or rings an alarm. They are used in the, control of a counting device which records every interruption of the light beam caused by a, person or object passing across the beam. So photocells help count the persons entering an, auditorium, provided they enter the hall one by one. They are used for detection of traffic, law defaulters: an alarm may be sounded whenever a beam of (invisible) radiation is, intercepted., In burglar alarm, (invisible) ultraviolet light is continuously made to fall on a photocell, installed at the doorway. A person entering the door interrupts the beam falling on the, photocell. The abrupt change in photocurrent is used to start an electric bell ringing. In fire, alarm, a number of photocells are installed at suitable places in a building. In the event of, breaking out of fire, light radiations fall upon the photocell. This completes the electric, circuit through an electric bell or a siren which starts operating as a warning signal., Photocells are used in the reproduction of sound in motion pictures and in the television, camera for scanning and telecasting scenes. They are used in industries for detecting minor, flaws or holes in metal sheets., , p = m v = 0.12 kg × 20 m s–1 = 2.40 kg m s–1, , O=, , 6.63 u 1034 J s, h, =, = 2.76 × 10–34 m, 2.40 kg m s 1, p, , 399

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LOUIS VICTOR DE BROGLIE (1892 – 1987), , Physics, This wavelength is so small that it is beyond any, measurement. This is the reason why macroscopic objects, in our daily life do not show wave-like properties. On the, other hand, in the sub-atomic domain, the wave character, of particles is significant and measurable., Consider an electron (mass m, charge e) accelerated, from rest through a potential V. The kinetic energy K, of the electron equals the work done (eV ) on it by the, electric field:, K=eV, (11.8), Now, K =, Louis Victor de Broglie, (1892 – 1987) French, physicist who put forth, revolutionary idea of wave, nature of matter. This idea, was developed by Erwin, Schródinger into a fullfledged theory of quantum, mechanics, commonly, known as wave mechanics., In 1929, he was awarded the, Nobel Prize in Physics for his, discovery of the wave nature, of electrons., , p=, , 1, p2, m v2 =, , so that, 2, 2m, , 2m K, , 2 m eV, , (11.9), , The de Broglie wavelength Oof the electron is then, O , , h, p, , h, 2 mK, , h, 2 m eV, , (11.10), , Substituting the numerical values of h, m, e,, we get, , 1.227, nm, (11.11), V, where V is the magnitude of accelerating potential in, volts. For a 120 V accelerating potential, Eq. (11.11) gives, O = 0.112 nm. This wavelength is of the same order as, the spacing between the atomic planes in crystals. This, suggests that matter waves associated with an electron could be verified, by crystal diffraction experiments analogous to X-ray diffraction. We, describe the experimental verification of the de Broglie hypothesis in the, next section. In 1929, de Broglie was awarded the Nobel Prize in Physics, for his discovery of the wave nature of electrons., The matter–wave picture elegantly incorporated the Heisenberg’s, uncertainty principle. According to the principle, it is not possible to, measure both the position and momentum of an electron (or any other, particle) at the same time exactly. There is always some uncertainty ('x ), in the specification of position and some uncertainty ('p) in the specification, of momentum. The product of 'x and 'p is of the order of = *, (with = = h/2S), i.e.,, , O, , (11.12), 'x 'p | =, Equation (11.12) allows the possibility that 'x is zero; but then 'p, must be infinite in order that the product is non-zero. Similarly, if 'p is, zero, 'x must be infinite. Ordinarily, both 'x and 'p are non-zero such, that their product is of the order of = ., Now, if an electron has a definite momentum p, (i.e.'p = 0), by the de, Broglie relation, it has a definite wavelength O. A wave of definite (single), , 400, , * A more rigorous treatment gives 'x 'p t = /2.

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Dual Nature of Radiation, and Matter, wavelength extends all over space. By Born’s, probability interpretation this means that the, electron is not localised in any finite region of, space. That is, its position uncertainty is infinite, ('x o f), which is consistent with the, uncertainty principle., In general, the matter wave associated with, the electron is not extended all over space. It is, a wave packet extending over some finite region, of space. In that case 'x is not infinite but has, some finite value depending on the extension, of the wave packet. Also, you must appreciate, that a wave packet of finite extension does not, have a single wavelength. It is built up of, wavelengths spread around some central, wavelength., By de Broglie’s relation, then, the, momentum of the electron will also have a, spread – an uncertainty 'p. This is as expected, from the uncertainty principle. It can be shown, that the wave packet description together with, de Broglie relation and Born’s probability, interpretation reproduce the Heisenberg’s, uncertainty principle exactly., In Chapter 12, the de Broglie relation will, be seen to justify Bohr’s postulate on, quantisation of angular momentum of electron, in an atom., Figure 11.6 shows a schematic diagram of, (a) a localised wave packet, and (b) an extended, wave with fixed wavelength., , FIGURE 11.6 (a) The wave packet description of, an electron. The wave packet corresponds to a, spread of wavelength around some central, wavelength (and hence by de Broglie relation,, a spread in momentum). Consequently, it is, associated with an uncertainty in position, ('x) and an uncertainty in momentum ('p)., (b) The matter wave corresponding to a, definite momentum of an electron, extends all over space. In this case,, ' p = 0 and ' x o f, , Example 11.4 What is the de Broglie wavelength associated with (a) an, electron moving with a speed of 5.4u106 m/s, and (b) a ball of mass 150 g, travelling at 30.0 m/s?, Solution, (a) For the electron:, Mass m = 9.11u10–31 kg, speed v = 5.4u106 m/s. Then, momentum, p = m v = 9.11u10–31 (kg) u 5.4 u 106 (m/s), p = 4.92 u 10–24 kg m/s, de Broglie wavelength, O = h/p, =, , 6.63 u 10 –34 J s, 4.92 u10 –24 kg m/s, , EXAMPLE 11.4, , O= 0.135 nm, (b) For the ball:, Mass m ’ = 0.150 kg, speed v ’ = 30.0 m/s., Then momentum p ’ = m’ v ’ = 0.150 (kg) u 30.0 (m/s), p ’= 4.50 kg m/s, de Broglie wavelength O’ = h/p’, , 401

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EXAMPLE 11.4, , Physics, 6.63 u 10±34 J s, 4.50 u kg m/s, , O ’ = 1.47 u10–34 m, The de Broglie wavelength of electron is comparable with X-ray, wavelengths. However, for the ball it is about 10–19 times the size of, the proton, quite beyond experimental measurement., Example 11.5 An electron, an D-particle, and a proton have the same, kinetic energy. Which of these particles has the shortest de Broglie, wavelength?, , EXAMPLE 11.5, , Solution, For a particle, de Broglie wavelength, O = h/p, Kinetic energy, K = p2/2m, Then, O h / 2mK, For the same kinetic energy K, the de Broglie wavelength associated, with the particle is inversely proportional to the square root of their, masses. A proton, , 1, 1, , H, , is 1836 times massive than an electron and, , 4, 2, , an D-particle He four times that of a proton., Hence, D – particle has the shortest de Broglie wavelength., , PROBABILITY, , INTERPRETATION TO MATTER WAVES, , 402, , EXAMPLE 11.6, , It is worth pausing here to reflect on just what a matter wave associated with a particle,, say, an electron, means. Actually, a truly satisfactory physical understanding of the, dual nature of matter and radiation has not emerged so far. The great founders of, quantum mechanics (Niels Bohr, Albert Einstein, and many others) struggled with this, and related concepts for long. Still the deep physical interpretation of quantum, mechanics continues to be an area of active research. Despite this, the concept of, matter wave has been mathematically introduced in modern quantum mechanics with, great success. An important milestone in this connection was when Max Born (18821970) suggested a probability interpretation to the matter wave amplitude. According, to this, the intensity (square of the amplitude) of the matter wave at a point determines, the probability density of the particle at that point. Probability density means probability, per unit volume. Thus, if A is the amplitude of the wave at a point, |A|2 'V is the, probability of the particle being found in a small volume 'V around that point. Thus,, if the intensity of matter wave is large in a certain region, there is a greater probability, of the particle being found there than where the intensity is small., , Example 11.6 A particle is moving three times as fast as an electron., The ratio of the de Broglie wavelength of the particle to that of the, electron is 1.813 × 10–4. Calculate the particle’s mass and identify the, particle., Solution, de Broglie wavelength of a moving particle, having mass m and, velocity v:

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Dual Nature of Radiation, and Matter, O, , h, p, , h, mv, , Mass, m = h/Ov, For an electron, mass m e = h/Oe ve, Now, we have v/ve = 3 and, O/Oe = 1.813 × 10– 4, , m = (9.11×10–31 kg) × (1/3) × (1/1.813 × 10–4), m = 1.675 × 10–27 kg., Thus, the particle, with this mass could be a proton or a neutron., , EXAMPLE 11.6, , § Oe · § ve ·, , Then, mass of the particle, m = me ¨© ¸¹ ¨© ¸¹, v, O, , Example 11.7 What is the de Broglie wavelength associated with an, electron, accelerated through a potential differnece of 100 volts?, Solution Accelerating potential V = 100 V. The de Broglie wavelength, O is, , 1.227, nm, V, , 1.227, nm = 0.123 nm, 100, The de Broglie wavelength associated with an electron in this case is of, the order of X-ray wavelengths., , O, , 11.9 DAVISSON, , AND, , EXAMPLE 11.7, , O = h /p, , GERMER EXPERIMENT, , The wave nature of electrons was first experimentally verified by C.J., Davisson and L.H. Germer in 1927 and independently by G.P. Thomson,, in 1928, who observed, diffraction effects with beams of, electrons scattered by crystals., Davisson and Thomson shared, the Nobel Prize in 1937 for their, experimental discovery of, diffraction of electrons by, crystals., The experimental arrangement used by Davisson and, Germer is schematically shown, in Fig. 11.7. It consists of an, electron gun which comprises of, a tungsten filament F, coated, with barium oxide and heated, by a low voltage power supply, (L.T. or battery). Electrons, FIGURE 11.7 Davisson-Germer electron, emitted by the filament are, diffraction arrangement., accelerated to a desired velocity, , 403

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http://nobelprize.org, , Development of electron microscope, , Physics, by applying suitable potential/voltage from a high voltage power supply, (H.T. or battery). They are made to pass through a cylinder with fine, holes along its axis, producing a fine collimated beam. The beam is made, to fall on the surface of a nickel crystal. The electrons are scattered in all, directions by the atoms of the crystal. The intensity of the electron beam,, scattered in a given direction, is measured by the electron detector, (collector). The detector can be moved on a circular scale and is connected, to a sensitive galvanometer, which records the current. The deflection of, the galvanometer is proportional to the intensity of the electron beam, entering the collector. The apparatus is enclosed in an evacuated chamber., By moving the detector on the circular scale at different positions, the, intensity of the scattered electron beam is measured for different values, of angle of scatteringT which is the angle between the incident and the, scattered electron beams. The variation of the intensity (I ) of the scattered, electrons with the angle of scattering T is obtained for different accelerating, voltages., The experiment was performed by varying the accelarating voltage, from 44 V to 68 V. It was noticed that a strong peak appeared in the, intensity (I ) of the scattered electron for an accelarating voltage of 54V at, a scattering angle T= 50º, The appearance of the peak in a particular direction is due to the, constructive interference of electrons scattered from different layers of the, regularly spaced atoms of the crystals. From the electron diffraction, measurements, the wavelength of matter waves was found to be, 0.165 nm., The de Broglie wavelength O associated with electrons, using, Eq. (11.11), for V = 54 V is given by, , O = h /p, , O, , 404, , 1.227, nm, V, , 1.227, nm = 0.167 nm, 54, , Thus, there is an excellent agreement between the theoretical value, and the experimentally obtained value of de Broglie wavelength. DavissonGermer experiment thus strikingly confirms the wave nature of electrons, and the de Broglie relation. More recently, in 1989, the wave nature of a, beam of electrons was experimentally demonstrated in a double-slit, experiment, similar to that used for the wave nature of light. Also, in an, experiment in 1994, interference fringes were obtained with the beams of, iodine molecules, which are about a million times more massive than, electrons., The de Broglie hypothesis has been basic to the development of modern, quantum mechanics. It has also led to the field of electron optics. The, wave properties of electrons have been utilised in the design of electron, microscope which is a great improvement, with higher resolution, over, the optical microscope.

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Dual Nature of Radiation, and Matter, SUMMARY, 1., , The minimum energy needed by an electron to come out from a metal, surface is called the work function of the metal. Energy (greater than, the work function IR) required for electron emission from the metal, surface can be supplied by suitably heating or applying strong electric, field or irradiating it by light of suitable frequency., , 2., , Photoelectric effect is the phenomenon of emission of electrons by metals, when illuminated by light of suitable frequency. Certain metals respond, to ultraviolet light while others are sensitive even to the visible light., Photoelectric effect involves conversion of light energy into electrical, energy. It follows the law of conservation of energy. The photoelectric, emission is an instantaneous process and possesses certain special, features., , 3., , Photoelectric current depends on (i) the intensity of incident light, (ii), the potential difference applied between the two electrodes, and (iii), the nature of the emitter material., , 4., , The stopping potential (Vo) depends on (i) the frequency of incident, light, and (ii) the nature of the emitter material. For a given frequency, of incident light, it is independent of its intensity. The stopping potential, is directly related to the maximum kinetic energy of electrons emitted:, e V0 = (1/2) m v 2max = Kmax., , 5., , Below a certain frequency (threshold frequency) Q 0 , characteristic of, the metal, no photoelectric emission takes place, no matter how large, the intensity may be., , 6., , The classical wave theory could not explain the main features of, photoelectric effect. Its picture of continuous absorption of energy from, radiation could not explain the independence of Kmax on intensity, the, existence of Qo and the instantaneous nature of the process. Einstein, explained these features on the basis of photon picture of light., According to this, light is composed of discrete packets of energy called, quanta or photons. Each photon carries an energy E (= h Q) and, momentum p (= h/O), which depend on the frequency (Q ) of incident, light and not on its intensity. Photoelectric emission from the metal, surface occurs due to absorption of a photon by an electron., , 7., , Einstein’s photoelectric equation is in accordance with the energy, conservation law as applied to the photon absorption by an electron in, the metal. The maximum kinetic energy (1/2)m v 2max is equal to, the photon energy (hQ ) minus the work function I0 (= hQ0 ) of the, target metal:, , 1, m v 2max = V0 e = hQ – I0 = h (Q – Q0 ), 2, This photoelectric equation explains all the features of the photoelectric, effect. Millikan’s first precise measurements confirmed the Einstein’s, photoelectric equation and obtained an accurate value of Planck’s, constant h . This led to the acceptance of particle or photon description, (nature) of electromagnetic radiation, introduced by Einstein., 8., , Radiation has dual nature: wave and particle. The nature of experiment, determines whether a wave or particle description is best suited for, understanding the experimental result. Reasoning that radiation and, matter should be symmetrical in nature, Louis Victor de Broglie, , 405

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Physics, attributed a wave-like character to matter (material particles). The waves, associated with the moving material particles are called matter waves, or de Broglie waves., 9. The de Broglie wavelength (O) associated with a moving particle is related, to its momentum p as: O= h/p. The dualism of matter is inherent in the, de Broglie relation which contains a wave concept (O) and a particle concept, (p). The de Broglie wavelength is independent of the charge and nature of, the material particle. It is significantly measurable (of the order of the, atomic-planes spacing in crystals) only in case of sub-atomic particles, like electrons, protons, etc. (due to smallness of their masses and hence,, momenta). However, it is indeed very small, quite beyond measurement,, in case of macroscopic objects, commonly encountered in everyday life., 10. Electron diffraction experiments by Davisson and Germer, and by G. P., Thomson, as well as many later experiments, have verified and confirmed, the wave-nature of electrons. The de Broglie hypothesis of matter waves, supports the Bohr ’s concept of stationary orbits., , Physical, Quantity, , Symbol, , Dimensions, , Unit, , Remarks, , Planck’s, , h, , [ML2 T –1], , Js, , E = hQ, , Stopping, potential, , V0, , [ML2 T –3A–1], , V, , e V0= Kmax, , Work, function, , I, , [ML2 T –2], , J; eV, , Kmax = E –I, , Threshold, frequency, , Q, , [T –1], , Hz, , Q0=I0 /h, , de Broglie, wavelength, , O, , [L], , m, , O = h/p, , constant, , POINTS TO PONDER, 1., , 2., , 3., , 406, , Free electrons in a metal are free in the sense that they move inside the, metal in a constant potential (This is only an approximation). They are, not free to move out of the metal. They need additional energy to get, out of the metal., Free electrons in a metal do not all have the same energy. Like molecules, in a gas jar, the electrons have a certain energy distribution at a given, temperature. This distribution is different from the usual Maxwell’s, distribution that you have learnt in the study of kinetic theory of gases., You will learn about it in later courses, but the difference has to do, with the fact that electrons obey Pauli’s exclusion principle., Because of the energy distribution of free electrons in a metal, the, energy required by an electron to come out of the metal is different for, different electrons. Electrons with higher energy require less additional, energy to come out of the metal than those with lower energies. Work, function is the least energy required by an electron to come out of the, metal.

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Dual Nature of Radiation, and Matter, 4., , 5., , 6., , Observations on photoelectric effect imply that in the event of matterlight interaction, absorption of energy takes place in discrete units of hQ., This is not quite the same as saying that light consists of particles,, each of energy hQ., Observations on the stopping potential (its independence of intensity, and dependence on frequency) are the crucial discriminator between, the wave-picture and photon-picture of photoelectric effect., The wavelength of a matter wave given by O, , h, p, , has physical, , significance; its phase velocity vp has no physical significance. However,, the group velocity of the matter wave is physically meaningful and, equals the velocity of the particle., , EXERCISES, 11.1, , 11.2, , 11.3, 11.4, , Find the, (a) maximum frequency, and, (b) minimum wavelength of X-rays produced by 30 kV electrons., The work function of caesium metal is 2.14 eV. When light of, frequency 6 ×1014Hz is incident on the metal surface, photoemission, of electrons occurs. What is the, (a) maximum kinetic energy of the emitted electrons,, (b) Stopping potential, and, (c) maximum speed of the emitted photoelectrons?, The photoelectric cut-off voltage in a certain experiment is 1.5 V., What is the maximum kinetic energy of photoelectrons emitted?, Monochromatic light of wavelength 632.8 nm is produced by a, helium-neon laser. The power emitted is 9.42 mW., (a) Find the energy and momentum of each photon in the light beam,, (b) How many photons per second, on the average, arrive at a target, irradiated by this beam? (Assume the beam to have uniform, cross-section which is less than the target area ), and, (c) How fast does a hydrogen atom have to travel in order to have, the same momentum as that of the photon?, , 11.5, , The energy flux of sunlight reaching the surface of the earth is, 1.388 × 103 W/m2. How many photons (nearly) per square metre are, incident on the Earth per second ? Assume that the photons in the, sunlight have an average wavelength of 550 nm., , 11.6, , In an experiment on photoelectric effect, the slope of the cut-off, voltage versus frequency of incident light is found to be 4.12 × 10–15 V s., Calculate the value of Planck’s constant., , 11.7, , A 100W sodium lamp radiates energy uniformly in all directions., The lamp is located at the centre of a large sphere that absorbs all, the sodium light which is incident on it. The wavelength of the, sodium light is 589 nm. (a) What is the energy per photon associated, , 407

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Physics, with the sodium light? (b) At what rate are the photons delivered to, the sphere?, 11.8, , The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light, of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission., , 11.9, , The work function for a certain metal is 4.2 eV. Will this metal give, photoelectric emission for incident radiation of wavelength 330 nm?, , 11.10 Light of frequency 7.21 × 1014 Hz is incident on a metal surface., Electrons with a maximum speed of 6.0 × 105 m/s are ejected from, the surface. What is the threshold frequency for photoemission of, electrons?, 11.11 Light of wavelength 488 nm is produced by an argon laser which is, used in the photoelectric effect. When light from this spectral line is, incident on the emitter, the stopping (cut-of f) potential of, photoelectrons is 0.38 V. Find the work function of the material, from which the emitter is made., 11.12 Calculate the, (a) momentum, and, (b) de Broglie wavelength of the electrons accelerated through a, potential difference of 56 V., 11.13 What is the, (a) momentum,, (b) speed, and, (c) de Broglie wavelength of an electron with kinetic energy of, 120 eV., 11.14 The wavelength of light from the spectral emission line of sodium is, 589 nm. Find the kinetic energy at which, (a) an electron, and, (b) a neutron, would have the same de Broglie wavelength., 11.15 What is the de Broglie wavelength of, (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and, (c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of, 2.2 m/s ?, 11.16 An, (a), (b), (c), , electron and a photon each have a wavelength of 1.00 nm. Find, their momenta,, the energy of the photon, and, the kinetic energy of electron., , 11.17 (a) For what kinetic energy of a neutron will the associated de Broglie, wavelength be 1.40 × 10–10 m?, (b) Also find the de Broglie wavelength of a neutron, in thermal, equilibrium with matter, having an average kinetic energy of, (3/2) k T at 300 K., 11.18 Show that the wavelength of electromagnetic radiation is equal to, the de Broglie wavelength of its quantum (photon)., , 408, , 11.19 What is the de Broglie wavelength of a nitrogen molecule in air at, 300 K ? Assume that the molecule is moving with the root-meansquare speed of molecules at this temperature. (Atomic mass of, nitrogen = 14.0076 u)

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Dual Nature of Radiation, and Matter, , ADDITIONAL EXERCISES, 11.20 (a) Estimate the speed with which electrons emitted from a heated, emitter of an evacuated tube impinge on the collector maintained, at a potential difference of 500 V with respect to the emitter., Ignore the small initial speeds of the electrons. The, specific charge of the electron, i.e., its e/m is given to be, 1.76 × 1011 C kg–1., (b) Use the same formula you employ in (a) to obtain electron speed, for an collector potential of 10 MV. Do you see what is wrong ? In, what way is the formula to be modified?, 11.21 (a) A monoenergetic electron beam with electron speed of, 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 T, normal to the beam velocity. What is the radius of the circle traced, by the beam, given e/m for electron equals 1.76 × 1011C kg–1., (b) Is the formula you employ in (a) valid for calculating radius of, the path of a 20 MeV electron beam? If not, in what way is it, modified ?, [Note: Exercises 11.20(b) and 11.21(b) take you to relativistic, mechanics which is beyond the scope of this book. They have been, inserted here simply to emphasise the point that the formulas you, use in part (a) of the exercises are not valid at very high speeds or, energies. See answers at the end to know what ‘very high speed or, energy’ means.], 11.22 An electron gun with its collector at a potential of 100 V fires out, electrons in a spherical bulb containing hydrogen gas at low, pressure (a10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves, the path of the electrons in a circular orbit of radius 12.0 cm. (The, path can be viewed because the gas ions in the path focus the beam, by attracting electrons, and emitting light by electron capture; this, method is known as the ‘fine beam tube’ method.) Determine, e/m from the data., 11.23 (a) An X-ray tube produces a continuous spectrum of radiation with, its short wavelength end at 0.45 Å. What is the maximum energy, of a photon in the radiation ?, (b) From your answer to (a), guess what order of accelerating voltage, (for electrons) is required in such a tube ?, 11.24 In an accelerator experiment on high-energy collisions of electrons, with positrons, a certain event is interpreted as annihilation of an, electron-positron pair of total energy 10.2 BeV into two J-rays of, equal energy. What is the wavelength associated with each J-ray?, (1BeV = 109 eV), 11.25 Estimating the following two numbers should be interesting. The, first number will tell you why radio engineers do not need to worry, much about photons! The second number tells you why our eye can, never ‘count photons’, even in barely detectable light., (a) The number of photons emitted per second by a Medium wave, transmitter of 10 kW power, emitting radiowaves of wavelength, 500 m., (b) The number of photons entering the pupil of our eye per second, corresponding to the minimum intensity of white light that we, , 409

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Physics, humans can perceive (a10–10 W m–2). Take the area of the pupil, to be about 0.4 cm2, and the average frequency of white light to, be about 6 × 1014 Hz., 11.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source, irradiates a photo-cell made of molybdenum metal. If the stopping, potential is –1.3 V, estimate the work function of the metal. How, would the photo-cell respond to a high intensity (a105 W m–2) red, light of wavelength 6328 Å produced by a He-Ne laser ?, 11.27 Monochromatic radiation of wavelength 640.2 nm (1nm = 10–9 m), from a neon lamp irradiates photosensitive material made of caesium, on tungsten. The stopping voltage is measured to be 0.54 V. The, source is replaced by an iron source and its 427.2 nm line irradiates, the same photo-cell. Predict the new stopping voltage., 11.28 A mercury lamp is a convenient source for studying frequency, dependence of photoelectric emission, since it gives a number of, spectral lines ranging from the UV to the red end of the visible, spectrum. In our experiment with rubidium photo-cell, the following, lines from a mercury source were used:, O1= 3650 Å, O2= 4047 Å, O3= 4358 Å, O4= 5461 Å, O5= 6907 Å,, The stopping voltages, respectively, were measured to be:, V01 = 1.28 V, V02 = 0.95 VV03 = 0.74 V, V04 = 0.16 V, V05 = 0 V, Determine the value of Planck’s constant h, the threshold frequency, and work function for the material., [Note: You will notice that to get h from the data, you will need to, know e (which you can take to be 1.6 × 10–19 C). Experiments of this, kind on Na, Li, K, etc. were performed by Millikan, who, using his, own value of e (from the oil-drop experiment) confirmed Einstein’s, photoelectric equation and at the same time gave an independent, estimate of the value of h.], 11.29 The work function for the following metals is given:, Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these, metals will not give photoelectric emission for a radiation of, wavelength 3300 Å from a He-Cd laser placed 1 m away from the, photocell? What happens if the laser is brought nearer and placed, 50 cm away ?, 11.30 Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface, area 2 cm2. Assuming that the top 5 layers of sodium absorb the, incident energy, estimate time required for photoelectric emission, in the wave-picture of radiation. The work function for the metal is, given to be about 2 eV. What is the implication of your answer ?, 11.31 Crystal diffraction experiments can be performed using X-rays, or, electrons accelerated through appropriate voltage. Which probe has, greater energy? (For quantitative comparison, take the wavelength, of the probe equal to 1 Å, which is of the order of inter-atomic spacing, in the lattice) (me=9.11 × 10–31 kg)., , 410, , 11.32 (a) Obtain the de Broglie wavelength of a neutron of kinetic energy, 150 eV. As you have seen in Exercise 11.31, an electron beam of, this energy is suitable for crystal diffraction experiments. Would, a neutron beam of the same energy be equally suitable ? Explain., (mn = 1.675 × 10–27 kg)

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Dual Nature of Radiation, and Matter, , 11.33, , 11.34, , 11.35, , 11.36, , 11.37, , (b) Obtain the de Broglie wavelength associated with thermal, neutrons at room temperature (27 ºC). Hence explain why a fast, neutron beam needs to be thermalised with the environment, before it can be used for neutron diffraction experiments., An electron microscope uses electrons accelerated by a voltage of, 50 kV. Determine the de Broglie wavelength associated with the, electrons. If other factors (such as numerical aperture, etc.) are, taken to be roughly the same, how does the resolving power of an, electron microscope compare with that of an optical microscope, which uses yellow light?, The wavelength of a probe is roughly a measure of the size of a, structure that it can probe in some detail. The quark structure, of protons and neutrons appears at the minute length-scale of, 10–15 m or less. This structure was first probed in early 1970’s using, high energy electron beams produced by a linear accelerator at, Stanford, USA. Guess what might have been the order of energy of, these electron beams. (Rest mass energy of electron = 0.511 MeV.), Find the typical de Broglie wavelength associated with a He atom in, helium gas at room temperature (27 ºC) and 1 atm pressure; and, compare it with the mean separation between two atoms under these, conditions., Compute the typical de Broglie wavelength of an electron in a metal, at 27 ºC and compare it with the mean separation between two, electrons in a metal which is given to be about 2 × 10–10 m., [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets, associated with gaseous molecules under ordinary conditions are, non-overlapping, the electron wave-packets in a metal strongly, overlap with one another. This suggests that whereas molecules in, an ordinary gas can be distinguished apart, electrons in a metal, cannot be distintguished apart fr om one another. This, indistinguishibility has many fundamental implications which you, will explore in more advanced Physics courses.], Answer the following questions:, (a) Quarks inside protons and neutrons are thought to carry, fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up, in Millikan’s oil-drop experiment?, (b) What is so special about the combination e/m? Why do we not, simply talk of e and m separately?, (c) Why should gases be insulators at ordinary pressures and start, conducting at very low pressures?, (d) Every metal has a definite work function. Why do all, photoelectrons not come out with the same energy if incident, radiation is monochromatic? Why is there an energy distribution, of photoelectrons?, (e) The energy and momentum of an electron are related to the, frequency and wavelength of the associated matter wave by the, relations:, E = h Q, p =, , h, , O, , But while the value of O is physically significant, the value of Q, (and therefore, the value of the phase speed QO) has no physical, significance. Why ?, , 411

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Physics, APPENDIX, , 11.1 The history of wave-particle flip-flop, What is light? This question has haunted mankind for a long time. But systematic experiments were done by, scientists since the dawn of the scientific and industrial era, about four centuries ago. Around the same time,, theoretical models about what light is made of were developed. While building a model in any branch of, science, it is essential to see that it is able to explain all the experimental observations existing at that time., It is therefore appropriate to summarize some observations about light that were known in the seventeenth, century., The properties of light known at that time included (a) rectilinear propagation of light, (b) reflection from, plane and curved surfaces, (c) refraction at the boundary of two media, (d) dispersion into various colours, (e), high speed. Appropriate laws were formulated for the first four phenomena. For example, Snell formulated his, laws of refraction in 1621. Several scientists right from the days of Galileo had tried to measure the speed of, light. But they had not been able to do so. They had only concluded that it was higher than the limit of their, measurement., Two models of light were also proposed in the seventeenth century. Descartes, in early decades of seventeenth, century, proposed that light consists of particles, while Huygens, around 1650-60, proposed that light consists, of waves. Descartesc proposal was merely a philosophical model, devoid of any experiments or scientific, arguments. Newton soon after, around 1660-70, extended Descartesc particle model, known as corpuscular, theory, built it up as a scientific theory, and explained various known properties with it. These models, light, as waves and as particles, in a sense, are quite opposite of each other. But both models could explain all the, known properties of light. There was nothing to choose between them., The history of the development of these models over the next few centuries is interesting. Bartholinus, in, 1669, discovered double refraction of light in some crystals, and Huygens, in 1678, was quick to explain it on, the basis of his wave theory of light. In spite of this, for over one hundred years, Newton’s particle model was, firmly believed and preferred over the wave model. This was partly because of its simplicity and partly because, of Newton’s influence on contemporary physics., Then in 1801, Young performed his double-slit experiment and observed interference fringes. This, phenomenon could be explained only by wave theory. It was realized that diffraction was also another, phenomenon which could be explained only by wave theory. In fact, it was a natural consequence of Huygens, idea of secondary wavelets emanating from every point in the path of light. These experiments could not be, explained by assuming that light consists of particles. Another phenomenon of polarisation was discovered, around 1810, and this too could be naturally explained by the wave theory. Thus wave theory of Huygens, came to the forefront and Newton’s particle theory went into the background. This situation again continued, for almost a century., Better experiments were performed in the nineteenth century to determine the speed of light. With more, accurate experiments, a value of 3×108 m/s for speed of light in vacuum was arrived at. Around 1860, Maxwell, proposed his equations of electromagnetism and it was realized that all electromagnetic phenomena known at, that time could be explained by Maxwell’s four equations. Soon Maxwell showed that electric and magnetic, fields could propagate through empty space (vacuum) in the form of electromagnetic waves. He calculated the, speed of these waves and arrived at a theoretical value of 2.998×108 m/s. The close agreement of this value, with the experimental value suggested that light consists of electromagnetic waves. In 1887 Hertz demonstrated, the generation and detection of such waves. This established the wave theory of light on a firm footing. We, might say that while eighteenth century belonged to the particle model, the nineteenth century belonged to, the wave model of light., Vast amounts of experiments were done during the period 1850-1900 on heat and related phenomena, an, altogether different area of physics. Theories and models like kinetic theory and thermodynamics were developed, which quite successfully explained the various phenomena, except one., , 412

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Dual Nature of Radiation, and Matter, Every body at any temperature emits radiation of all wavelengths. It also absorbs radiation falling on it., A body which absorbs all the radiation falling on it is called a black body. It is an ideal concept in physics, like, concepts of a point mass or uniform motion. A graph of the intensity of radiation emitted by a body versus, wavelength is called the black body spectrum. No theory in those days could explain the complete black body, spectrum!, In 1900, Planck hit upon a novel idea. If we assume, he said, that radiation is emitted in packets of energy, instead of continuously as in a wave, then we can explain the black body spectrum. Planck himself regarded, these quanta, or packets, as a property of emission and absorption, rather than that of light. He derived a, formula which agreed with the entire spectrum. This was a confusing mixture of wave and particle pictures –, radiation is emitted as a particle, it travels as a wave, and is again absorbed as a particle! Moreover, this put, physicists in a dilemma. Should we again accept the particle picture of light just to explain one phenomenon?, Then what happens to the phenomena of interference and diffraction which cannot be explained by the, particle model?, But soon in 1905, Einstein explained the photoelectric effect by assuming the particle picture of light., In 1907, Debye explained the low temperature specific heats of solids by using the particle picture for lattice, vibrations in a crystalline solid. Both these phenomena belonging to widely diverse areas of physics could be, explained only by the particle model and not by the wave model. In 1923, Compton’s x-ray scattering experiments, from atoms also went in favour of the particle picture. This increased the dilemma further., Thus by 1923, physicists faced with the following situation. (a) There were some phenomena like rectilinear, propagation, reflection, refraction, which could be explained by either particle model or by wave model. (b), There were some phenomena such as diffraction and interference which could be explained only by the wave, model but not by the particle model. (c) There were some phenomena such as black body radiation, photoelectric, effect, and Compton scattering which could be explained only by the particle model but not by the wave model., Somebody in those days aptly remarked that light behaves as a particle on Mondays, Wednesdays and Fridays,, and as a wave on Tuesdays, Thursdays and Saturdays, and we don’t talk of light on Sundays!, In 1924, de Broglie proposed his theory of wave-particle duality in which he said that not only photons, of light but also ‘particles’ of matter such as electrons and atoms possess a dual character, sometimes, behaving like a particle and sometimes as a wave. He gave a formula connecting their mass, velocity, momentum, (particle characteristics), with their wavelength and frequency (wave characteristics)! In 1927 Thomson, and, Davisson and Germer, in separate experiments, showed that electrons did behave like waves with a wavelength, which agreed with that given by de Broglie’s formula. Their experiment was on diffraction of electrons through, crystalline solids, in which the regular arrangement of atoms acted like a grating. Very soon, diffraction, experiments with other ‘particles’ such as neutrons and protons were performed and these too confirmed with, de Broglie’s formula. This confirmed wave-particle duality as an established principle of physics. Here was a, principle, physicists thought, which explained all the phenomena mentioned above not only for light but also, for the so-called particles., But there was no basic theoretical foundation for wave-particle duality. De Broglie’s proposal was, merely a qualitative argument based on symmetry of nature. Wave-particle duality was at best a principle, not, an outcome of a sound fundamental theory. It is true that all experiments whatever agreed with de Broglie, formula. But physics does not work that way. On the one hand, it needs experimental confirmation, while on, the other hand, it also needs sound theoretical basis for the models proposed. This was developed over the, next two decades. Dirac developed his theory of radiation in about 1928, and Heisenberg and Pauli gave it a, firm footing by 1930. Tomonaga, Schwinger, and Feynman, in late 1940s, produced further refinements and, cleared the theory of inconsistencies which were noticed. All these theories mainly put wave-particle duality, on a theoretical footing., Although the story continues, it grows more and more complex and beyond the scope of this note. But, we have here the essential structure of what happened, and let us be satisfied with it at the moment. Now it, is regarded as a natural consequence of present theories of physics that electromagnetic radiation as well as, particles of matter exhibit both wave and particle properties in different experiments, and sometimes even in, the different parts of the same experiment., , 413

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Chapter Twelve, , ATOMS, , 12.1 INTRODUCTION, By the nineteenth century, enough evidence had accumulated in favour of, atomic hypothesis of matter. In 1897, the experiments on electric discharge, through gases carried out by the English physicist J. J. Thomson (1856 –, 1940) revealed that atoms of different elements contain negatively charged, constituents (electrons) that are identical for all atoms. However, atoms on a, whole are electrically neutral. Therefore, an atom must also contain some, positive charge to neutralise the negative charge of the electrons. But what, is the arrangement of the positive charge and the electrons inside the atom?, In other words, what is the structure of an atom?, The first model of atom was proposed by J. J. Thomson in 1898., According to this model, the positive charge of the atom is uniformly, distributed throughout the volume of the atom and the negatively charged, electrons are embedded in it like seeds in a watermelon. This model was, picturesquely called plum pudding model of the atom. However, subsequent studies on atoms, as described in this chapter, showed that, the distribution of the electrons and positive charges are very different, from that proposed in this model., We know that condensed matter (solids and liquids) and dense gases at, all temperatures emit electromagnetic radiation in which a continuous, distribution of several wavelengths is present, though with different, intensities. This radiation is considered to be due to oscillations of atoms

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Atoms, , ERNST RUTHERFORD (1871 – 1937), , and molecules, governed by the interaction of each atom or, molecule with its neighbours. In contrast, light emitted from, rarefied gases heated in a flame, or excited electrically in a, glow tube such as the familiar neon sign or mercury vapour, light has only certain discrete wavelengths. The spectrum, appears as a series of bright lines. In such gases, the, average spacing between atoms is large. Hence, the, radiation emitted can be considered due to individual atoms, rather than because of interactions between atoms or, molecules., In the early nineteenth century it was also established, that each element is associated with a characteristic, spectrum of radiation, for example, hydrogen always gives, a set of lines with fixed relative position between the lines., Ernst Rutherford (1871 –, 1937) British physicist, This fact suggested an intimate relationship between the, who did pioneering work on, internal structure of an atom and the spectrum of, radioactive radiation. He, radiation emitted by it. In 1885, Johann Jakob Balmer, discovered alpha-rays and, (1825 – 1898) obtained a simple empirical formula which, beta-rays. Along with, gave the wavelengths of a group of lines emitted by atomic, Federick Soddy, he created, hydrogen. Since hydrogen is simplest of the elements, the modern theory of, known, we shall consider its spectrum in detail in this, radioactivity. He studied, chapter., the ‘emanation’ of thorium, Ernst Rutherford (1871–1937), a former research, and discovered a new noble, student of J. J. Thomson, was engaged in experiments on, gas, an isotope of radon,, D-particles emitted by some radioactive elements. In 1906,, now known as thoron. By, he proposed a classic experiment of scattering of these, scattering alpha-rays from, D-particles by atoms to investigate the atomic structure., the, metal, foils,, he, This experiment was later performed around 1911 by Hans, discovered the atomic, Geiger (1882–1945) and Ernst Marsden (1889–1970, who, nucleus and proposed the, was 20 year-old student and had not yet earned his, plenatery model of the, bachelor’s degree). The details are discussed in Section, atom. He also estimated the, 12.2. The explanation of the results led to the birth of, approximate size of the, Rutherford’s planetary model of atom (also called the, nucleus., nuclear model of the atom). According to this the entire, positive charge and most of the mass of the atom is concentrated in a small, volume called the nucleus with electrons revolving around the nucleus just, as planets revolve around the sun., Rutherford’s nuclear model was a major step towards how we see, the atom today. However, it could not explain why atoms emit light of, only discrete wavelengths. How could an atom as simple as hydrogen,, consisting of a single electron and a single proton, emit a complex, spectrum of specific wavelengths? In the classical picture of an atom, the, electron revolves round the nucleus much like the way a planet revolves, round the sun. However, we shall see that there are some serious, difficulties in accepting such a model., , 12.2 ALPHA-PARTICLE SCATTERING AND, RUTHERFORD’S NUCLEAR MODEL OF ATOM, At the suggestion of Ernst Rutherford, in 1911, H. Geiger and E. Marsden, performed some experiments. In one of their experiments, as shown in, , 415

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Physics, , FIGURE 12.1 Geiger-Marsden scattering experiment., The entire apparatus is placed in a vacuum chamber, (not shown in this figure)., , Fig. 12.1, they directed a beam of, 5.5 MeV D-particles emitted from a, 214, radioactive source at a thin metal, 83 Bi, foil made of gold. Figure 12.2 shows a, schematic diagram of this experiment., Alpha-particles emitted by a 214, 83 Bi, radioactive source were collimated into, a narrow beam by their passage, through lead bricks. The beam was, allowed to fall on a thin foil of gold of, thickness 2.1 × 10–7 m. The scattered, alpha-particles were observed through, a rotatable detector consisting of zinc, sulphide screen and a microscope. The, scattered alpha-particles on striking, the screen produced brief light flashes, or scintillations. These flashes may be, viewed through a microscope and the, distribution of the number of scattered, particles may be studied as a function, of angle of scattering., , FIGURE 12.2 Schematic arrangement of the Geiger-Marsden experiment., , 416, , A typical graph of the total number of D-particles scattered at different, angles, in a given interval of time, is shown in Fig. 12.3. The dots in this, figure represent the data points and the solid curve is the theoretical, prediction based on the assumption that the target atom has a small,, dense, positively charged nucleus. Many of the D-particles pass through, the foil. It means that they do not suffer any collisions. Only about 0.14%, of the incident D-particles scatter by more than 1º; and about 1 in 8000, deflect by more than 90º. Rutherford argued that, to deflect the D-particle, backwards, it must experience a large repulsive force. This force could

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Atoms, be provided if the greater part of the, mass of the atom and its positive charge, were concentrated tightly at its centre., Then the incoming D-particle could get, very close to the positive charge without, penetrating it, and such a close, encounter would result in a large, deflection. This agreement supported, the hypothesis of the nuclear atom. This, is why Rutherford is credited with the, discovery of the nucleus., In Rutherford’s nuclear model of, the atom, the entire positive charge and, most of the mass of the atom are, concentrated in the nucleus with the, electrons some distance away. The, electrons would be moving in orbits, FIGURE 12.3 The dots are D-particle scattering data, about the nucleus just as the planets, for a thin gold foil obtained by Geiger and Marsden, do around the sun. Rutherford’s, using the experiment shown in Figs. 12.1 and, experiments suggested the size of, 12.2. The solid curve is the theoretical, the nucleus to be about 10–15 m to, prediction based on the assumption that, 10–14 m. From kinetic theory, the size, atom has a small, dense, positively, charged nucleus., of an atom was known to be 10–10 m,, about 10,000 to 100,000 times larger, than the size of the nucleus (see Chapter 11, Section 11.6 in Class XI, Physics textbook). Thus, the electrons would seem to be at a distance, from the nucleus of about 10,000 to 100,000 times the size of the nucleus, itself. Thus, most of an atom is empty space. With the atom being largely, empty space, it is easy to see why most D-particles go right through a, thin metal foil. However, when D-particle happens to come near a nucleus,, the intense electric field there scatters it through a large angle. The atomic, electrons, being so light, do not appreciably affect the D-particles., The scattering data shown in Fig. 12.3 can be analysed by employing, Rutherford’s nuclear model of the atom. As the gold foil is very thin, it, can be assumed that D-particles will suffer not more than one scattering, during their passage through it. Therefore, computation of the trajectory, of an alpha-particle scattered by a single nucleus is enough. Alphaparticles are nuclei of helium atoms and, therefore, carry two units, 2e,, of positive charge and have the mass of the helium atom. The charge of, the gold nucleus is Ze, where Z is the atomic number of the atom; for, gold Z = 79. Since the nucleus of gold is about 50 times heavier than an, D-particle, it is reasonable to assume that it remains stationary, throughout the scattering process. Under these assumptions, the, trajectory of an alpha-particle can be computed employing Newton’s, second law of motion and the Coulomb’s law for electrostatic, force of repulsion between the alpha-particle and the positively, 417, charged nucleus.

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Physics, The magnitude of this force is, 1 (2e )( Ze ), F, (12.1), 4 SH 0, r2, where r is the distance between the D-particle and the nucleus. The force, is directed along the line joining the D-particle and the nucleus. The, magnitude and direction of the force on an D-particle continuously, changes as it approaches the nucleus and recedes away from it., , 12.2.1 Alpha-particle trajectory, The trajectory traced by an D-particle depends on the impact parameter,, b of collision. The impact parameter is the perpendicular distance of the, initial velocity vector of the D-particle from the centre of the nucleus (Fig., 12.4). A given beam of D-particles has a, distribution of impact parameters b, so that, the beam is scattered in various directions, with different probabilities (Fig. 12.4). (In, a beam, all particles have nearly same, kinetic energy.) It is seen that an D-particle, close to the nucleus (small impact, parameter) suffers large scattering. In case, of head-on collision, the impact parameter, is minimum and the D-particle rebounds, back (T#S). For a large impact parameter,, the D-particle goes nearly undeviated and, has a small deflection (T# 0)., FIGURE 12.4 Trajectory of D-particles in the, The fact that only a small fraction of the, coulomb field of a target nucleus. The impact, number of incident particles rebound back, parameter, b and scattering angle T, indicates that the number of D-particles, are also depicted., undergoing head on collision is small. This,, in turn, implies that the mass of the atom is concentrated in a small, volume. Rutherford scattering therefore, is a powerful way to determine, an upper limit to the size of the nucleus., , 418, , EXAMPLE 12.1, , Example 12.1 In the Rutherford’s nuclear model of the atom, the, nucleus (radius about 10–15 m) is analogous to the sun about which, the electron move in orbit (radius | 10–10 m) like the earth orbits, around the sun. If the dimensions of the solar system had the same, proportions as those of the atom, would the earth be closer to or, farther away from the sun than actually it is ? The radius of earth’s, orbit is about 1.5 u 1011 m. The radius of sun is taken as 7 u 108 m., Solution The ratio of the radius of electron’s orbit to the radius of, nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s, orbit is 105 times larger than the radius of nucleus. If the radius of, the earth’s orbit around the sun were 105 times larger than the radius, of the sun, the radius of the earth’s orbit would be 105 u 7 u 108 m =, 7 u 1013 m. This is more than 100 times greater than the actual orbital, radius of earth. Thus, the earth would be much farther away from, the sun., It implies that an atom contains a much greater fraction of empty, space than our solar system does.

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Atoms, Example 12.2 In a Geiger-Marsden experiment, what is the distance, of closest approach to the nucleus of a 7.7 MeV D-particle before it, comes momentarily to rest and reverses its direction?, , 2Ze 2, 4 SH 0d, , Thus the distance of closest approach d is given by, d, , 2 Ze 2, 4 SH 0 K, , The maximum kinetic energy found in D-particles of natural origin is, 7.7 MeV or 1.2 × 10–12 J. Since 1/4SH= 9.0 × 109 N m2/C2. Therefore, with e = 1.6 × 10–19 C, we have,, (2)(9.0 u 10 9 Nm 2 / C 2 )(1.6 u 10 –19 C )2 Z, 1.2 u 10 12 J, –16, = 3.84 × 10 Z m, The atomic number of foil material gold is Z = 79, so that, d, , d (Au) = 3.0 × 10–14 m = 30 fm. (1 fm (i.e. fermi) = 10–15 m.), , EXAMPLE 12.2, , The radius of gold nucleus is, therefore, less than 3.0 × 10–14 m. This, is not in very good agreement with the observed result as the actual, radius of gold nucleus is 6 fm. The cause of discrepancy is that the, distance of closest approach is considerably larger than the sum of, the radii of the gold nucleus and the D-particle. Thus, the D-particle, reverses its motion without ever actually touching the gold nucleus., , Simulate Rutherford scattering experiment, , 1 (2e )( Ze ), d, 4SH 0, , K, , http://www outreach.phy.cam.ac.uk/camphy/nucleus/nucleus_exp.htm, , Solution The key idea here is that throughout the scattering process,, the total mechanical energy of the system consisting of an D-particle, and a gold nucleus is conserved. The system’s initial mechanical, energy is Ei, before the particle and nucleus interact, and it is equal, to its mechanical energy Ef when the D-particle momentarily stops., The initial energy Ei is just the kinetic energy K  of the incoming, D- particle. The final energy Ef is just the electric potential energy U, of the system. The potential energy U can be calculated from, Eq. (12.1)., Let d be the centre-to-centre distance between the D-particle and, the gold nucleus when the D-particle is at its stopping point. Then, we can write the conservation of energy Ei = Ef as, , 12.2.2 Electron orbits, The Rutherford nuclear model of the atom which involves classical, concepts, pictures the atom as an electrically neutral sphere consisting, of a very small, massive and positively charged nucleus at the centre, surrounded by the revolving electrons in their respective dynamically, stable orbits. The electrostatic force of attraction, Fe between the revolving, electrons and the nucleus provides the requisite centripetal force (Fc ) to, keep them in their orbits. Thus, for a dynamically stable orbit in a, hydrogen atom, Fe = Fc, , mv 2, r, , 1 e2, 4 SH 0 r 2, , (12.2), , 419

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Physics, Thus the relation between the orbit radius and the electron, velocity is, , r, , e2, 4SH 0mv 2, , (12.3), , The kinetic energy (K ) and electrostatic potential energy (U ) of the electron, in hydrogen atom are, , 1, mv 2, 2, , K, , e2, and U, 8SH 0r, , , , e2, 4SH 0r, , (The negative sign in U signifies that the electrostatic force is in the –r, direction.) Thus the total energy E of the electron in a hydrogen atom is, , K U, , E, , e2, e2, , 8SH 0r 4SH 0r, , , , e2, 8SH 0r, , (12.4), , The total energy of the electron is negative. This implies the fact that, the electron is bound to the nucleus. If E were positive, an electron will, not follow a closed orbit around the nucleus., Example 12.3 It is found experimentally that 13.6 eV energy is, required to separate a hydrogen atom into a proton and an electron., Compute the orbital radius and the velocity of the electron in a, hydrogen atom., Solution Total energy of the electron in hydrogen atom is –13.6 eV =, –13.6 × 1.6 × 10–19 J = –2.2 ×10–18 J. Thus from Eq. (12.4), we have, , , , e2, 8SH 0r, , 2.2 u 10 18 J, , This gives the orbital radius, , EXAMPLE 12.3, , r, , , , e2, 8SH 0 E, , , , (9 u 109 N m 2/C2 )(1.6 u 10 19 C)2, (2)(–2.2 u 1018 J), , = 5.3 × 10–11 m., The velocity of the revolving electron can be computed from Eq. (12.3), with m = 9.1 × 10–31 kg,, , v, , e, 4SH 0mr, , 2.2 u 106 m/s., , 12.3 ATOMIC SPECTRA, , 420, , As mentioned in Section 12.1, each element has a characteristic spectrum, of radiation, which it emits. When an atomic gas or vapour is excited at, low pressure, usually by passing an electric current through it, the emitted, radiation has a spectrum which contains certain specific wavelengths, only. A spectrum of this kind is termed as emission line spectrum and it

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Atoms, consists of bright lines on a, dark, background., The, spectrum emitted by atomic, hydrogen is shown in, Fig. 12.5. Study of emission, line spectra of a material can, therefore serve as a type of, “fingerprint” for identification, of the gas. When white light, passes through a gas and we, analyse the transmitted light, using a spectrometer we find, some dark lines in the, FIGURE 12.5 Emission lines in the spectrum of hydrogen., spectrum. These dark lines, correspond precisely to those wavelengths which were found in the, emission line spectrum of the gas. This is called the absorption spectrum, of the material of the gas., , 12.3.1 Spectral series, We might expect that the frequencies of the light emitted by a particular, element would exhibit some regular pattern. Hydrogen is the simplest, atom and therefore, has the simplest spectrum. In the observed spectrum,, however, at first sight, there does not seem to be, any resemblance of order or regularity in spectral, lines. But the spacing between lines within certain, sets of the hydrogen spectrum decreases in a, regular way (Fig. 12.5). Each of these sets is called, a spectral series. In 1885, the first such series was, observed by a Swedish school teacher Johann Jakob, Balmer (1825–1898) in the visible region of the, hydrogen spectrum. This series is called Balmer, series (Fig. 12.6). The line with the longest, wavelength, 656.3 nm in the red is called HD; the, FIGURE 12.6 Balmer series in the, next line with wavelength 486.1 nm in the blueemission spectrum of hydrogen., green is called HE, the third line 434.1 nm in the, violet is called HJ; and so on. As the wavelength, decreases, the lines appear closer together and are weaker in intensity., Balmer found a simple empirical formula for the observed wavelengths, 1, , 1·, § 1, R¨ 2  2¸, (12.5), ©2 n ¹, where O is the wavelength, R is a constant called the Rydberg constant,, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 107 m–1., This equation is also called Balmer formula., Taking n = 3 in Eq. (12.5), one obtains the wavelength of the HD line:, , O, , 1, , 1·, § 1, 1.097 u 107 ¨ 2  2 ¸ m –1, ©2, 3 ¹, = 1.522 × 106 m–1, i.e., O = 656.3 nm, , O, , 421

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Physics, For n = 4, one obtains the wavelength of HE line, etc. For n = f, one obtains, the limit of the series, at O = 364.6 nm. This is the shortest wavelength in, the Balmer series. Beyond this limit, no further distinct lines appear,, instead only a faint continuous spectrum is seen., Other series of spectra for hydrogen were subsequently discovered., These are known, after their discoverers, as Lyman, Paschen, Brackett,, and Pfund series. These are represented by the formulae:, Lyman series:, 1, , §1 1·, R¨ 2  2¸, ©1 n ¹, O, Paschen series:, 1, , 1·, § 1, R¨ 2  2¸, ©3 n ¹, O, Brackett series:, 1, , 1·, § 1, R¨ 2  2¸, ©4 n ¹, Pfund series:, , O, , n = 2,3,4..., , (12.6), , n = 4,5,6..., , (12.7), , n = 5,6,7..., , (12.8), , 1, , 1·, § 1, R¨ 2  2¸, n = 6,7,8..., (12.9), ©5 n ¹, The Lyman series is in the ultraviolet, and the Paschen and Brackett, series are in the infrared region., The Balmer formula Eq. (12.5) may be written in terms of frequency, of the light, recalling that, c = QO, , O, , or, , 1, , Q, , O c, Thus, Eq. (12.5) becomes, 1·, § 1, , (12.10), © 22 n 2 ¸¹, There are only a few elements (hydrogen, singly ionised helium, and, doubly ionised lithium) whose spectra can be represented by simple, formula like Eqs. (12.5) – (12.9)., Equations (12.5) – (12.9) are useful as they give the wavelengths that, hydrogen atoms radiate or absorb. However, these results are empirical, and do not give any reasoning why only certain frequencies are observed, in the hydrogen spectrum., , Q = Rc ¨, , 12.4 BOHR MODEL, , 422, , OF THE, , HYDROGEN ATOM, , The model of the atom proposed by Rutherford assumes that the atom,, consisting of a central nucleus and revolving electron is stable much like, sun-planet system which the model imitates. However, there are some, fundamental differences between the two situations. While the planetary, system is held by gravitational force, the nucleus-electron system being, charged objects, interact by Coulomb’s Law of force. We know that an

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Atoms, , Niels Henrik David Bohr, (1885 – 1962) Danish, physicist who explained the, spectrum of hydrogen atom, based on quantum ideas., He gave a theory of nuclear, fission based on the liquiddrop model of nucleus., Bohr contributed to the, clarification of conceptual, problems in quantum, mechanics, in particular by, proposing the complementary principle., , NIELS HENRIK DAVID BOHR (1885 – 1962), , object which moves in a circle is being constantly, accelerated – the acceleration being centripetal in nature., According to classical electromagnetic theory, an, accelerating charged particle emits radiation in the form, of electromagnetic waves. The energy of an accelerating, electron should therefore, continuously decrease. The, electron would spiral inward and eventually fall into the, nucleus (Fig. 12.7). Thus, such an atom can not be stable., Further, according to the classical electromagnetic theory,, the frequency of the electromagnetic waves emitted by, the revolving electrons is equal to the frequency of, revolution. As the electrons spiral inwards, their angular, velocities and hence their frequencies would change, continuously, and so will the frequency of the light, emitted. Thus, they would emit a continuous spectrum,, in contradiction to the line spectrum actually observed., Clearly Rutherford model tells only a part of the story, implying that the classical ideas are not sufficient to, explain the atomic structure., , FIGURE 12.7 An accelerated atomic electron must spiral into the, nucleus as it loses energy., Example 12.4 According to the classical electromagnetic theory,, calculate the initial frequency of the light emitted by the electron, revolving around a proton in hydrogen atom., Solution From Example 12.3 we know that velocity of electron moving, around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m, is 2.2 × 10–6 m/s. Thus, the frequency of the electron moving around, the proton is, , Q, , 2.2 u 106 m s 1, 2S 5.3 u 1011 m, , | 6.6 × 1015 Hz., According to the classical electromagnetic theory we know that the, frequency of the electromagnetic waves emitted by the revolving, electrons is equal to the frequency of its revolution around the nucleus., Thus the initial frequency of the light emitted is 6.6 × 1015 Hz., , EXAMPLE 12.4, , v, 2Sr, , 423

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Physics, It was Niels Bohr (1885 – 1962) who made certain modifications in, this model by adding the ideas of the newly developing quantum, hypothesis. Niels Bohr studied in Rutherford’s laboratory for several, months in 1912 and he was convinced about the validity of Rutherford, nuclear model. Faced with the dilemma as discussed above, Bohr, in, 1913, concluded that in spite of the success of electromagnetic theory in, explaining large-scale phenomena, it could not be applied to the processes, at the atomic scale. It became clear that a fairly radical departure from, the established principles of classical mechanics and electromagnetism, would be needed to understand the structure of atoms and the relation, of atomic structure to atomic spectra. Bohr combined classical and early, quantum concepts and gave his theory in the form of three postulates., These are :, (i) Bohr’s first postulate was that an electron in an atom could revolve, in certain stable orbits without the emission of radiant energy,, contrary to the predictions of electromagnetic theory. According to, this postulate, each atom has certain definite stable states in which it, can exist, and each possible state has definite total energy. These are, called the stationary states of the atom., (ii) Bohr’s second postulate defines these stable orbits. This postulate, states that the electron revolves around the nucleus only in those, orbits for which the angular momentum is some integral multiple of, h/2S where h is the Planck’s constant (= 6.6 u 10–34 J s). Thus the, angular momentum (L) of the orbiting electron is quantised. That is, (12.11), , L = nh/2S, , (iii) Bohr’s third postulate incorporated into atomic theory the early, quantum concepts that had been developed by Planck and Einstein., It states that an electron might make a transition from one of its, specified non-radiating orbits to another of lower energy. When it, does so, a photon is emitted having energy equal to the energy, difference between the initial and final states. The frequency of the, emitted photon is then given by, hQ = Ei – Ef, (12.12), where Ei and Ef are the energies of the initial and final states and Ei > Ef ., For a hydrogen atom, Eq. (12.4) gives the expression to determine, the energies of different energy states. But then this equation requires, the radius r of the electron orbit. To calculate r, Bohr’s second postulate, about the angular momentum of the electron–the quantisation, condition – is used. The angular momentum L is given by, L = mvr, Bohr’s second postulate of quantisation [Eq. (12.11)] says that the, allowed values of angular momentum are integral multiples of h/2S., Ln = mvnrn =, , 424, , nh, 2S, , (12.13), , where n is an integer, rn is the radius of nth possible orbit and vn is the, speed of moving electron in the nth orbit. The allowed orbits are numbered

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Atoms, 1, 2, 3 ..., according to the values of n, which is called the principal, quantum number of the orbit., From Eq. (12.3), the relation between vn and rn is, e, vn, 4SH 0mrn, Combining it with Eq. (12.13), we get the following expressions for vn, and rn,, vn, , 1 e2, 1, n 4 SH 0 h 2S, , (12.14), , and, 2, , rn, , § n 2 · § h · 4 SH 0, ¨© m ¸¹ ¨© 2 S ¸¹ e 2, , (12.15), , Eq. (12.14) depicts that the orbital speed in the nth orbit falls by a factor, of n. Using Eq. (12.15), the size of the innermost orbit (n = 1) can be, obtained as, , r1, , h 2 H0, S me 2, , This is called the Bohr radius, represented by the symbol a0. Thus,, h 2H0, (12.16), Sme 2, Substitution of values of h, m, H0 and e gives a 0 = 5.29 u 10–11 m. From, Eq. (12.15), it can also be seen that the radii of the orbits increase as n2., The total energy of the electron in the stationary states of the hydrogen, atom can be obtained by substituting the value of orbital radius in Eq., (12.4) as, a0, , 2, , En, , or En, , § e 2 · § m · § 2S · § e 2 ·, ¨, ¨ ¸¨ ¸, © 8SH 0 ¸¹ © n 2 ¹ © h ¹ ¨© 4 SH 0 ¸¹, , , , me 4, 8n 2 H 02h 2, , (12.17), , Substituting values, Eq. (12.17) yields, 2.18 u 10 18, J, (12.18), n2, Atomic energies are often expressed in electron volts (eV) rather than, joules. Since 1 eV = 1.6 u 10–19 J, Eq. (12.18) can be rewritten as, En, , , , 13.6, eV, (12.19), n2, The negative sign of the total energy of an electron moving in an orbit, means that the electron is bound with the nucleus. Energy will thus be, required to remove the electron from the hydrogen atom to a distance, infinitely far away from its nucleus (or proton in hydrogen atom)., En, , , , 425

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Physics, The derivation of Eqs. (12.17) – (12.19) involves the assumption that, the electronic orbits are circular, though orbits under inverse square, force are, in general elliptical. (Planets move in elliptical orbits under the, inverse square gravitational force of the sun.) However, it was shown by, the German physicist Arnold Sommerfeld (1868 – 1951) that, when the, restriction of circular orbit is relaxed, these equations continue to hold, even for elliptic orbits., , ORBIT, , VS STATE (ORBITAL PICTURE) OF ELECTRON IN ATOM, , 426, , EXAMPLE 12.5, , We are introduced to the Bohr Model of atom one time or the other in the course of, physics. This model has its place in the history of quantum mechanics and particularly, in explaining the structure of an atom. It has become a milestone since Bohr introduced, the revolutionary idea of definite energy orbits for the electrons, contrary to the classical, picture requiring an accelerating particle to radiate. Bohr also introduced the idea of, quantisation of angular momentum of electrons moving in definite orbits. Thus it was a, semi-classical picture of the structure of atom., Now with the development of quantum mechanics, we have a better understanding, of the structure of atom. Solutions of the Schrödinger wave equation assign a wave-like, description to the electrons bound in an atom due to attractive forces of the protons., An orbit of the electron in the Bohr model is the circular path of motion of an electron, around the nucleus. But according to quantum mechanics, we cannot associate a definite, path with the motion of the electrons in an atom. We can only talk about the probability, of finding an electron in a certain region of space around the nucleus. This probability, can be inferred from the one-electron wave function called the orbital. This function, depends only on the coordinates of the electron., It is therefore essential that we understand the subtle differences that exist in the two, models:, z, Bohr model is valid for only one-electron atoms/ions; an energy value, assigned to, each orbit, depends on the principal quantum number n in this model. We know, that energy associated with a stationary state of an electron depends on n only, for, one-electron atoms/ions. For a multi-electron atom/ion, this is not true., z, The solution of the Schrödinger wave equation, obtained for hydrogen-like atoms/, ions, called the wave function, gives information about the probability of finding an, electron in various regions around the nucleus. This orbital has no resemblance, whatsoever with the orbit defined for an electron in the Bohr model., , Example 12.5 A 10 kg satellite circles earth once every 2 h in an, orbit having a radius of 8000 km. Assuming that Bohr’s angular, momentum postulate applies to satellites just as it does to an electron, in the hydrogen atom, find the quantum number of the orbit of the, satellite., Solution, From Eq. (12.13), we have, m vn rn = nh/2S

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Atoms, , EXAMPLE 12.5, , Here m = 10 kg and rn = 8 u 106 m. We have the time period T of the, circling satellite as 2 h. That is T = 7200 s., Thus the velocity vn = 2S rn/T., The quantum number of the orbit of satellite, n = (2S rn)2 u m/(T u h)., Substituting the values,, n = (2S u 8 u 106 m)2 u 10/(7200 s u 6.64 u 10–34 J s), = 5.3 u 1045, Note that the quantum number for the satellite motion is extremely, large! In fact for such large quantum numbers the results of, quantisation conditions tend to those of classical physics., , 12.4.1 Energy levels, The energy of an atom is the least (largest negative value), when its electron is revolving in an orbit closest to the, nucleus i.e., the one for which n = 1. For n = 2, 3, ... the, absolute value of the energy E is smaller, hence the energy, is progressively larger in the outer orbits. The lowest state, of the atom, called the ground state, is that of the lowest, energy, with the electron revolving in the orbit of smallest, radius, the Bohr radius, a 0. The energy of this state (n = 1),, E1 is –13.6 eV. Therefore, the minimum energy required to, free the electron from the ground state of the hydrogen atom, is 13.6 eV. It is called the ionisation energy of the hydrogen, atom. This prediction of the Bohr’s model is in excellent, agreement with the experimental value of ionisation energy., At room temperature, most of the hydrogen atoms are, in ground state. When a hydrogen atom receives energy, by processes such as electron collisions, the atom may, acquire sufficient energy to raise the electron to higher, energy states. The atom is then said to be in an excited, state. From Eq. (12.19), for n = 2; the energy E2 is, –3.40 eV. It means that the energy required to excite an, electron in hydrogen atom to its first excited state, is an, energy equal to E2 – E1 = –3.40 eV – (–13.6) eV = 10.2 eV., Similarly, E3 = –1.51 eV and E3 – E1 = 12.09 eV, or to excite, the hydrogen atom from its ground state (n = 1) to second, excited state (n = 3), 12.09 eV energy is required, and so, on. From these excited states the electron can then fall back, to a state of lower energy, emitting a photon in the process., Thus, as the excitation of hydrogen atom increases (that is, as n increases) the value of minimum energy required to, free the electron from the excited atom decreases., The energy level diagram* for the stationary states of a, hydrogen atom, computed from Eq. (12.19), is given in, , FIGURE 12.8 The energy level, diagram for the hydrogen atom., The electron in a hydrogen atom, at room temperature spends, most of its time in the ground, state. To ionise a hydrogen, atom an electron from the, ground state, 13.6 eV of energy, must be supplied. (The horizontal, lines specify the presence of, allowed energy states.), , * An electron can have any total energy above E = 0 eV. In such situations the, electron is free. Thus there is a continuum of energy states above E = 0 eV, as, shown in Fig. 12.8., , 427

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Physics, Fig. 12.8. The principal quantum number n labels the stationary, states in the ascending order of energy. In this diagram, the highest, energy state corresponds to n = f in Eq, (12.19) and has an energy, of 0 eV. This is the energy of the atom when the electron is, completely removed (r = f) from the nucleus and is at rest. Observe how, the energies of the excited states come closer and closer together as n, increases., , FRANCK – HERTZ, , EXPERIMENT, , The existence of discrete energy levels in an atom was directly verified in 1914 by James, Franck and Gustav Hertz. They studied the spectrum of mercury vapour when electrons, having different kinetic energies passed through the vapour. The electron energy was, varied by subjecting the electrons to electric fields of varying strength. The electrons, collide with the mercury atoms and can transfer energy to the mercury atoms. This can, only happen when the energy of the electron is higher than the energy difference between, an energy level of Hg occupied by an electron and a higher unoccupied level (see Figure)., For instance, the difference between an occupied energy level of Hg and a higher, unoccupied level is 4.9 eV. If an electron of having an energy of 4.9 eV or more passes, through mercury, an electron in mercury atom can absorb energy from the bombarding, electron and get excited to the higher level [Fig (a)]. The colliding electron’s kinetic energy, would reduce by this amount., , The excited electron would subsequently fall back to the ground state by emission of, radiation [Fig. (b)]. The wavelength of emitted radiation is:, , O, , hc, E, , 6.625 u 1034 u 3 u 108, = 253 nm, 4.9 u 1.6 u 1019, , By direct measurement, Franck and Hertz found that the emission spectrum of, mercury has a line corresponding to this wavelength. For this experimental verification, of Bohr’s basic ideas of discrete energy levels in atoms and the process of photon emission,, Frank and Hertz were awarded the Nobel prize in 1925., , 12.5 THE LINE SPECTRA, , 428, , OF THE, , HYDROGEN ATOM, , According to the third postulate of Bohr’s model, when an atom makes a, transition from the higher energy state with quantum number ni to the, lower energy state with quantum number nf (nf < ni ), the difference of, energy is carried away by a photon of frequency Qif such that

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Atoms, hQif = Eni – Enf, , (12.20), , Using Eq. (12.16), for Enf and Eni, we get, hQif =, , me 4 § 1, 1·,  2¸, 2 2 ¨ 2, 8H 0h © n f n i ¹, , or Qif =, , me 4, 8H 02h 3, , § 1, 1·, ¨ n2  n2 ¸, © f, i ¹, , (12.21), , (12.22), , Equation (12.21) is the Rydberg formula, for the spectrum of the, hydrogen atom. In this relation, if we take nf = 2 and ni = 3, 4, 5..., it, reduces to a form similar to Eq. (12.10) for the Balmer series. The Rydberg, constant R is readily identified to be, R=, , me 4, 8H2h 3c, , (12.23), , If we insert the values of various constants in Eq. (12.23), we get, R = 1.03 × 107 m–1, This is a value very close to the value (1.097 × 107 m–1) obtained from the, empirical Balmer formula. This agreement between the theoretical and, experimental values of the Rydberg constant provided a direct and, striking confirmation of the Bohr’s model., Since both nf and ni are integers,, this immediately shows that in, transitions between different atomic, levels, light is radiated in various, discrete frequencies. For hydrogen, spectrum, the Balmer formula, corresponds to nf = 2 and ni = 3, 4, 5,, etc. The results of the Bohr’s model, suggested the presence of other series, spectra for hydrogen atom–those, corresponding to transitions resulting, from nf = 1 and ni = 2, 3, etc.; nf = 3, and ni = 4, 5, etc., and so on. Such, series were identified in the course of, spectroscopic investigations and are, known as the L yman, Balmer,, Paschen, Brackett, and Pfund series., The, electronic, transitions, corresponding to these series are, shown in Fig. 12.9., The various lines in the atomic, spectra are produced when electrons, jump from higher energy state to a, lower energy state and photons are, emitted. These spectral lines are called, emission lines. But when an atom, FIGURE 12.9 Line spectra originate in, transitions between energy levels., absorbs a photon that has precisely, , 429

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Physics, the same energy needed by the electron in a lower energy state to make, transitions to a higher energy state, the process is called absorption., Thus if photons with a continuous range of frequencies pass through a, rarefied gas and then are analysed with a spectrometer, a series of dark, spectral absorption lines appear in the continuous spectrum. The dark, lines indicate the frequencies that have been absorbed by the atoms of, the gas., The explanation of the hydrogen atom spectrum provided by Bohr’s, model was a brilliant achievement, which greatly stimulated progress, towards the modern quantum theory. In 1922, Bohr was awarded Nobel, Prize in Physics., Example 12.6 Using the Rydberg formula, calculate the wavelengths, of the first four spectral lines in the Lyman series of the hydrogen, spectrum., Solution The Rydberg formula is, hc/Oif =, , me 4 § 1, 1·,  2¸, 2 2 ¨ 2, 8H  h © n f n i ¹, , The wavelengths of the first four lines in the Lyman series correspond, to transitions from ni = 2,3,4,5 to nf = 1. We know that, , me 4, = 13.6 eV = 21.76 ×10–19 J, 8H2h 2, Therefore,, , Oi 1, , hc, 21.76 u 10, , 19, , §1 1 ·, ¨© 1  n 2 ¸¹, , m, , EXAMPLE 12.6, , i, , =, , 6.625 u 1034 u 3 u 108 u n i2, m =, 21.76 u 1019 u (n i2  1), , 0.9134 n i2, u 107 m, (n i2  1), , = 913.4 ni2/(ni2 –1) Å, Substituting ni = 2,3,4,5, we get O21 = 1218 Å, O31 = 1028 Å, O41 = 974.3 Å,, and O51 = 951.4 Å., , 12.6 DE BROGLIE’S EXPLANATION OF BOHR’S, SECOND POSTULATE OF QUANTISATION, , 430, , Of all the postulates, Bohr made in his model of the atom, perhaps the, most puzzling is his second postulate. It states that the angular, momentum of the electron orbiting around the nucleus is quantised (that, is, Ln = nh/2S; n = 1, 2, 3 …). Why should the angular momentum have, only those values that are integral multiples of h/2S ? The French physicist, Louis de Broglie explained this puzzle in 1923, ten years after Bohr, proposed his model., We studied, in Chapter 11, about the de Broglie’s hypothesis that, material particles, such as electrons, also have a wave nature. C. J. Davisson, and L. H. Germer later experimentally verified the wave nature of electrons

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Atoms, in 1927. Louis de Broglie argued that the electron in its, circular orbit, as proposed by Bohr, must be seen as a particle, wave. In analogy to waves travelling on a string, particle waves, too can lead to standing waves under resonant conditions., From Chapter 15 of Class XI Physics textbook, we know that, when a string is plucked, a vast number of wavelengths are, excited. However only those wavelengths survive which have, nodes at the ends and form the standing wave in the string. It, means that in a string, standing waves are formed when the, total distance travelled by a wave down the string and back is, one wavelength, two wavelengths, or any integral number of, wavelengths. Waves with other wavelengths interfere with, themselves upon reflection and their amplitudes quickly drop, to zero. For an electron moving in nth circular orbit of radius, rn, the total distance is the circumference of the orbit, 2Srn., Thus, , FIGURE 12.10 A standing, wave is shown on a circular, orbit where four de Broglie, wavelengths fit into the, circumference of the orbit., , n = 1, 2, 3..., (12.24), 2S rn = nO,, Figure 12.10 illustrates a standing particle wave on a, circular orbit for n = 4, i.e., 2Srn = 4O, where O is the de Broglie, wavelength of the electron moving in nth orbit. From Chapter, 11, we have O = h/p, where p is the magnitude of the electron’s, momentum. If the speed of the electron is much less than the speed of, light, the momentum is mvn. Thus, O = h/mvn. From Eq. (12.24), we have, 2S rn = n h/mvn, or m vn rn = nh/2S, This is the quantum condition proposed by Bohr for the angular, momentum of the electron [Eq. (12.13)]. In Section 12.5, we saw that, this equation is the basis of explaining the discrete orbits and energy, levels in hydrogen atom. Thus de Broglie hypothesis provided an, explanation for Bohr’s second postulate for the quantisation of angular, momentum of the orbiting electron. The quantised electron orbits and, energy states are due to the wave nature of the electron and only resonant, standing waves can persist., Bohr’s model, involving classical trajectory picture (planet-like electron, orbiting the nucleus), correctly predicts the gross features of the, hydrogenic atoms*, in particular, the frequencies of the radiation emitted, or selectively absorbed. This model however has many limitations., Some are:, (i) The Bohr model is applicable to hydrogenic atoms. It cannot be, extended even to mere two electron atoms such as helium. The analysis, of atoms with more than one electron was attempted on the lines of, Bohr’s model for hydrogenic atoms but did not meet with any success., Difficulty lies in the fact that each electron interacts not only with the, positively charged nucleus but also with all other electrons., , * Hydrogenic atoms are the atoms consisting of a nucleus with positive charge, +Ze and a single electron, where Z is the proton number. Examples are hydrogen, atom, singly ionised helium, doubly ionised lithium, and so forth. In these, atoms more complex electron-electron interactions are nonexistent., , 431

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Physics, The formulation of Bohr model involves electrical force between, positively charged nucleus and electron. It does not include the, electrical forces between electrons which necessarily appear in, multi-electron atoms., (ii) While the Bohr’s model correctly predicts the frequencies of the light, emitted by hydrogenic atoms, the model is unable to explain the, relative intensities of the frequencies in the spectrum. In emission, spectrum of hydrogen, some of the visible frequencies have weak, intensity, others strong. Why? Experimental observations depict that, some transitions are more favoured than others. Bohr’s model is, unable to account for the intensity variations., Bohr’s model presents an elegant picture of an atom and cannot be, generalised to complex atoms. For complex atoms we have to use a new, and radical theory based on Quantum Mechanics, which provides a more, complete picture of the atomic structure., , LASER, , LIGHT, , Imagine a crowded market place or a railway platform with people entering a gate and, going towards all directions. Their footsteps are random and there is no phase correlation, between them. On the other hand, think of a large number of soldiers in a regulated march., Their footsteps are very well correlated. See figure here., This is similar to the difference between light emitted by, an ordinary source like a candle or a bulb and that emitted, by a laser. The acronym LASER stands for Light Amplification, by Stimulated Emission of Radiation. Since its development, in 1960, it has entered into all areas of science and technology., It has found applications in physics, chemistry, biology,, medicine, surgery, engineering, etc. There are low power, lasers, with a power of 0.5 mW, called pencil lasers, which, serve as pointers. There are also lasers of different power,, suitable for delicate surgery of eye or glands in the stomach., Finally, there are lasers which can cut or weld steel., Light is emitted from a source in the form of packets of, waves. Light coming out from an ordinary source contains a mixture of many wavelengths., There is also no phase relation between the various waves. Therefore, such light, even if it is, passed through an aperture, spreads very fast and the beam size increases rapidly with, distance. In the case of laser light, the wavelength of each packet is almost the same. Also, the average length of the packet of waves is much larger. This means that there is better, phase correlation over a longer duration of time. This results in reducing the divergence of, a laser beam substantially., If there are N atoms in a source, each emitting light with intensity I, then the total, intensity produced by an ordinary source is proportional to NI, whereas in a laser source,, it is proportional to N2I. Considering that N is very large, we see that the light from a laser, can be much stronger than that from an ordinary source., When astronauts of the Apollo missions visited the moon, they placed a mirror on its, surface, facing the earth. Then scientists on the earth sent a strong laser beam, which was, reflected by the mirror on the moon and received back on the earth. The size of the reflected, laser beam and the time taken for the round trip were measured. This allowed a very, accurate determination of (a) the extremely small divergence of a laser beam and (b) the, distance of the moon from the earth., , 432

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Atoms, SUMMARY, 1., 2., 3., , 4., , 5., , 6., , Atom, as a whole, is electrically neutral and therefore contains equal, amount of positive and negative charges., In Thomson’s model, an atom is a spherical cloud of positive charges, with electrons embedded in it., In Rutherford’s model, most of the mass of the atom and all its positive, charge are concentrated in a tiny nucleus (typically one by ten thousand, the size of an atom), and the electrons revolve around it., Rutherford nuclear model has two main difficulties in explaining the, structure of atom: (a) It predicts that atoms are unstable because the, accelerated electrons revolving around the nucleus must spiral into, the nucleus. This contradicts the stability of matter. (b) It cannot, explain the characteristic line spectra of atoms of different elements., Atoms of each element are stable and emit characteristic spectrum., The spectrum consists of a set of isolated parallel lines termed as line, spectrum. It provides useful information about the atomic structure., The atomic hydrogen emits a line spectrum consisting of various series., The frequency of any line in a series can be expressed as a difference, of two terms;, Rc ¨, , Balmer series: Q, , Rc ¨, , 1·, § 1, ; n = 3, 4, 5,..., , © 22 n 2 ¸¹, 1·, § 1, ; n = 4, 5, 6,..., , © 32 n 2 ¸¹, , Paschen series: Q, , Rc ¨, , Brackett series: Q, , Rc ¨, , 1 ·, § 1, ; n = 5, 6, 7,..., , © 4 2 n 2 ¸¹, , § 1 1·, ; n = 6, 7, 8,..., , © 52 n 2 ¸¹, To explain the line spectra emitted by atoms, as well as the stability, of atoms, Niel’s Bohr proposed a model for hydrogenic (single elctron), atoms. He introduced three postulates and laid the foundations of, quantum mechanics:, (a) In a hydrogen atom, an electron revolves in certain stable orbits, (called stationary orbits) without the emission of radiant energy., (b) The stationary orbits are those for which the angular momentum, is some integral multiple of h/2S. (Bohr’s quantisation condition.), That is L = nh/2S, where n is an integer called a quantum number., (c) The third postulate states that an electron might make a transition, from one of its specified non-radiating orbits to another of lower, energy. When it does so, a photon is emitted having energy equal, to the energy difference between the initial and final states. The, frequency (Q of the emitted photon is then given by, hQ = Ei – Ef, An atom absorbs radiation of the same frequency the atom emits,, in which case the electron is transferred to an orbit with a higher, value of n., Ei + hQ = Ef, Pfund series: Q, , 7., , §1 1·, ; n = 2, 3, 4,..., , © 12 n 2 ¸¹, , Lyman series: Q, , Rc ¨, , 433

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Physics, 8., , As a result of the quantisation condition of angular momentum, the, electron orbits the nucleus at only specific radii. For a hydrogen atom, it is given by, 2, , rn, , § n 2 · § h · 4 SH 0, ¨© m ¸¹ ¨© 2S ¸¹ e 2, , The total energy is also quantised:, , En, , , , me 4, 8n 2 H 02h 2, , = –13.6 eV/n2, The n = 1 state is called ground state. In hydrogen atom the ground, state energy is –13.6 eV. Higher values of n correspond to excited, states (n > 1). Atoms are excited to these higher states by collisions, with other atoms or electrons or by absorption of a photon of right, frequency., 9. de Broglie’s hypothesis that electrons have a wavelength O h/mv gave, an explanation for Bohr’s quantised orbits by bringing in the waveparticle duality. The orbits correspond to circular standing waves in, which the circumference of the orbit equals a whole number of, wavelengths., 10. Bohr’s model is applicable only to hydrogenic (single electron) atoms., It cannot be extended to even two electron atoms such as helium., This model is also unable to explain for the relative intensities of the, frequencies emitted even by hydrogenic atoms., , POINTS TO PONDER, 1., , 2., , 3., , 4., , 434, , 5., , Both the Thomson’s as well as the Rutherford’s models constitute an, unstable system. Thomson’s model is unstable electrostatically, while, Rutherford’s model is unstable because of electromagnetic radiation, of orbiting electrons., What made Bohr quantise angular momentum (second postulate) and, not some other quantity? Note, h has dimensions of angular, momentum, and for circular orbits, angular momentum is a very, relevant quantity. The second postulate is then so natural!, The orbital picture in Bohr’s model of the hydrogen atom was, inconsistent with the uncertainty principle. It was replaced by modern, quantum mechanics in which Bohr’s orbits are regions where the, electron may be found with large probability., Unlike the situation in the solar system, where planet-planet, gravitational forces are very small as compared to the gravitational, force of the sun on each planet (because the mass of the sun is so, much greater than the mass of any of the planets), the electron-electron, electric force interaction is comparable in magnitude to the electronnucleus electrical force, because the charges and distances are of the, same order of magnitude. This is the reason why the Bohr’s model, with its planet-like electron is not applicable to many electron atoms., Bohr laid the foundation of the quantum theory by postulating specific, orbits in which electrons do not radiate. Bohr’s model include only

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Atoms, , 6., , 7., , one quantum number n. The new theory called quantum mechanics, supportes Bohr’s postulate. However in quantum mechanics (more, generally accepted), a given energy level may not correspond to just, one quantum state. For example, a state is characterised by four, quantum numbers (n, l, m, and s), but for a pure Coulomb potential, (as in hydrogen atom) the energy depends only on n., In Bohr model, contrary to ordinary classical expectation, the, frequency of revolution of an electron in its orbit is not connected to, the frequency of spectral line. The later is the difference between two, orbital energies divided by h. For transitions between large quantum, numbers (n to n – 1, n very large), however, the two coincide as expected., Bohr’s semiclassical model based on some aspects of classical physics, and some aspects of modern physics also does not provide a true, picture of the simplest hydrogenic atoms. The true picture is quantum, mechanical affair which differs from Bohr model in a number of, fundamental ways. But then if the Bohr model is not strictly correct,, why do we bother about it? The reasons which make Bohr’s model, still useful are:, (i) The model is based on just three postulates but accounts for almost, all the general features of the hydrogen spectrum., (ii) The model incorporates many of the concepts we have learnt in, classical physics., (iii) The model demonstrates how a theoretical physicist occasionally, must quite literally ignore certain problems of approach in hopes, of being able to make some predictions. If the predictions of the, theory or model agree with experiment, a theoretician then must, somehow hope to explain away or rationalise the problems that, were ignored along the way., , EXERCISES, 12.1, , 12.2, , Choose the correct alternative from the clues given at the end of, the each statement:, (a) The size of the atom in Thomson’s model is .......... the atomic, size in Rutherford’s model. (much greater than/no different, from/much less than.), (b) In the ground state of .......... electrons are in stable equilibrium,, while in .......... electrons always experience a net force., (Thomson’s model/ Rutherford’s model.), (c) A classical atom based on .......... is doomed to collapse., (Thomson’s model/ Rutherford’s model.), (d) An atom has a nearly continuous mass distribution in a .........., but has a highly non-uniform mass distribution in .........., (Thomson’s model/ Rutherford’s model.), (e) The positively charged part of the atom possesses most of the, mass in .......... (Rutherford’s model/both the models.), Suppose you are given a chance to repeat the alpha-particle, scattering experiment using a thin sheet of solid hydrogen in place, of the gold foil. (Hydrogen is a solid at temperatures below 14 K.), What results do you expect?, , 435

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Physics, What is the shortest wavelength present in the Paschen series of, spectral lines?, 12.4 A difference of 2.3 eV separates two energy levels in an atom. What, is the frequency of radiation emitted when the atom make a, transition from the upper level to the lower level?, 12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the, kinetic and potential energies of the electron in this state?, 12.6 A hydrogen atom initially in the ground level absorbs a photon,, which excites it to the n = 4 level. Determine the wavelength and, frequency of photon., 12.7 (a) Using the Bohr’s model calculate the speed of the electron in a, hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital, period in each of these levels., 12.8 The radius of the innermost electron orbit of a hydrogen atom is, 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?, 12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at, room temperature. What series of wavelengths will be emitted?, 12.10 In accordance with the Bohr’s model, find the quantum number, that characterises the earth’s revolution around the sun in an orbit, of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth, = 6.0 × 1024 kg.), 12.3, , ADDITIONAL EXERCISES, , 436, , 12.11 Answer the following questions, which help you understand the, difference between Thomson’s model and Rutherford’s model better., (a) Is the average angle of deflection of D-particles by a thin gold foil, predicted by Thomson’s model much less, about the same, or, much greater than that predicted by Rutherford’s model?, (b) Is the probability of backward scattering (i.e., scattering of, D-particles at angles greater than 90°) predicted by Thomson’s, model much less, about the same, or much greater than that, predicted by Rutherford’s model?, (c) Keeping other factors fixed, it is found experimentally that for, small thickness t, the number of D -particles scattered at, moderate angles is proportional to t. What clue does this linear, dependence on t provide?, (d) In which model is it completely wrong to ignore multiple, scattering for the calculation of average angle of scattering of, D-particles by a thin foil?, 12.12 The gravitational attraction between electron and proton in a, hydrogen atom is weaker than the coulomb attraction by a factor of, about 10–40. An alternative way of looking at this fact is to estimate, the radius of the first Bohr orbit of a hydrogen atom if the electron, and proton were bound by gravitational attraction. You will find the, answer interesting., 12.13 Obtain an expression for the frequency of radiation emitted when a, hydrogen atom de-excites from level n to level (n–1). For large n,, show that this frequency equals the classical frequency of revolution, of the electron in the orbit.

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Atoms, 12.14 Classically, an electron can be in any orbit around the nucleus of, an atom. Then what determines the typical atomic size? Why is an, atom not, say, thousand times bigger than its typical size? The, question had greatly puzzled Bohr before he arrived at his famous, model of the atom that you have learnt in the text. To simulate what, he might well have done before his discovery, let us play as follows, with the basic constants of nature and see if we can get a quantity, with the dimensions of length that is roughly equal to the known, size of an atom (~ 10–10m)., (a) Construct a quantity with the dimensions of length from the, fundamental constants e, me, and c. Determine its numerical, value., (b) You will find that the length obtained in (a) is many orders of, magnitude smaller than the atomic dimensions. Further, it, involves c. But energies of atoms are mostly in non-relativistic, domain where c is not expected to play any role. This is what, may have suggested Bohr to discard c and look for ‘something, else’ to get the right atomic size. Now, the Planck’s constant h, had already made its appearance elsewhere. Bohr’s great insight, lay in recognising that h, me, and e will yield the right atomic, size. Construct a quantity with the dimension of length from h,, me, and e and confirm that its numerical value has indeed the, correct order of magnitude., 12.15 The total energy of an electron in the first excited state of the, hydrogen atom is about –3.4 eV., (a) What is the kinetic energy of the electron in this state?, (b) What is the potential energy of the electron in this state?, (c) Which of the answers above would change if the choice of the, zero of potential energy is changed?, 12.16 If Bohr’s quantisation postulate (angular momentum = nh/2S) is a, basic law of nature, it should be equally valid for the case of planetary, motion also. Why then do we never speak of quantisation of orbits, of planets around the sun?, 12.17 Obtain the first Bohr’s radius and the ground state energy of a, muonic hydrogen atom [i.e., an atom in which a negatively charged, muon (P–) of mass about 207me orbits around a proton]., , 437

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Chapter Thirteen, , NUCLEI, , 13.1 INTRODUCTION, In the previous chapter, we have learnt that in every atom, the positive, charge and mass are densely concentrated at the centre of the atom, forming its nucleus. The overall dimensions of a nucleus are much smaller, than those of an atom. Experiments on scattering of D-particles, demonstrated that the radius of a nucleus was smaller than the radius, of an atom by a factor of about 104. This means the volume of a nucleus, is about 10–12 times the volume of the atom. In other words, an atom is, almost empty. If an atom is enlarged to the size of a classroom, the nucleus, would be of the size of pinhead. Nevertheless, the nucleus contains most, (more than 99.9%) of the mass of an atom., Does the nucleus have a structure, just as the atom does? If so, what, are the constituents of the nucleus? How are these held together? In this, chapter, we shall look for answers to such questions. We shall discuss, various properties of nuclei such as their size, mass and stability, and, also associated nuclear phenomena such as radioactivity, fission and fusion., , 13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS, The mass of an atom is very small, compared to a kilogram; for example,, the mass of a carbon atom, 12C, is 1.992647 × 10–26 kg. Kilogram is not, a very convenient unit to measure such small quantities. Therefore, a

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Nuclei, different mass unit is used for expressing atomic masses. This unit is the, atomic mass unit (u), defined as 1/12th of the mass of the carbon (12C), atom. According to this definition, 1u =, , mass of one, 12, , 12, , C atom, , 1.992647 u 10 26 kg, 12, (13.1), 1.660539 u 1027 kg, The atomic masses of various elements expressed in atomic mass, unit (u) are close to being integral multiples of the mass of a hydrogen, atom. There are, however, many striking exceptions to this rule. For, example, the atomic mass of chlorine atom is 35.46 u., Accurate measurement of atomic masses is carried out with a mass, spectrometer, The measurement of atomic masses reveals the existence, of different types of atoms of the same element, which exhibit the same, chemical properties, but differ in mass. Such atomic species of the same, element differing in mass are called isotopes. (In Greek, isotope means, the same place, i.e. they occur in the same place in the periodic table of, elements.) It was found that practically every element consists of a mixture, of several isotopes. The relative abundance of different isotopes differs, from element to element. Chlorine, for example, has two isotopes having, masses 34.98 u and 36.98 u, which are nearly integral multiples of the, mass of a hydrogen atom. The relative abundances of these isotopes are, 75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine, atom is obtained by the weighted average of the masses of the two, isotopes, which works out to be, 75.4 u 34.98  24.6 u 36.98, 100, = 35.47 u, which agrees with the atomic mass of chlorine., Even the lightest element, hydrogen has three isotopes having masses, 1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of, hydrogen, which has a relative abundance of 99.985%, is called the, proton. The mass of a proton is, , =, , mp, , 1.00727 u, , 1.67262 u 10 27 kg, , (13.2), , This is equal to the mass of the hydrogen atom (= 1.00783u), minus, the mass of a single electron (me = 0.00055 u). The other two isotopes of, hydrogen are called deuterium and tritium. Tritium nuclei, being, unstable, do not occur naturally and are produced artificially in, laboratories., The positive charge in the nucleus is that of the protons. A proton, carries one unit of fundamental charge and is stable. It was earlier thought, that the nucleus may contain electrons, but this was ruled out later using, arguments based on quantum theory. All the electrons of an atom are, outside the nucleus. We know that the number of these electrons outside, the nucleus of the atom is Z, the atomic number. The total charge of the, , 439

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Physics, atomic electrons is thus (–Ze), and since the atom is neutral, the charge, of the nucleus is (+Ze). The number of protons in the nucleus of the atom, is, therefore, exactly Z, the atomic number., Discovery of Neutron, Since the nuclei of deuterium and tritium are isotopes of hydrogen, they, must contain only one proton each. But the masses of the nuclei of, hydrogen, deuterium and tritium are in the ratio of 1:2:3. Therefore, the, nuclei of deuterium and tritium must contain, in addition to a proton,, some neutral matter. The amount of neutral matter present in the nuclei, of these isotopes, expressed in units of mass of a proton, is approximately, equal to one and two, respectively. This fact indicates that the nuclei of, atoms contain, in addition to protons, neutral matter in multiples of a, basic unit. This hypothesis was verified in 1932 by James Chadwick, who observed emission of neutral radiation when beryllium nuclei were, bombarded with alpha-particles. (D-particles are helium nuclei, to be, discussed in a later section). It was found that this neutral radiation, could knock out protons from light nuclei such as those of helium, carbon, and nitrogen. The only neutral radiation known at that time was photons, (electromagnetic radiation). Application of the principles of conservation, of energy and momentum showed that if the neutral radiation consisted, of photons, the energy of photons would have to be much higher than is, available from the bombardment of beryllium nuclei with D-particles., The clue to this puzzle, which Chadwick satisfactorily solved, was to, assume that the neutral radiation consists of a new type of neutral, particles called neutrons. From conservation of energy and momentum,, he was able to determine the mass of new particle ‘as very nearly the, same as mass of proton’., The mass of a neutron is now known to a high degree of accuracy. It is, (13.3), m n = 1.00866 u = 1.6749×10–27 kg, Chadwick was awarded the 1935 Nobel Prize in Physics for his, discovery of the neutron., A free neutron, unlike a free proton, is unstable. It decays into a, proton, an electron and a antineutrino (another elementary particle), and, has a mean life of about 1000s. It is, however, stable inside the nucleus., The composition of a nucleus can now be described using the following, terms and symbols:, Z - atomic number = number of protons, , [13.4(a)], , N - neutron number = number of neutrons, , [13.4(b)], , A - mass number = Z + N, , 440, , = total number of protons and neutrons [13.4(c)], One also uses the term nucleon for a proton or a neutron. Thus the, number of nucleons in an atom is its mass number A., Nuclear species or nuclides are shown by the notation ZA X where X is, the chemical symbol of the species. For example, the nucleus of gold is, . It contains 197 nucleons, of which 79 are protons, denoted by 197, 79 Au, and the rest118 are neutrons.

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Nuclei, The composition of isotopes of an element can now be readily, explained. The nuclei of isotopes of a given element contain the same, number of protons, but differ from each other in their number of neutrons., Deuterium, 12 H , which is an isotope of hydrogen, contains one proton, and one neutron. Its other isotope tritium, 13 H , contains one proton and, two neutrons. The element gold has 32 isotopes, ranging from A =173 to, A = 204. We have already mentioned that chemical properties of elements, depend on their electronic structure. As the atoms of isotopes have, identical electronic structure they have identical chemical behaviour and, are placed in the same location in the periodic table., All nuclides with same mass number A are called isobars. For, example, the nuclides 13 H and 32 He are isobars. Nuclides with same, neutron number N but different atomic number Z, for example 198, 80 Hg, and 197, ,, are, called, isotones., Au, 79, , 13.3 SIZE, , OF THE, , NUCLEUS, , As we have seen in Chapter 12, Rutherford was the pioneer who, postulated and established the existence of the atomic nucleus. At, Rutherford’s suggestion, Geiger and Marsden performed their classic, experiment: on the scattering of D-particles from thin gold foils. Their, experiments revealed that the distance of closest approach to a gold, nucleus of an D-particle of kinetic energy 5.5 MeV is about 4.0 × 10–14 m., The scattering of D-particle by the gold sheet could be understood by, Rutherford by assuming that the coulomb repulsive force was solely, responsible for scattering. Since the positive charge is confined to the, nucleus, the actual size of the nucleus has to be less than 4.0 × 10–14 m., If we use D-particles of higher energies than 5.5 MeV, the distance of, closest approach to the gold nucleus will be smaller and at some point, the scattering will begin to be affected by the short range nuclear forces,, and differ from Rutherford’s calculations. Rutherford’s calculations are, based on pure coulomb repulsion between the positive charges of the Dparticle and the gold nucleus. From the distance at which deviations set, in, nuclear sizes can be inferred., By performing scattering experiments in which fast electrons, instead, of D-particles, are projectiles that bombard targets made up of various, elements, the sizes of nuclei of various elements have been accurately, measured., It has been found that a nucleus of mass number A has a radius, R = R 0 A1/3, (13.5), , where R 0 = 1.2 × 10–15 m. This means the volume of the nucleus, which, is proportional to R 3 is proportional to A. Thus the density of nucleus is, a constant, independent of A, for all nuclei. Different nuclei are likes, drop of liquid of constant density. The density of nuclear matter is, approximately 2.3 × 1017 kg m–3. This density is very large compared to, ordinary matter, say water, which is 103 kg m–3. This is understandable,, as we have already seen that most of the atom is empty. Ordinary matter, consisting of atoms has a large amount of empty space., , 441

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Physics, Example 13.1 Given the mass of iron nucleus as 55.85u and A=56,, find the nuclear density?, Solution, mFe = 55.85, u = 9.27 × 10–26 kg, , EXAMPLE 13.1, , 26, , 9.27 u 10, 1, mass, u, =, (4S /3)(1.2 u 10 15 )3 56, volume, = 2.29 × 1017 kg m–3, The density of matter in neutron stars (an astrophysical object) is, comparable to this density. This shows that matter in these objects, has been compressed to such an extent that they resemble a big, nucleus., , Nuclear density =, , 13.4.1 Mass – Energy, Einstein showed from his theory of special relativity that it is necessary, to treat mass as another form of energy. Before the advent of this theory, of special relativity it was presumed that mass and energy were conserved, separately in a reaction. However, Einstein showed that mass is another, form of energy and one can convert mass-energy into other forms of, energy, say kinetic energy and vice-versa., Einstein gave the famous mass-energy equivalence relation, (13.6), E = mc 2, Here the energy equivalent of mass m is related by the above equation, and c is the velocity of light in vacuum and is approximately equal to, 3×108 m s–1., , EXAMPLE 13.2, , Example 13.2 Calculate the energy equivalent of 1 g of substance., Solution, Energy, E = 10–3 × ( 3 × 108)2 J, E = 10–3 × 9 × 1016 = 9 × 1013 J, Thus, if one gram of matter is converted to energy, there is a release, of enormous amount of energy., , Experimental verification of the Einstein’s mass-energy relation has, been achieved in the study of nuclear reactions amongst nucleons, nuclei,, electrons and other more recently discovered particles. In a reaction the, conservation law of energy states that the initial energy and the final, energy are equal provided the energy associated with mass is also, included. This concept is important in understanding nuclear masses, and the interaction of nuclei with one another. They form the subject, matter of the next few sections., , 13.4.2 Nuclear binding energy, 442, , In Section 13.2 we have seen that the nucleus is made up of neutrons, and protons. Therefore it may be expected that the mass of the nucleus, is equal to the total mass of its individual protons and neutrons. However,

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Nuclei, the nuclear mass M is found to be always less than this. For example, let, us consider 168 O ; a nucleus which has 8 neutrons and 8 protons. We, have, Mass of 8 neutrons = 8 × 1.00866 u, Mass of 8 protons = 8 × 1.00727 u, Mass of 8 electrons = 8 × 0.00055 u, Therefore the expected mass of 168 O nucleus, = 8 × 2.01593 u = 16.12744 u., The atomic mass of 168 O found from mass spectroscopy experiments, is seen to be 15.99493 u. Substracting the mass of 8 electrons (8 × 0.00055 u), from this, we get the experimental mass of 168 O nucleus to be 15.99053 u., Thus, we find that the mass of the 168 O nucleus is less than the total, mass of its constituents by 0.13691u. The difference in mass of a nucleus, and its constituents, 'M, is called the mass defect, and is given by, 'M [ Zm p  ( A  Z ) m n ]  M, , (13.7), , What is the meaning of the mass defect? It is here that Einstein’s, equivalence of mass and energy plays a role. Since the mass of the oxygen, nucleus is less that the sum of the masses of its constituents (8 protons, and 8 neutrons, in the unbound state), the equivalent energy of the oxygen, nucleus is less than that of the sum of the equivalent energies of its, constituents. If one wants to break the oxygen nucleus into 8 protons, and 8 neutrons, this extra energy 'M c2, has to supplied. This energy, required Eb is related to the mass defect by, Eb = ' M c 2, , (13.8), , Example 13.3 Find the energy equivalent of one atomic mass unit,, first in Joules and then in MeV. Using this, express the mass defect, of 168 O in MeV/c 2., Solution, 1u = 1.6605 × 10–27 kg, To convert it into energy units, we multiply it by c 2 and find that, energy equivalent = 1.6605 × 10–27 × (2.9979 × 108)2 kg m2/s2, = 1.4924 × 10–10 J, =, , 1.4924 u 10 10, eV, 1.602 u 10 19, , The energy needed to separate, , 16, 8, , O into its constituents is thus, , 2, , 127.5 MeV/c ., , If a certain number of neutrons and protons are brought together to, form a nucleus of a certain charge and mass, an energy Eb will be released, , EXAMPLE 13.3, , = 0.9315 × 109 eV, = 931.5 MeV, or, 1u = 931.5 MeV/c 2, For 168 O , 'M = 0.13691 u = 0.13691×931.5 MeV/c 2, = 127.5 MeV/c 2, , 443

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Physics, in the process. The energy Eb is called the binding energy of the nucleus., If we separate a nucleus into its nucleons, we would have to supply a, total energy equal to Eb, to those particles. Although we cannot tear, apart a nucleus in this way, the nuclear binding energy is still a convenient, measure of how well a nucleus is held together. A more useful measure, of the binding between the constituents of the nucleus is the binding, energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a, nucleus to the number of the nucleons, A, in that nucleus:, Ebn = Eb / A, (13.9), We can think of binding energy per nucleon as the average energy, per nucleon needed to separate a nucleus into its individual nucleons., Figure 13.1 is a plot of the, binding energy per nucleon Ebn, versus the mass number A for a, large number of nuclei. We notice, the following main features of, the plot:, (i) the binding energy per, nucleon, Ebn, is practically, constant, i.e. practically, independent of the atomic, number for nuclei of middle, mass number ( 30 < A < 170)., The curve has a maximum of, about 8.75 MeV for A = 56, and has a value of 7.6 MeV, FIGURE 13.1 The binding energy per nucleon, for A = 238., as a function of mass number., (ii) Ebn is lower for both light, nuclei (A<30) and heavy, nuclei (A>170)., We can draw some conclusions from these two observations:, (i) The force is attractive and sufficiently strong to produce a binding, energy of a few MeV per nucleon., (ii) The constancy of the binding energy in the range 30 < A < 170 is a, consequence of the fact that the nuclear force is short-ranged. Consider, a particular nucleon inside a sufficiently large nucleus. It will be under, the influence of only some of its neighbours, which come within the, range of the nuclear force. If any other nucleon is at a distance more, than the range of the nuclear force from the particular nucleon it will, have no influence on the binding energy of the nucleon under, consideration. If a nucleon can have a maximum of p neighbours, within the range of nuclear force, its binding energy would be, proportional to p. Let the binding energy of the nucleus be pk, where, k is a constant having the dimensions of energy. If we increase A by, adding nucleons they will not change the binding energy of a nucleon, inside. Since most of the nucleons in a large nucleus reside inside it, and not on the surface, the change in binding energy per nucleon, would be small. The binding energy per nucleon is a constant and is, 444, approximately equal to pk. The property that a given nucleon

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Nuclei, influences only nucleons close to it is also referred to as saturation, property of the nuclear force., (iii) A very heavy nucleus, say A = 240, has lower binding energy per, nucleon compared to that of a nucleus with A = 120. Thus if a, nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more, tightly bound. This implies energy would be released in the process., It has very important implications for energy production through, fission, to be discussed later in Section 13.7.1., (iv) Consider two very light nuclei (A d 10) joining to form a heavier, nucleus. The binding energy per nucleon of the fused heavier nuclei, is more than the binding energy per nucleon of the lighter nuclei., This means that the final system is more tightly bound than the initial, system. Again energy would be released in such a process of, fusion. This is the energy source of sun, to be discussed later in, Section 13.7.3., , 13.5 NUCLEAR FORCE, The force that determines the motion of atomic electrons is the familiar, Coulomb force. In Section 13.4, we have seen that for average mass, nuclei the binding energy per nucleon is approximately 8 MeV, which is, much larger than the binding energy in atoms. Therefore, to bind a, nucleus together there must be a strong attractive force of a totally, different kind. It must be strong enough to overcome the repulsion, between the (positively charged) protons and to bind both protons and, neutrons into the tiny nuclear volume. We have already seen, that the constancy of binding energy per nucleon can be, understood in terms of its short-range. Many features of the, nuclear binding force are summarised below. These are, obtained from a variety of experiments carried out during 1930, to 1950., (i) The nuclear force is much stronger than the Coulomb force, acting between charges or the gravitational forces between, masses. The nuclear binding force has to dominate over, the Coulomb repulsive force between protons inside the, nucleus. This happens only because the nuclear force is, much stronger than the coulomb force. The gravitational, force is much weaker than even Coulomb force., FIGURE 13.2 Potential energy, (ii) The nuclear force between two nucleons falls rapidly to, of a pair of nucleons as a, zero as their distance is more than a few femtometres. This, function of their separation., leads to saturation of forces in a medium or a large-sized, For a separation greater, nucleus, which is the reason for the constancy of the, than r0, the force is attractive, binding energy per nucleon., and for separations less, than r0, the force is, A rough plot of the potential energy between two nucleons, strongly repulsive., as a function of distance is shown in the Fig. 13.2. The, potential energy is a minimum at a distance r0 of about, 0.8 fm. This means that the force is attractive for distances larger, than 0.8 fm and repulsive if they are separated by distances less, 445, than 0.8 fm.

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Physics, (iii) The nuclear force between neutron-neutron, proton-neutron and, proton-proton is approximately the same. The nuclear force does not, depend on the electric charge., Unlike Coulomb’s law or the Newton’s law of gravitation there is no, simple mathematical form of the nuclear force., , 13.6 RADIOACTIVITY, A. H. Becquerel discovered radioactivity in 1896 purely by accident. While, studying the fluorescence and phosphorescence of compounds irradiated, with visible light, Becquerel observed an interesting phenomenon. After, illuminating some pieces of uranium-potassium sulphate with visible, light, he wrapped them in black paper and separated the package from a, photographic plate by a piece of silver. When, after several hours of, exposure, the photographic plate was developed, it showed blackening, due to something that must have been emitted by the compound and, was able to penetrate both black paper and the silver., Experiments performed subsequently showed that radioactivity was, a nuclear phenomenon in which an unstable nucleus undergoes a decay., This is referred to as radioactive decay. Three types of radioactive decay, occur in nature :, (i) D-decay in which a helium nucleus 42 He is emitted;, (ii) E-decay in which electrons or positrons (particles with the same mass, as electrons, but with a charge exactly opposite to that of electron), are emitted;, (iii) J-decay in which high energy (hundreds of keV or more) photons are, emitted., Each of these decay will be considered in subsequent sub-sections., , 13.6.1 Law of radioactive decay, In any radioactive sample, which undergoes D, E or J-decay, it is found, that the number of nuclei undergoing the decay per unit time is, proportional to the total number of nuclei in the sample. If N is the, number of nuclei in the sample and 'N undergo decay in time 't then, 'N, vN, 't, , or, 'N/'t = ON,, (13.10), where Ois called the radioactive decay constant or disintegration constant., The change in the number of nuclei in the sample* is dN = – 'N in, time 't. Thus the rate of change of N is (in the limit 't o 0), dN, dt, , 446, , – ON, , * 'N is the number of nuclei that decay, and hence is always positive. dN is the, change in N, which may have either sign. Here it is negative, because out of, original N nuclei, 'N have decayed, leaving (N–'N ) nuclei.

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Nuclei, dN, – O dt, N, Now, integrating both sides of the above equation,we get,, N, t, dN, , O, ³ N, ³ dt, (13.11), N0, t0, or, ln N ln N0 = O(t – t0), (13.12), Here N0 is the number of radioactive nuclei in the sample at some, arbitrary time t0 and N is the number of radioactive nuclei at any, subsequent time t. Setting t0 = 0 and rearranging Eq. (13.12) gives us, , or,, , ln, , N, N0, ,  Ot, , (13.13), , which gives, , (13.14), N(t) = N0 e O t, Note, for example, the light bulbs follow no such exponential decay law., If we test 1000 bulbs for their life (time span before they burn out or, fuse), we expect that they will ‘decay’ (that is, burn out) at more or less, the same time. The decay of radionuclides follows quite a different law,, the law of radioactive decay represented by Eq. (13.14)., We are quite often interested in the decay rate R (= –dN/dt) than in N, itself. It gives us the number of nuclei decaying per unit time. For example,, suppose we have some amount of a radioactive substance. We need not, know the number of nuclei present in it. But we may measure the number, of emissions of D, E or J particles in a given time interval, say 10 seconds, or 20 seconds. Suppose we make measurements for a time interval dt, and get a decay count in our instruments equal to 'N (=–dN). Then the, decay rate R is defined as, R=–, , dN, dt, , Differentiating Eq. (13.14), we find, dN, R =–, = ON0 e Ot, dt, or, R = R0 e –Ot, , (13.15), , which is an alternative form of the law of radioactive, decay [Eq. (13.14)]. Here R0 is the radioactive decay, rate at time t = 0, and R is the rate at any subsequent, time t. We can now write Eq. (13.10) in terms of the, decay rate R of the sample as, R = ON, , (13.16), , FIGURE 13.3 Exponential decay of a, radioactive species. After a lapse of, T1/2 , population of the given species, drops by a factor of 2., , where R and the number of radioactive nuclei that have, not yet undergone decay must be evaluated at the same, instant., The total decay rate R of a sample of one or more radionuclides, is called the activity of that sample. The SI unit for activity is, becquerel, named after the discoverer of radioactivity, Henry Becquerel., It is defined as, , 447

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MARIE SKLODOWSKA CURIE (1867-1934), , Physics, 1 becquerel = 1Bq = 1 decay per second, An older unit, the curie, is still in common use:, 1 curie = 1 Ci = 3.7 × 1010 Bq (decays per second), There are two common time measures of how long, any given type of radionuclide lasts. One measure is the, half-life T1/2 of a radionuclide, which is the time at which, both N and R have been reduced to one-half their initial, values. The other measure is the mean life W, which is the, time at which both N and R have been reduced to e–1of, their initial values., To relate T1/2 to the disintegration constant O, we put, R = (1/2) R0 and t = T1/2in Eq. (13.15) and solve for T1/2,, we find, T1/2 =, =, , Marie Sklodowska Curie, (1867-1934) Born in, Poland. She is recognised, both as a physicist and as, a chemist. The discovery of, radioactivity by Henri, Becquerel in 1896 inspired, Marie and her husband, Pierre Curie in their, researches and analyses, which led to the isolation of, radium and polonium, elements. She was the first, person to be awarded two, Nobel Prizes- for Physics in, 1903 and for Chemistry, in 1911., , ln 2, , O, 0.693, , (13.17), O, The average life or mean life, W can also be obtained, from Eq. (13.14). The number of nuclei which decay in, the time interval t to t +'t is R(t)'t = (ON0e–Ot't). Each of, them has lived for time t. Thus the total life of all these, nuclei would be t ON0e–Ot 't. It is clear that some nuclei, may live for a short time while others may live longer., Therefore to obtain the meanlife, we have to sum (or, integrate) this expression over all times from 0 to f , and, divide by the total number N0 of nuclei at t = 0. Thus,, f, , O N 0 ³ te – Ot dt, W, , 0, , f, , O ³ te – Ot dt, , N0, 0, One can show by performing this integral that, W = 1/O, We summarise these results with the following:, ln 2, , = W ln 2, (13.18), O, Those radioactive elements whose half-life is short compared to the, age of the universe (13.7 billion years) are not found in observable, quantities in nature today. They have, however, been seen in the, laboratory in nuclear reactions. Tritium and plutonium belong to this, category., , 448, , EXAMPLE 13.4, , T1/2 =, , undergoing D-decay is 4.5 × 109, Example 13.4 The half-life of 238, 92 U, years. What is the activity of 1g sample of 238, ?, 92 U, Solution, T1/2 = 4.5 × 109 y, = 4.5 × 109 y x 3.16 x 107 s/y, = 1.42 × 1017s

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Nuclei, One k mol of any isotope contains Avogadro’s number of atoms, and, so 1g of 238, contains, 92 U, , =, , EXAMPLE 13.4, , 1, kmol × 6.025 × 1026 atoms/kmol, 238 u 10 –3, = 25.3 × 1020 atoms., The decay rate R is, R = ON, , 0.693, 0.693 u 25.3 u 1020 1, N =, s, T1/ 2, 1.42 u 1017, , = 1.23 × 104 s–1, = 1.23 × 104 Bq, , Solution, By definition of half-life, half of the initial sample will remain, undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei, would have decayed. Hence, one fourth of the sample of the initial, pure tritium will remain undecayed., , EXAMPLE 13.5, , Example 13.5 Tritium has a half-life of 12.5 y undergoing beta decay., What fraction of a sample of pure tritium will remain undecayed, after 25 y., , 13.6.2 Alpha decay, When a nucleus undergoes alpha-decay, it transforms to a different, nucleus by emitting an alpha-particle (a helium nucleus, 42 He ). For, undergoes alpha-decay, it transforms to 234, example, when 238, 92 U, 90 Th, 238, 92, , Uo, , 234, 90, , Th + 24 He, , (13.19), , In this process, it is observed that since 42 He contains two protons, and two neutrons, the mass number and the atomic number of the, daughter nucleus decreases by four and two, respectively. Thus, the, transformation of a nucleus ZA X into a nucleus AZ42 Y can be expressed as, A, Z, , X o, , A 4, Z 2, , Y + 24 He, , A, Z, , (13.20), A 4, Z 2, , where X is the parent nucleus and, Y is the daughter nucleus., The alpha-decay of 238, can, occur, spontaneously, (without an external, 92 U, and, source of energy) because the total mass of the decay products 234, 90 Th, 4, 238, is, less, than, the, mass, of, the, original, ., Thus,, the, total, mass, energy, He, U, 2, 92, of the decay products is less than the mass energy of the original nuclide., The difference between the initial mass energy and the final mass energy of, the decay products is called the Q value of the process or the disintegration, energy. Thus, the Q value of an alpha decay can be expressed as, Q = (mX – mY – mHe) c2, , (13.21), A 4, Z 2, , This energy is shared by the daughter nucleus, Y and the alphaparticle, 42 He in the form of kinetic energy. Alpha-decay obeys the, radioactive law as given in Eqs. (13.14) and (13.15)., , 449

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Physics, Example 13.6 We are given the following atomic masses:, 238, 4, = 238.05079 u, = 4.00260 u, 92 U, 2 He, 234, 1, =, 234.04363, u, =, 1.00783 u, Th, H, 90, 1, 237, 91, , Pa = 237.05121 u, Here the symbol Pa is for the element protactinium (Z = 91)., (a) Calculate the energy released during the alpha decay of 238, ., 92 U, 238, (b) Show that 92 U can not spontaneously emit a proton., Solution, (a) The alpha decay of 238, is given by Eq. (13.20). The energy released, 92 U, in this process is given by, Q = (MU – MTh – MHe) c 2, Substituting the atomic masses as given in the data, we find, Q = (238.05079 – 234.04363 – 4.00260)u × c 2, = (0.00456 u) c 2, = (0.00456 u) (931.5 MeV/u), = 4.25 MeV., (b) If 238, spontaneously emits a proton, the decay process would be, 92 U, + 11 H, U o 237, 91 Pa, The Q for this process to happen is, = (MU – MPa – MH ) c 2, = (238.05079 – 237.05121 – 1.00783) u × c 2, = (– 0.00825 u) c 2, = – (0.00825 u)(931.5 MeV/u), = – 7.68 MeV, Thus, the Q of the process is negative and therefore it cannot proceed, spontaneously. We will have to supply an energy of 7.68 MeV to a, 238, nucleus to make it emit a proton., 92 U, , EXAMPLE 13.6, , 238, 92, , 13.6.3 Beta decay, A nucleus that decays spontaneously by emitting an electron or a positron, is said to undergo beta decay. Like alpha decay, this is a spontaneous, process, with a definite disintegration energy and half-life. Again like alpha, decay, beta decay is a statistical process governed by Eqs. (13.14) and, (13.15). In beta minus (E ) decay, an electron is emitted by the nucleus,, 32, as in the decay of 15, P, 32, (T1/2 = 14.3 d), (13.22), P o 16, S  e  Q, In beta plus (E ) decay, a positron is emitted by the nucleus, as in the, 32, 15, , decay of, , Na, , 22, (13.23), Na o 10, Ne  e   Q (T1/2 = 2.6 y), The symbols Q and Q represent antineutrino and neutrino,, respectively; both are neutral particles, with very little or no mass. These, particles are emitted from the nucleus along with the electron or positron, during the decay process. Neutrinos interact only very weakly with matter;, they can even penetrate the earth without being absorbed. It is for this, reason that their detection is extremely difficult and their presence went, unnoticed for long., 22, 11, , 450, , 22, 11

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Nuclei, In beta-minus decay, a neutron transforms into a proton within the, nucleus according to, n o p + e– + Q, , (13.24), , whereas in beta-plus decay, a proton transforms into neutron (inside the, nucleus) via, (13.25), p o n + e+ + Q, These processes show why the mass number A of a nuclide undergoing, beta decay does not change; one of its constituent nucleons simply changes, its character according to Eq. (13.24) or (13.25)., , 13.6.4 Gamma decay, There are energy levels in a nucleus, just like there are energy levels in, atoms. When a nucleus is in an excited state, it can make a transition to, a lower energy state by the emission of electromagnetic radiation. As the, energy differences between levels in a nucleus are, of the order of MeV, the photons emitted by, the nuclei have MeV energies and are called, gamma rays., Most radionuclides after an alpha decay or a beta, decay leave the daughter nucleus in an excited state., The daughter nucleus reaches the ground state by, a single transition or sometimes by successive, transitions by emitting one or more gamma rays. A, well-known example of such a process is that of, 60, . By beta emission, the 60, nucleus, 27 Co, 27 Co, 60, transforms into 28 Ni nucleus in its excited state., FIGURE 13.4 Energy level diagram, The excited 60, nucleus so formed then de-excites, 28 Ni, showing the emission of J rays by a, to its ground state by successive emission of, 60, nucleus subsequent to beta decay., Co, 27, 1.17 MeV and 1.33 MeV gamma rays. This, process is depicted in Fig.13.4 through an energy, level diagram., , 13.7 NUCLEAR ENERGY, The curve of binding energy per nucleon Ebn, given in Fig. 13.1, has, a long flat middle region between A = 30 and A = 170. In this region, the binding energy per nucleon is nearly constant (8.0 MeV). For, the lighter nuclei region, A < 30, and for the heavier nuclei region,, A > 170, the binding energy per nucleon is less than 8.0 MeV, as we, have noted earlier. This feature of the binding energy curve means, that nuclei in the middle region 30 d A d 170 are more tightly, bound than nuclei with A < 30 and A > 170 . Energy then can be, released if less tightly bound nuclei are transmuted into more tightly, bound nuclei. Two such processes, which we have already referred to,, are fission and fusion., In conventional energy sources like coal or petroleum, energy is, released through chemical reactions. The energies involved are of the, order of electron volts per atom. As we have seen, energies involved in, , 451

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Physics, nuclear processes are million times larger (in MeVs per nucleon). This, means that for the same quantity of matter, nuclear sources will give a, million times larger energy than conventional sources. One kilogram of, coal on burning gives 107 J of energy, whereas 1 kg of uranium, which, undergoes fission, will generate on fission 1014 J of energy., , 13.7.1 Fission, Soon after the discovery of neutron by Chadwick, Enrico Fermi found, that when neutrons bombard various elements, new radioactive elements, are produced. However, when a neutron was bombared on a uranium, target, the uranium nucleus broke into two nearly equal fragments, releasing great amount of energy. An example of such a reaction is, 1, 0, , 236, 144, 89, 1, (13.26), n  235, 92 U o 92 U o 56 Ba  36 Kr  3 0 n, Fission does not always produce barium and krypton. A different, pair can be produced, for example, 1, 0, , 236, 133, 99, 1, n  235, 92 U o 92 U o 51 Sb  41 Nb  4 0 n, Still another example is, , (13.27), , 1, 0, , 140, 94, 1, (13.28), n  235, 92 U o 54 Xe  38 Sr  2 0 n, The fragment nuclei produced in fission are highly neutron-rich and, unstable. They are radioactive and emit beta particles in succession until, each reaches a stable end product., The energy released (the Q value ) in the fission reaction of nuclei like, uranium is of the order of 200 MeV per fissioning nucleus. This is, estimated as follows:, Let us take a nucleus with A = 240 breaking into two fragments each, of A = 120. Then, Ebn for A = 240 nucleus is about 7.6 MeV,, , Ebn for the two A = 120 fragment nuclei is about 8.5 MeV., ? Gain in binding energy for nucleon is about 0.9 MeV., Hence the total gain in binding energy is 240×0.9 or 216 MeV, The disintegration energy in fission events first appears as the kinetic, energy of the fragments and neutrons. Eventually it is transferred to the, surrounding matter appearing as heat. The source of energy in nuclear, reactors, which produce electricity, is nuclear fission. The enormous, energy released in an atom bomb comes from uncontrolled nuclear fission., We discuss some details in the next section how a nuclear reactor, functions., , 13.7.2 Nuclear reactor, , 452, , When 235, undergoes a fission after bombarded by a neutron, it also, 92 U, releases an extra neutron. This extra neutron is then available for initiating, fission of another 235, nucleus. In fact, on an average, 2½ neutrons per, 92 U, fission of uranium nucleus are released. The fact that more neutrons are, produced in fission than are consumed raises the possibility of a chain, reaction with each neutron that is produced triggering another fission.

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Nuclei, INDIA’S ATOMIC ENERGY PROGRAMME, The atomic energy programme in India was launched around the time of independence, under the leadership of Homi J. Bhabha (1909-1966). An early historic achievement, was the design and construction of the first nuclear reactor in India (named Apsara), which went critical on August 4, 1956. It used enriched uranium as fuel and water as, moderator. Following this was another notable landmark: the construction of CIRUS, (Canada India Research U.S.) reactor in 1960. This 40 MW reactor used natural uranium, as fuel and heavy water as moderator. Apsara and CIRUS spurred research in a wide, range of areas of basic and applied nuclear science. An important milestone in the first, two decades of the programme was the indigenous design and construction of the, plutonium plant at Trombay, which ushered in the technology of fuel reprocessing, (separating useful fissile and fertile nuclear materials from the spent fuel of a reactor) in, India. Research reactors that have been subsequently commissioned include ZERLINA,, PURNIMA (I, II and III), DHRUVA and KAMINI. KAMINI is the country’s first large research, reactor that uses U-233 as fuel. As the name suggests, the primary objective of a research, reactor is not generation of power but to provide a facility for research on different aspects, of nuclear science and technology. Research reactors are also an excellent source for, production of a variety of radioactive isotopes that find application in diverse fields:, industry, medicine and agriculture., The main objectives of the Indian Atomic Energy programme are to provide safe and, reliable electric power for the country’s social and economic progress and to be selfreliant in all aspects of nuclear technology. Exploration of atomic minerals in India, undertaken since the early fifties has indicated that India has limited reserves of uranium,, but fairly abundant reserves of thorium. Accordingly, our country has adopted a threestage strategy of nuclear power generation. The first stage involves the use of natural, uranium as a fuel, with heavy water as moderator. The Plutonium-239 obtained from, reprocessing of the discharged fuel from the reactors then serves as a fuel for the second, stage — the fast breeder reactors. They are so called because they use fast neutrons for, sustaining the chain reaction (hence no moderator is needed) and, besides generating, power, also breed more fissile species (plutonium) than they consume. The third stage,, most significant in the long term, involves using fast breeder reactors to produce fissile, Uranium-233 from Thorium-232 and to build power reactors based on them., India is currently well into the second stage of the programme and considerable, work has also been done on the third — the thorium utilisation — stage. The country, has mastered the complex technologies of mineral exploration and mining, fuel, fabrication, heavy water production, reactor design, construction and operation, fuel, reprocessing, etc. Pressurised Heavy Water Reactors (PHWRs) built at different sites in, the country mark the accomplishment of the first stage of the programme. India is now, more than self-sufficient in heavy water production. Elaborate safety measures both in, the design and operation of reactors, as also adhering to stringent standards of, radiological protection are the hallmark of the Indian Atomic Energy Programme., , Enrico Fermi first suggested such a possibility in 1939. The chain reaction, can be either uncontrolled and rapid (as in a nuclear bomb) or controlled, and steady (as in a nuclear reactor). The first leads to destruction while, the latter can be harnessed to generate electric power., , 453

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http://www.npcil.nic.in/plantsInoperation.asp, , Nuclear power plants in India, , Physics, But soon it was discovered that neutrons liberated in fission of a, uranium nucleus were so energetic that they would escape instead of, triggering another fission reaction. Also, it turns out that slow neutrons, have a much higher intrinsic probability of inducing fission in 235, than, 92 U, fast neutrons., is 2 MeV., The average energy of a neutron produced in fission of 235, 92 U, These neutrons unless slowed down will escape from the reactor without, interacting with the uranium nuclei, unless a very large amount of, fissionable material is used for sustaining the chain reaction. What one, needs to do is to slow down the fast neutrons by elastic scattering with, light nuclei. In fact, Chadwick’s experiments showed that in an elastic, collision with hydrogen the neutron almost comes to rest and proton, carries away the energy. This is the same situation as when a marble hits, head-on an identical marble at rest. Therefore, in reactors, light nuclei, called moderators are provided along with the fissionable nuclei for slowing, down fast neutrons. The moderators commonly used are water, heavy, water (D2O) and graphite. The Apsara reactor at the Bhabha Atomic, Research Centre (BARC), Mumbai, uses water as moderator. The other, Indian reactors, which are used for power production, use heavy water, as moderator., Because of the use of moderator, it is possible that the ratio, K, of, number of fission produced by a given generation of neutrons to the, number of fission of the preceeding generation may be greater than one., This ratio is called the multiplication factor; it is the measure of the growth, rate of the neutrons in the reactor. For K = 1, the operation of the reactor, is said to be critical, which is what we wish it to be for steady power, operation. If K becomes greater than one, the reaction rate and the reactor, power increases exponentially. Unless the factor K is brought down very, close to unity, the reactor will become supercritical and can even explode., The explosion of the Chernobyl reactor in Ukraine in 1986 is a sad, reminder that accidents in a nuclear reactor can be catastrophic., The reaction rate is controlled through control-rods made out of, neutron-absorbing material such as cadmium. In addition to control rods,, reactors are provided with safety rods which, when required, can be, inserted into the reactor and K can be reduced rapidly to less than unity., isotope, which does not fission, on capturing a, The abundant 238, 92 U, neutron leads to the formation of plutonium. The series of reactions, involved is, 238, 92, 239, 93, , 454, , 239, –, U + n o 239, 92 U o 93 Np + e  Q, , –, (13.29), Np o 239, 94 Pu + e  Q, Plutonium is highly radioactive and can also undergo fission under, bombardment by slow neutrons., The broad outlines of a typical nuclear power plant based on, pressurised-water reactor are shown in Fig. 13.5. In such a reactor, water, is used both as the moderator and as the heat transfer medium. In the, primary-loop, water is circulated through the reactor vessel and transfers, energy at high temperature and pressure (at about 600 K and 150 atm), to the steam generator, which is part of the secondary-loop. In the steam, generator, evaporation provides high-pressure steam to operate the

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Nuclei, , A simplified online simulation of a nuclear reactor, , turbine that drives the electric generator. The low-pressure steam from, the turbine is cooled and condensed to water and forced back into the, steam generator., The energy released in nuclear reactions is a million times larger than, in chemical reactions. Therefore, the nuclear reactors need fuel a million, times less in mass than used in the chemical reactors of the same, on complete fission generates about, power capacity. A kilogram of 235, 92 U, 3 u 104 MW. However, in nuclear reactions highly radioactive elements, are continuously produced. Therefore, an unavoidable feature of reactor, operation is the accumulation of radioactive waste, including both fission, products and heavy transuranic elements such as plutonium and, americium., Historically energy has been produced by using chemical reactions,, such as burning coal, wood, gas and petroleum products. The, environmental pollution produced by these is creating serious problems, due to green house effect leading to global warming. The problem, encountered with the nuclear power station is that the spent fuel is highly, radioactive and extremely hazardous to all forms of life on earth. Elaborate, safety measures, both for reactor operation as well as handling and, reprocessing the spent fuel, are required. These safety measures are a, distinguishing feature of the Indian Atomic Energy programme. An, appropriate plan is being evolved to study the possibility of converting, radioactive waste into less active and short-lived material., , http://www.kernenergie.net/datf/en/interactive/reactor.php, , FIGURE 13.5 Simplified outlines of a nuclear power plant., , 13.7.3 Nuclear fusion – energy generation in stars, The binding energy curve shown in Fig.13.1 shows that energy can be, released if two light nuclei combine to form a single larger nucleus, a, process called nuclear fusion. Some examples of such energy liberating, reactions are, 1, 1, , +, H  11 H o 21 H + e + Q + 0.42 MeV, , [13.29(a)], , 455

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Physics, 2, 1, , H  21 H o 32 He + n + 3.27 MeV, , [13.29(b)], , 2, 1, , [13.29(c)], H  21 H o 31 H  11 H + 4.03 MeV, In reaction (13.29a), two protons combine to form a deuteron and, a positron with a release of 0.42 MeV energy. In reaction [13.29(b)], two, deuterons combine to form the light isotope of helium. In reaction, (13.29c), two deuterons combine to form a triton and a proton. In all, these reactions, we find that two positively charged particles combine to, form a larger nucleus. It must be realised that such a process is hindered, by the Coulomb repulsion that acts to prevent the two positively charged, particles from getting close enough to be within the range of their, attractive nuclear forces and thus ‘fusing’. The height of this Coulomb, barrier depends on the charges and the radii of the two interacting, nuclei. For example, it can be easily shown that for two protons, the, barrier height is ~ 400 keV. The barrier height for more highly charged, nuclei is higher. The temperature at which protons in a proton gas, would have enough energy to overcome the coulomb’s barrier is given, by (3/2)k T = K  400 keV and is about 3 × 109 K., To generate useful amount of energy, nuclear fusion must occur in, bulk matter. What is needed is to raise the temperature of the material, until the particles have enough energy – due to their thermal motions, alone – to penetrate the coulomb barrier. This process is called, thermonuclear fusion., The temperature of the core of the sun is only about 1.5u107 K., Therefore, even in the sun if the fusion is to take place, it must involve, protons whose energies are far above the average energy., Thus, for thermonuclear fusion to take place, extreme conditions of, temperature and pressure are required, which are available only in the, interiors of stars including sun. The energy generation in stars takes place, via thermonuclear fusion., The fusion reaction in the sun is a multi-step process in which, hydrogen is burned into helium, hydrogen being the ‘fuel’ and helium, the ‘ashes’. The proton-proton (p, p) cycle by which this occurs is, represented by the following sets of reactions:, +, H  11 H o 21 H + e + Q + 0.42 MeV, e + + e – o J + J + 1.02 MeV, 1, 1, , 2, 1, 3, 2, , 456, , H  11 H o 32 He + J + 5.49 MeV, , (i), (ii), (iii), , (iv), (13.30), H  32 H o 42 He  11 H  11 H + 12.86 MeV, For the fourth reaction to occur, the first three reactions must occur, twice, in which case two light helium nuclei unite to form ordinary helium, or nucleus. If we consider the combination 2(i) + 2(ii) + 2(iii) +(iv), the net, effect is, 4 11 H  2e  o 42 He  2Q  6J  26.7 MeV, or (4 11 H  4e  ) o ( 42 He  2e  )  2Q  6J  26.7 MeV, (13.31), 4, Thus, four hydrogen atoms combine to form an 2 He atom with a, release of 26.7 MeV of energy., The burning of hydrogen in the sun’s core is alchemy on a grand, scale in the sense that one element is turned into another. It has been

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Nuclei, going on for about 5 × 109 y, and calculations show that there is enough, hydrogen to keep the sun going for about the same time into the future., In about 5 billion years, however, the sun’s core, which by that time will, be largely helium, will begin to cool and the sun will start to collapse, under its own gravity. This will raise the core temperature and cause the, outer envelope to expand, turning the sun into what is called a red giant., If the core temperature increases to 108 K again, energy can be, produced through fusion once more – this time by burning helium to, make carbon. As a star evolves further and becomes still hotter, other, elements can be formed by other fusion reactions. However, elements more, massive than those near the peak of the binding energy curve of Fig. 13.1, cannot be produced by further fusion., The energy generation in stars takes place via thermonuclear, fusion., , NUCLEAR, , HOLOCAUST, , In a single uranium fission about 0.9×235 MeV (|200 MeV) of energy is liberated. If, each nucleus of about 50 kg of 235U undergoes fission the amount of energy involved is, about 4 × 1015J. This energy is equivalent to about 20,000 tons of TNT, enough for a, superexplosion. Uncontrolled release of large nuclear energy is called an atomic explosion., On August 6, 1945 an atomic device was used in warfare for the first time. The US, dropped an atom bomb on Hiroshima, Japan. The explosion was equivalent to 20,000, tons of TNT. Instantly the radioactive products devastated 10 sq km of the city which, had 3,43,000 inhabitants. Of this number 66,000 were killed and 69,000 were injured;, more than 67% of the city’s structures were destroyed., High temperature conditions for fusion reactions can be created by exploding a fission, bomb. Super-explosions equivalent to 10 megatons of explosive power of TNT were tested, in 1954. Such bombs which involve fusion of isotopes of hydrogen, deuterium and tritium, are called hydrogen bombs. It is estimated that a nuclear arsenal sufficient to destroy, every form of life on this planet several times over is in position to be triggered by the, press of a button. Such a nuclear holocaust will not only destroy the life that exists now, but its radioactive fallout will make this planet unfit for life for all times. Scenarios based, on theoretical calculations predict a long nuclear winter, as the radioactive waste will, hang like a cloud in the earth’s atmosphere and will absorb the sun’s radiation., , 13.7.4 Controlled thermonuclear fusion, The first thermonuclear reaction on earth occurred at Eniwetok Atoll on, November 1, 1952, when USA exploded a fusion device, generating energy, equivalent to 10 million tons of TNT (one ton of TNT on explosion releases, 2.6 × 1022 MeV of energy)., A sustained and controllable source of fusion power is considerably, more difficult to achieve. It is being pursued vigorously in many countries, around the world (including India), because fusion reactor is regarded as, the future power source., , 457

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Physics, Example 13.7 Answer the following questions:, (a) Are the equations of nuclear reactions (such as those given in, Section 13.7) ‘balanced’ in the sense a chemical equation (e.g.,, 2H2 + O2o 2 H2O) is? If not, in what sense are they balanced on, both sides?, (b) If both the number of protons and the number of neutrons are, conserved in each nuclear reaction, in what way is mass converted, into energy (or vice-versa) in a nuclear reaction?, (c) A general impression exists that mass-energy interconversion, takes place only in nuclear reaction and never in chemical, reaction. This is strictly speaking, incorrect. Explain., Solution, (a) A chemical equation is balanced in the sense that the number of, atoms of each element is the same on both sides of the equation., A chemical reaction merely alters the original combinations of, atoms. In a nuclear reaction, elements may be transmuted. Thus,, the number of atoms of each element is not necessarily conserved, in a nuclear reaction. However, the number of protons and the, number of neutrons are both separately conserved in a nuclear, reaction. [Actually, even this is not strictly true in the realm of, very high energies – what is strictly conserved is the total charge, and total ‘baryon number’. We need not pursue this matter here.], In nuclear reactions (e.g., Eq. 13.26), the number of protons and, the number of neutrons are the same on the two sides of the equation., , 458, , EXAMPLE 13.7, , (b) We know that the binding energy of a nucleus gives a negative, contribution to the mass of the nucleus (mass defect). Now, since, proton number and neutron number are conserved in a nuclear, reaction, the total rest mass of neutrons and protons is the same, on either side of a reaction. But the total binding energy of nuclei, on the left side need not be the same as that on the right hand, side. The difference in these binding energies appears as energy, released or absorbed in a nuclear reaction. Since binding energy, contributes to mass, we say that the difference in the total mass, of nuclei on the two sides get converted into energy or vice-versa., It is in these sense that a nuclear reaction is an example of massenergy interconversion., (c) From the point of view of mass-energy interconversion, a chemical, reaction is similar to a nuclear reaction in principle. The energy, released or absorbed in a chemical reaction can be traced to the, difference in chemical (not nuclear) binding energies of atoms and, molecules on the two sides of a reaction. Since, strictly speaking,, chemical binding energy also gives a negative contribution (mass, defect) to the total mass of an atom or molecule, we can equally, well say that the difference in the total mass of atoms or molecules,, on the two sides of the chemical reaction gets converted into energy, or vice-versa. However, the mass defects involved in a chemical, reaction are almost a million times smaller than those in a nuclear, reaction.This is the reason for the general impression, (which is, incorrect ) that mass-energy interconversion does not take place, in a chemical reaction.

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Nuclei, SUMMARY, 1., , 2., , 3., 4., , An atom has a nucleus. The nucleus is positively charged. The radius, of the nucleus is smaller than the radius of an atom by a factor of, 104. More than 99.9% mass of the atom is concentrated in the nucleus., On the atomic scale, mass is measured in atomic mass units (u). By, definition, 1 atomic mass unit (1u) is 1/12th mass of one atom of 12C;, 1u = 1.660563 × 10–27 kg., A nucleus contains a neutral particle called neutron. Its mass is almost, the same as that of proton, The atomic number Z is the number of protons in the atomic nucleus, of an element. The mass number A is the total number of protons and, neutrons in the atomic nucleus; A = Z+N; Here N denotes the number, of neutrons in the nucleus., , A nuclear species or a nuclide is represented as Z X , where X is the, chemical symbol of the species., Nuclides with the same atomic number Z, but different neutron number, N are called isotopes. Nuclides with the same A are isobars and those, with the same N are isotones., Most elements are mixtures of two or more isotopes. The atomic mass, of an element is a weighted average of the masses of its isotopes. The, masses are the relative abundances of the isotopes., 5. A nucleus can be considered to be spherical in shape and assigned a, radius. Electron scattering experiments allow determination of the, nuclear radius; it is found that radii of nuclei fit the formula, R = R0 A1/3,, where R0 = a constant = 1.2 fm. This implies that the nuclear density, is independent of A. It is of the order of 1017 kg/m3., 6. Neutrons and protons are bound in a nucleus by the short-range strong, nuclear force. The nuclear force does not distinguish between neutron, and proton., 7. The nuclear mass M is always less than the total mass, 6m, of its, constituents. The difference in mass of a nucleus and its constituents, is called the mass defect,, 'M = (Z mp + ( A – Z )mn ) – M, Using Einstein’s mass energy relation, we express this mass difference, in terms of energy as, 'Eb = 'M c2, The energy 'Eb represents the binding energy of the nucleus. In the, mass number range A = 30 to 170, the binding energy per nucleon is, nearly constant, about 8 MeV/nucleon., 8. Energies associated with nuclear processes are about a million times, larger than chemical process., 9. The Q-value of a nuclear process is, Q = final kinetic energy – initial kinetic energy., Due to conservation of mass-energy, this is also,, Q = (sum of initial masses – sum of final masses)c2, 10. Radioactivity is the phenomenon in which nuclei of a given species, transform by giving out D or E or J rays; D-rays are helium nuclei;, A, , 459

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Physics, E-rays are electrons. J-rays are electromagnetic radiation of wavelengths, shorter than X-rays;, 11. Law of radioactive decay : N (t) = N(0) e–Ot, where O is the decay constant or disintegration constant., The half-life T1/2 of a radionuclide is the time in which N has been, reduced to one-half of its initial value. The mean life W is the time at, which N has been reduced to e–1 of its initial value, , T1/2, , ln 2, , O, , W ln 2, , 12. Energy is released when less tightly bound nuclei are transmuted into, more tightly bound nuclei. In fission, a heavy nucleus like, into two smaller fragments, e.g.,, , 235, 92, , 1, 0, , 133, 51, , U+ n o, , Sb , , 99, 41, , 235, 92, , U breaks, , Nb + 4 10 n, , 13. The fact that more neutrons are produced in fission than are consumed, gives the possibility of a chain reaction with each neutron that is, produced triggering another fission. The chain reaction is uncontrolled, and rapid in a nuclear bomb explosion. It is controlled and steady in, a nuclear reactor. In a reactor, the value of the neutron multiplication, factor k is maintained at 1., 14. In fusion, lighter nuclei combine to form a larger nucleus. Fusion of, hydrogen nuclei into helium nuclei is the source of energy of all stars, including our sun., , Physical Quantity, , Symbol, , Dimensions, , Units, , [M], , u, , O, , [T –1], , s–1, , T1/2, , [T ], , s, , Time taken for the decay, of one-half of the initial, number of nuclei present, in a radioactive sample., , Mean life, , W, , [T ], , s, , Time at which number of, nuclei has been reduced to, e–1 of its initial value, , Activity of a radioactive sample, , R, , [ T–1 ], , Bq, , Measure of the activity, of a radioactive source., , Atomic mass unit, , Disintegration or, decay constant, Half-life, , Remarks, Unit of mass for, expressing atomic or, nuclear masses. One, atomic mass unit equals, 1/12th of the mass of 12C, atom., , POINTS TO PONDER, 1., 2., , 460, , The density of nuclear matter is independent of the size of the nucleus., The mass density of the atom does not follow this rule., The radius of a nucleus determined by electron scattering is found to, be slightly different from that determined by alpha-particle scattering.

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Nuclei, This is because electron scattering senses the charge distribution of, the nucleus, whereas alpha and similar particles sense the nuclear, matter., 3., , After Einstein showed the equivalence of mass and energy, E = mc 2,, we cannot any longer speak of separate laws of conservation of mass, and conservation of energy, but we have to speak of a unified law of, conservation of mass and energy. The most convincing evidence that, this principle operates in nature comes from nuclear physics. It is, central to our understanding of nuclear energy and harnessing it as a, source of power. Using the principle, Q of a nuclear process (decay or, reaction) can be expressed also in terms of initial and final masses., , 4., , The nature of the binding energy (per nucleon) curve shows that, exothermic nuclear reactions are possible, when two light nuclei fuse, or when a heavy nucleus undergoes fission into nuclei with intermediate, mass., , 5., , For fusion, the light nuclei must have sufficient initial energy to, overcome the coulomb potential barrier. That is why fusion requires, very high temperatures., , 6., , Although the binding energy (per nucleon) curve is smooth and slowly, varying, it shows peaks at nuclides like 4He, 16O etc. This is considered, as evidence of atom-like shell structure in nuclei., , 7., , Electrons and positron are a particle-antiparticle pair. They are, identical in mass; their charges are equal in magnitude and opposite., ( It is found that when an electron and a positron come together, they, annihilate each other giving energy in the form of gamma-ray photons.), , 8., , In â--decay (electron emission), the particle emitted along with electron, is anti-neutrino ( Q ). On the other hand, the particle emitted in E+decay (positron emission) is neutrino (Q). Neutrino and anti-neutrino, are a particle-antiparticle pair. There are anti particles associated, with every particle. What should be antiproton which is the anti, particle of the proton?, , 9., , A free neutron is unstable ( n o p  e –  Q ). But a similar free proton, decay is not possible, since a proton is (slightly) lighter than a neutron., , 10. Gamma emission usually follows alpha or beta emission. A nucleus, in an excited (higher) state goes to a lower state by emitting a gamma, photon. A nucleus may be left in an excited state after alpha or beta, emission. Successive emission of gamma rays from the same nucleus, (as in case of 60Ni, Fig. 13.4) is a clear proof that nuclei also have, discrete energy levels as do the atoms., 11. Radioactivity is an indication of the instability of nuclei. Stability, requires the ratio of neutron to proton to be around 1:1 for light, nuclei. This ratio increases to about 3:2 for heavy nuclei. (More, neutrons are required to overcome the effect of repulsion among the, protons.) Nuclei which are away from the stability ratio, i.e., nuclei, which have an excess of neutrons or protons are unstable. In fact,, only about 10% of knon isotopes (of all elements), are stable. Others, have been either artificially produced in the laboratory by bombarding, D, p, d, n or other particles on targets of stable nuclear species or, identified in astronomical observations of matter in the universe., , 461

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Physics, EXERCISES, You may find the following data useful in solving the exercises:, e = 1.6×10–19C, N = 6.023×1023 per mole, 9, 2, 2, 1/(4SH0) = 9 × 10 N m /C, k = 1.381×10–23J 0K–1, 1 MeV = 1.6×10–13J, 1 year = 3.154×107 s, , 1 u = 931.5 MeV/c2, , mH = 1.007825 u, , mn = 1.008665 u, , m( He ) = 4.002603 u, , me = 0.000548 u, , 4, 2, , 13.1, , (a) Two stable isotopes of lithium 63 Li and 73 Li have respective, abundances of 7.5% and 92.5%. These isotopes have masses, 6.01512 u and 7.01600 u, respectively. Find the atomic mass, of lithium., (b) Boron has two stable isotopes, 105 B and 115B . Their respective, masses are 10.01294 u and 11.00931 u, and the atomic mass of, boron is 10.811 u. Find the abundances of 105 B and 11, ., 5 B, , 13.2, , 20, 21, 22, The three stable isotopes of neon: 10, Ne, 10, Ne and 10, Ne have, respective abundances of 90.51%, 0.27% and 9.22%. The atomic, masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u,, respectively. Obtain the average atomic mass of neon., , 13.3, , Obtain the binding energy (in MeV) of a nitrogen nucleus, given m 147 N =14.00307 u, , 13.4, , Obtain the binding energy of the nuclei, MeV from the following data:, m ( 56, ) = 55.934939 u, 26 Fe, , Fe and, , 209, 83, , N ,, , Bi in units of, , m ( 209, ) = 208.980388 u, 83 Bi, , 13.5, , A given coin has a mass of 3.0 g. Calculate the nuclear energy that, would be required to separate all the neutrons and protons from, each other. For simplicity assume that the coin is entirely made of, 63, atoms (of mass 62.92960 u)., 29 Cu, , 13.6, , Write nuclear reaction equations for, (i) D-decay of, , 226, 88, , –, , (iii) E -decay of, +, , (v) E -decay of, , Ra, , 32, 15, , 11, 6, , P, , C, , (ii) D-decay of, , 13.7, 13.8, , 242, 94, , Pu, , –, , 210, 83, , +, , 97, 43, , (iv) E -decay of, (vi) E -decay of, , (vii) Electron capture of, , 462, , 56, 26, , 14, 7, , 120, 54, , Bi, , Tc, , Xe, A radioactive isotope has a half-life of T years. How long will it take, the activity to reduce to a) 3.125%, b) 1% of its original value?, The normal activity of living carbon-containing matter is found to, be about 15 decays per minute for every gram of carbon. This activity, arises from the small proportion of radioactive 146 C present with the, stable carbon isotope 126 C . When the organism is dead, its interaction, with the atmosphere (which maintains the above equilibrium activity), ceases and its activity begins to drop. From the known half-life (5730, years) of 146 C , and the measured activity, the age of the specimen, can be approximately estimated. This is the principle of 146 C dating

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Nuclei, used in archaeology. Suppose a specimen from Mohenjodaro gives, an activity of 9 decays per minute per gram of carbon. Estimate the, approximate age of the Indus-Valley civilisation., 13.9, , Obtain the amount of 60, necessary to provide a radioactive source, 27 Co, of 8.0 mCi strength. The half-life of 60, is 5.3 years., 27 Co, , 13.10 The half-life of 90, is 28 years. What is the disintegration rate of, 38 Sr, 15 mg of this isotope?, 13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope, 197, and the silver isotope 107, ., 79 Au, 47 Ag, 13.12 Find the Q-value and the kinetic energy of the emitted D-particle in, the D-decay of (a) 226, and (b) 220, ., 88 Ra, 86 Rn, Given m ( 226, ) = 226.02540 u,, 88 Ra, m ( 222, ) = 220.01137 u,, 86 Rn, 13.13 The radionuclide, 11, 6, , 11, 5, , Co, , m ( 222, ) = 222.01750 u,, 86 Rn, m ( 216, ) = 216.00189 u., 84 Po, , 11, , C decays according to, , +, , B + e +Q : T1/2 =20.3 min, , The maximum energy of the emitted positron is 0.960 MeV., Given the mass values:, m ( 116 C ) = 11.011434 u and m ( 116 B ) = 11.009305 u,, calculate Q and compare it with the maximum energy of the positron, emitted., –, 23, 13.14 The nucleus 10, Ne decays by E emission. Write down the E-decay, equation and determine the maximum kinetic energy of the, electrons emitted. Given that:, 23, m ( 10, Ne ) = 22.994466 u, 23, m ( 11, Na ) = 22.089770 u., , 13.15 The Q value of a nuclear reaction A + b o C + d is defined by, Q = [ mA + mb – mC – md]c2, where the masses refer to the respective nuclei. Determine from the, given data the Q-value of the following reactions and state whether, the reactions are exothermic or endothermic., (i) 11 H+13 H o12 H+12 H, (ii), , 12, 6, , 20, C+126 C o10, Ne+24 He, , Atomic masses are given to be, m, m, m, m, , ( 12 H ) = 2.014102 u, ( 13 H ) = 3.016049 u, ( 126 C ) = 12.000000 u, 20, ( 10, Ne ) = 19.992439 u, , nucleus into two equal, 13.16 Suppose, we think of fission of a 56, 26 Fe, 28, fragments, 13, Al . Is the fission energetically possible? Argue by, ) = 55.93494 u and, working out Q of the process. Given m ( 56, 26 Fe, 28, m ( 13, Al ) = 27.98191 u., 13.17 The fission properties of 239, are very similar to those of 235, . The, 94 Pu, 92 U, average energy released per fission is 180 MeV. How much energy,, , 463

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Physics, in MeV, is released if all the atoms in 1 kg of pure 239, undergo, 94 Pu, fission?, 13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How, much 235, did it contain initially? Assume that the reactor operates, 92 U, 80% of the time, that all the energy generated arises from the fission, of 235, and that this nuclide is consumed only by the fission process., 92 U, 13.19 How long can an electric lamp of 100W be kept glowing by fusion of, 2.0 kg of deuterium? Take the fusion reaction as, 2, 1, , H+ 21 H o 23 He +n +3.27 MeV, , 13.20 Calculate the height of the potential barrier for a head on collision, of two deuterons. (Hint: The height of the potential barrier is given, by the Coulomb repulsion between the two deuterons when they, just touch each other. Assume that they can be taken as hard, spheres of radius 2.0 fm.), 13.21 From the relation R = R0 A1/3, where R 0 is a constant and A is the, mass number of a nucleus, show that the nuclear matter density is, nearly constant (i.e. independent of A)., 13.22 For the E+ (positron) emission from a nucleus, there is another, competing process known as electron capture (electron from an inner, orbit, say, the K–shell, is captured by the nucleus and a neutrino is, emitted)., , e   ZA X o Z A1Y  Q, Show that if E+ emission is energetically allowed, electron capture, is necessarily allowed but not vice–versa., , ADDITIONAL EXERCISES, 13.23 In a periodic table the average atomic mass of magnesium is given, as 24.312 u. The average value is based on their relative natural, abundance on earth. The three isotopes and their masses are 24, 12 Mg, 26, (23.98504u), 25, (24.98584u), and, (25.98259u)., The, natural, Mg, Mg, 12, 12, is, 78.99%, by, mass., Calculate the abundances, abundance of 24, Mg, 12, of other two isotopes., 13.24 The neutron separation energy is defined as the energy required to, remove a neutron from the nucleus. Obtain the neutron separation, 41, energies of the nuclei 20, from the following data:, Ca and 27, 13 Al, m ( 40, ) = 39.962591 u, 20 Ca, 41, m ( 20, Ca ) = 40.962278 u, , ) = 25.986895 u, m ( 26, 13 Al, m ( 27, ) = 26.981541 u, 13 Al, 13.25 A source contains two phosphorous radio nuclides 32, (T1/2 = 14.3d), 15 P, and 33, (T, =, 25.3d)., Initially,, 10%, of, the, decays, come, from 33, ., P, 15, 15 P, 1/2, How long one must wait until 90% do so?, 13.26 Under certain circumstances, a nucleus can decay by emitting a, particle more massive than an D-particle. Consider the following, decay processes:, , 464, , 223, 88, , 14, Ra o 209, 82 Pb  6 C

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Nuclei, 223, 88, , 4, Ra o 219, 86 Rn  2 He, , Calculate the Q-values for these decays and determine that both, are energetically allowed., 13.27 Consider the fission of 238, by fast neutrons. In one fission event,, 92 U, no neutrons are emitted and the final end products, after the beta, 99, decay of the primary fragments, are 140, and 44, Ru . Calculate Q, 58 Ce, for this fission process. The relevant atomic and particle masses, are, m ( 238, ) =238.05079 u, 92 U, m ( 140, ) =139.90543 u, 58 Ce, 99, m ( 44, Ru ) = 98.90594 u, , 13.28 Consider the D–T reaction (deuterium–tritium fusion), 2, 1, , H  31 H o 24 He  n, , (a) Calculate the energy released in MeV in this reaction from the, data:, m ( 21 H )=2.014102 u, m ( 31 H ) =3.016049 u, (b) Consider the radius of both deuterium and tritium to be, approximately 2.0 fm. What is the kinetic energy needed to, overcome the coulomb repulsion between the two nuclei? To what, temperature must the gas be heated to initiate the reaction?, (Hint: Kinetic energy required for one fusion event =average, thermal kinetic energy available with the interacting particles, = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.), 13.29 Obtain the maximum kinetic energy of E-particles, and the radiation, frequencies of J decays in the decay scheme shown in Fig. 13.6. You, are given that, m (198Au) = 197.968233 u, m (198Hg) =197.966760 u, , FIGURE13.6, , 465

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Physics, 13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of, hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a, fission reactor., 13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW, of electric power, ten percent of which was to be obtained from nuclear, power plants. Suppose we are given that, on an average, the efficiency, of utilization (i.e. conversion to electric energy) of thermal energy, produced in a reactor was 25%. How much amount of fissionable, uranium would our country need per year by 2020? Take the heat, energy per fission of 235U to be about 200MeV., , 466

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Chapter Fourteen, , SEMICONDUCTOR, ELECTRONICS:, MATERIALS, DEVICES, AND SIMPLE CIRCUITS, 14.1 INTRODUCTION, Devices in which a controlled flow of electrons can be obtained are the, basic building blocks of all the electronic circuits. Before the discovery of, transistor in 1948, such devices were mostly vacuum tubes (also called, valves) like the vacuum diode which has two electrodes, viz., anode (often, called plate) and cathode; triode which has three electrodes – cathode,, plate and grid; tetrode and pentode (respectively with 4 and 5 electrodes)., In a vacuum tube, the electrons are supplied by a heated cathode and, the controlled flow of these electrons in vacuum is obtained by varying, the voltage between its different electrodes. Vacuum is required in the, inter-electrode space; otherwise the moving electrons may lose their, energy on collision with the air molecules in their path. In these devices, the electrons can flow only from the cathode to the anode (i.e., only in one, direction). Therefore, such devices are generally referred to as valves., These vacuum tube devices are bulky, consume high power, operate, generally at high voltages (~100 V ) and have limited life and low reliability., The seed of the development of modern solid-state semiconductor, electronics goes back to 1930’s when it was realised that some solidstate semiconductors and their junctions offer the possibility of controlling, the number and the direction of flow of charge carriers through them., Simple excitations like light, heat or small applied voltage can change, the number of mobile charges in a semiconductor. Note that the supply

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Physics, and flow of charge carriers in the semiconductor devices are within the, solid itself, while in the earlier vacuum tubes/valves, the mobile electrons, were obtained from a heated cathode and they were made to flow in an, evacuated space or vacuum. No external heating or large evacuated space, is required by the semiconductor devices. They are small in size, consume, low power, operate at low voltages and have long life and high reliability., Even the Cathode Ray Tubes (CRT) used in television and computer, monitors which work on the principle of vacuum tubes are being replaced, by Liquid Crystal Display (LCD) monitors with supporting solid state, electronics. Much before the full implications of the semiconductor devices, was formally understood, a naturally occurring crystal of galena (Lead, sulphide, PbS) with a metal point contact attached to it was used as, detector of radio waves., In the following sections, we will introduce the basic concepts of, semiconductor physics and discuss some semiconductor devices like, junction diodes (a 2-electrode device) and bipolar junction transistor (a, 3-electrode device). A few circuits illustrating their applications will also, be described., , 14.2 CLASSIFICATION OF METALS, CONDUCTORS, SEMICONDUCTORS, , 468, , AND, , On the basis of conductivity, On the basis of the relative values of electrical conductivity (V ) or resistivity, ( U = 1/V ), the solids are broadly classified as:, (i) Metals: They possess very low resistivity (or high conductivity)., U ~ 10–2 – 10–8 : m, V~ 102 – 108 S m–1, (ii) Semiconductors: They have resistivity or conductivity intermediate, to metals and insulators., U ~ 10–5 – 106 : m, V ~ 105 – 10–6 S m–1, (iii)Insulators: They have high resistivity (or low conductivity)., U ~ 1011 – 1019 : m, V ~ 10–11 – 10–19 S m–1, The values of U and V given above are indicative of magnitude and, could well go outside the ranges as well. Relative values of the resistivity, are not the only criteria for distinguishing metals, insulators and, semiconductors from each other. There are some other differences, which, will become clear as we go along in this chapter., Our interest in this chapter is in the study of semiconductors which, could be:, (i) Elemental semiconductors: Si and Ge, (ii) Compound semiconductors: Examples are:, x Inorganic: CdS, GaAs, CdSe, InP, etc., x Organic: anthracene, doped pthalocyanines, etc., x Organic polymers: polypyrrole, polyaniline, polythiophene, etc., Most of the currently available semiconductor devices are based on, elemental semiconductors Si or Ge and compound inorganic

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, semiconductors. However, after 1990, a few semiconductor devices using, organic semiconductors and semiconducting polymers have been, developed signalling the birth of a futuristic technology of polymerelectronics and molecular-electronics. In this chapter, we will restrict, ourselves to the study of inorganic semiconductors, particularly, elemental semiconductors Si and Ge. The general concepts introduced, here for discussing the elemental semiconductors, by-and-large, apply, to most of the compound semiconductors as well., On the basis of energy bands, According to the Bohr atomic model, in an isolated atom the energy of, any of its electrons is decided by the orbit in which it revolves. But when, the atoms come together to form a solid they are close to each other. So, the outer orbits of electrons from neighbouring atoms would come very, close or could even overlap. This would make the nature of electron motion, in a solid very different from that in an isolated atom., Inside the crystal each electron has a unique position and no two, electrons see exactly the same pattern of surrounding charges. Because, of this, each electron will have a different energy level. These different, energy levels with continuous energy variation form what are called, energy bands. The energy band which includes the energy levels of the, valence electrons is called the valence band. The energy band above the, valence band is called the conduction band. With no external energy, all, the valence electrons will reside in the valence band. If the lowest level in, the conduction band happens to be lower than the highest level of the, valence band, the electrons from the valence band can easily move into, the conduction band. Normally the conduction band is empty. But when, it overlaps on the valence band electrons can move freely into it. This is, the case with metallic conductors., If there is some gap between the conduction band and the valence, band, electrons in the valence band all remain bound and no free electrons, are available in the conduction band. This makes the material an, insulator. But some of the electrons from the valence band may gain, external energy to cross the gap between the conduction band and the, valence band. Then these electrons will move into the conduction band., At the same time they will create vacant energy levels in the valence band, where other valence electrons can move. Thus the process creates the, possibility of conduction due to electrons in conduction band as well as, due to vacancies in the valence band., Let us consider what happens in the case of Si or Ge crystal containing, N atoms. For Si, the outermost orbit is the third orbit (n = 3), while for Ge, it is the fourth orbit (n = 4). The number of electrons in the outermost, orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons, in the crystal is 4N. The maximum possible number of electrons in the, outer orbit is 8 (2s + 6p electrons). So, for the 4N valence electrons there, are 8N available energy states. These 8N discrete energy levels can either, form a continuous band or they may be grouped in different bands, depending upon the distance between the atoms in the crystal (see box, on Band Theory of Solids)., At the distance between the atoms in the crystal lattices of Si and Ge,, the energy band of these 8N states is split apart into two which are, separated by an energy gap Eg (Fig. 14.1). The lower band which is, , 469

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Physics, completely occupied by the 4N valence electrons at temperature of absolute, zero is the valence band. The other band consisting of 4N energy states,, called the conduction band, is completely empty at absolute zero., , BAND, , THEORY OF SOLIDS, , Consider that the Si or Ge crystal, contains N atoms. Electrons of each, atom will have discrete energies in, different orbits. The electron energy, will be same if all the atoms are, isolated, i.e., separated from each, other by a large distance. However,, in a crystal, the atoms are close to, each other (2 to 3 Å) and therefore, the electrons interact with each, other and also with the, neighbouring atomic cores. The, overlap (or interaction) will be more, felt by the electrons in the, outermost orbit while the inner, orbit or core electron energies may, remain unaffected. Therefore, for understanding electron energies in Si or Ge crystal, we, need to consider the changes in the energies of the electrons in the outermost orbit only., For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit, (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence,, the total number of outer electrons in the crystal is 4N. The maximum possible number of, outer electrons in the orbit is 8 (2s + 6p electrons). So, out of the 4N electrons, 2N electrons, are in the 2N s-states (orbital quantum number l = 0) and 2N electrons are in the available, 6N p-states. Obviously, some p-electron states are empty as shown in the extreme right of, Figure. This is the case of well separated or isolated atoms [region A of Figure]., Suppose these atoms start coming nearer to each other to form a solid. The energies, of these electrons in the outermost orbit may change (both increase and decrease) due to, the interaction between the electrons of different atoms. The 6N states for l = 1, which, originally had identical energies in the isolated atoms, spread out and form an energy, band [region B in Figure]. Similarly, the 2N states for l = 0, having identical energies in, the isolated atoms, split into a second band (carefully see the region B of Figure) separated, from the first one by an energy gap., At still smaller spacing, however, there comes a region in which the bands merge with, each other. The lowest energy state that is a split from the upper atomic level appears to, drop below the upper state that has come from the lower atomic level. In this region (region, C in Figure), no energy gap exists where the upper and lower energy states get mixed., Finally, if the distance between the atoms further decreases, the energy bands again, split apart and are separated by an energy gap Eg (region D in Figure). The total number, of available energy states 8N has been re-apportioned between the two bands (4N states, each in the lower and upper energy bands). Here the significant point is that there are, exactly as many states in the lower band (4N ) as there are available valence electrons, from the atoms (4N )., Therefore, this band ( called the valence band ) is completely filled while the upper, band is completely empty. The upper band is called the conduction band., , 470

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, The lowest energy level in the, conduction band is shown as EC and, highest energy level in the valence band, is shown as EV . Above EC and below EV, there are a large number of closely spaced, energy levels, as shown in Fig. 14.1., The gap between the top of the valence, band and bottom of the conduction band, is called the energy band gap (Energy gap, E g ). It may be large, small, or zero,, depending upon the material. These, different situations, are depicted in Fig., 14.2 and discussed below:, Case I: This refers to a situation, as, shown in Fig. 14.2(a). One can have a, metal either when the conduction band, FIGURE 14.1 The energy band positions in a, is partially filled and the balanced band, semiconductor at 0 K. The upper band, called the, is partially empty or when the conduction, and valance bands overlap. When there conduction band, consists of infinitely large number, of closely spaced energy states. The lower band,, is overlap electrons from valence band can, called the valence band, consists of closely spaced, easily move into the conduction band., completely filled energy states., This situation makes a large number of, electrons available for electrical conduction. When the valence band is, partially empty, electrons from its lower level can move to higher level, making conduction possible. Therefore, the resistance of such materials, is low or the conductivity is high., , FIGURE 14.2 Difference between energy bands of (a) metals,, (b) insulators and (c) semiconductors., , 471

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Physics, Case II: In this case, as shown in Fig. 14.2(b), a large band gap Eg exists, (Eg > 3 eV). There are no electrons in the conduction band, and therefore, no electrical conduction is possible. Note that the energy gap is so large, that electrons cannot be excited from the valence band to the conduction, band by thermal excitation. This is the case of insulators., Case III: This situation is shown in Fig. 14.2(c). Here a finite but small, band gap (Eg < 3 eV) exists. Because of the small band gap, at room, temperature some electrons from valence band can acquire enough, energy to cross the energy gap and enter the conduction band. These, electrons (though small in numbers) can move in the conduction band., Hence, the resistance of semiconductors is not as high as that of the, insulators., In this section we have made a broad classification of metals,, conductors and semiconductors. In the section which follows you will, learn the conduction process in semiconductors., , 14.3 INTRINSIC SEMICONDUCTOR, We shall take the most common case of Ge and Si whose lattice structure, is shown in Fig. 14.3. These structures are called the diamond-like, structures. Each atom is surrounded by four nearest neighbours. We, know that Si and Ge have four valence electrons. In its crystalline, structure, every Si or Ge atom tends to share one of its four valence, electrons with each of its four nearest neighbour atoms, and also to take, share of one electron from each such neighbour. These shared electron, pairs are referred to as forming a covalent bond or simply a valence, bond. The two shared electrons can be assumed to shuttle back-andforth between the associated atoms holding them together strongly., Figure 14.4 schematically shows the 2-dimensional representation of Si, or Ge structure shown in Fig. 14.3 which overemphasises the covalent, bond. It shows an idealised picture in which no bonds are broken (all, bonds are intact). Such a situation arises at low, temperatures. As the temperature increases, more, thermal energy becomes available to these electrons, and some of these electrons may break–away, (becoming free electrons contributing to conduction)., The thermal energy effectively ionises only a few atoms, in the crystalline lattice and creates a vacancy in the, bond as shown in Fig. 14.5(a). The neighbourhood,, from which the free electron (with charge –q ) has come, out leaves a vacancy with an effective charge (+q ). This, vacancy with the effective positive electronic charge is, called a hole. The hole behaves as an apparent free, particle with effective positive charge., In intrinsic semiconductors, the number of free, FIGURE 14.3 Three-dimensional diaelectrons,, ne is equal to the number of holes, nh. That is, mond-like crystal structure for Carbon,, ne = nh = ni, (14.1), Silicon or Germanium with, where ni is called intrinsic carrier concentration., respective lattice spacing a equal, Semiconductors posses the unique property in, to 3.56, 5.43 and 5.66 Å., 472, which, apart from electrons, the holes also move.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, Suppose there is a hole at site 1 as shown in, Fig. 14.5(a). The movement of holes can be, visualised as shown in Fig. 14.5(b). An electron, from the covalent bond at site 2 may jump to, the vacant site 1 (hole). Thus, after such a jump,, the hole is at site 2 and the site 1 has now an, electron. Therefore, apparently, the hole has, moved from site 1 to site 2. Note that the electron, originally set free [Fig. 14.5(a)] is not involved, in this process of hole motion. The free electron, moves completely independently as conduction, electron and gives rise to an electron current, Ie, under an applied electric field. Remember that, the motion of hole is only a convenient way of, FIGURE 14.4 Schematic two-dimensional, describing the actual motion of bound electrons,, representation of Si or Ge structure showing, whenever there is an empty bond anywhere in, covalent bonds at low temperature, (all bonds intact). +4 symbol, the crystal. Under the action of an electric field,, indicates inner cores of Si or Ge., these holes move towards negative potential, giving the hole current, Ih. The total current, I is, thus the sum of the electron current Ie and the, hole current Ih:, (14.2), I = Ie + Ih, It may be noted that apart from the process of generation of conduction, electrons and holes, a simultaneous process of recombination occurs in, which the electrons recombine with the holes. At equilibrium, the rate of, generation is equal to the rate of recombination of charge carriers. The, recombination occurs due to an electron colliding with a hole., , FIGURE 14.5 (a) Schematic model of generation of hole at site 1 and conduction electron, due to thermal energy at moderate temperatures. (b) Simplified representation of, possible thermal motion of a hole. The electron from the lower left hand covalent bond, (site 2) goes to the earlier hole site1, leaving a hole at its site indicating an, 473, apparent movement of the hole from site 1 to site 2.

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Physics, , FIGURE 14.6 (a) An intrinsic semiconductor at T = 0 K, behaves like insulator. (b) At T > 0 K, four thermally generated, electron-hole pairs. The filled circles ( ) represent electrons, and empty fields ( ) represent holes., , An intrinsic semiconductor, will behave like an insulator at, T = 0 K as shown in Fig. 14.6(a)., It is the thermal energy at, higher temperatures (T > 0K),, which excites some electrons, from the valence band to the, conduction band. These, thermally excited electrons at, T > 0 K, partially occupy the, conduction band. Therefore,, the energy-band diagram of an, intrinsic semiconductor will be, as shown in Fig. 14.6(b). Here,, some electrons are shown in, the conduction band. These, have come from the valence, band leaving equal number of, holes there., , EXAMPLE 14.1, , Example 14.1 C, Si and Ge have same lattice structure. Why is C, insulator while Si and Ge intrinsic semiconductors?, Solution The 4 bonding electrons of C, Si or Ge lie, respectively, in, the second, third and fourth orbit. Hence, energy required to take, out an electron from these atoms (i.e., ionisation energy Eg ) will be, least for Ge, followed by Si and highest for C. Hence, number of free, electrons for conduction in Ge and Si are significant but negligibly, small for C., , 14.4 EXTRINSIC SEMICONDUCTOR, , 474, , The conductivity of an intrinsic semiconductor depends on its, temperature, but at room temperature its conductivity is very low. As, such, no important electronic devices can be developed using these, semiconductors. Hence there is a necessity of improving their, conductivity. This can be done by making use of impurities., When a small amount, say, a few parts per million (ppm), of a suitable, impurity is added to the pure semiconductor, the conductivity of the, semiconductor is increased manifold. Such materials are known as, extrinsic semiconductors or impurity semiconductors. The deliberate, addition of a desirable impurity is called doping and the impurity atoms, are called dopants. Such a material is also called a doped semiconductor., The dopant has to be such that it does not distort the original pure, semiconductor lattice. It occupies only a very few of the original, semiconductor atom sites in the crystal. A necessary condition to attain, this is that the sizes of the dopant and the semiconductor atoms should, be nearly the same., There are two types of dopants used in doping the tetravalent Si, or Ge:, (i) Pentavalent (valency 5); like Arsenic (As), Antimony (Sb), Phosphorous, (P), etc.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, (ii) Trivalent (valency 3); like Indium (In),, Boron (B), Aluminium (Al), etc., We shall now discuss how the doping, changes the number of charge carriers (and, hence the conductivity) of semiconductors., Si or Ge belongs to the fourth group in the, Periodic table and, therefore, we choose the, dopant element from nearby fifth or third, group, expecting and taking care that the, size of the dopant atom is nearly the same as, that of Si or Ge. Interestingly, the pentavalent, and trivalent dopants in Si or Ge give two, entirely different types of semiconductors as, discussed below., (i) n-type semiconductor, Suppose we dope Si or Ge with a pentavalent, element as shown in Fig. 14.7. When an atom, of +5 valency element occupies the position, of an atom in the crystal lattice of Si, four of, its electrons bond with the four silicon, neighbours while the fifth remains very, weakly bound to its parent atom. This is, because the four electrons participating in, bonding are seen as part of the effective core FIGURE 14.7 (a) Pentavalent donor atom (As, Sb,, P, etc.) doped for tetravalent Si or Ge giving nof the atom by the fifth electron. As a result, type semiconductor, and (b) Commonly used, the ionisation energy required to set this, schematic representation of n-type material, electron free is very small and even at room, which shows only the fixed cores of the, temperature it will be free to move in the, substituent donors with one additional effective, lattice of the semiconductor. For example, the positive charge and its associated extra electron., energy required is ~ 0.01 eV for germanium,, and 0.05 eV for silicon, to separate this, electron from its atom. This is in contrast to the energy required to jump, the forbidden band (about 0.72 eV for germanium and about 1.1 eV for, silicon) at room temperature in the intrinsic semiconductor. Thus, the, pentavalent dopant is donating one extra electron for conduction and, hence is known as donor impurity. The number of electrons made, available for conduction by dopant atoms depends strongly upon the, doping level and is independent of any increase in ambient temperature., On the other hand, the number of free electrons (with an equal number, of holes) generated by Si atoms, increases weakly with temperature., In a doped semiconductor the total number of conduction electrons, ne is due to the electrons contributed by donors and those generated, intrinsically, while the total number of holes nh is only due to the holes, from the intrinsic source. But the rate of recombination of holes would, increase due to the increase in the number of electrons. As a result, the, number of holes would get reduced further., Thus, with proper level of doping the number of conduction electrons, 475, can be made much larger than the number of holes. Hence in an extrinsic

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Physics, semiconductor doped with pentavalent impurity, electrons, become the majority carriers and holes the minority carriers., These semiconductors are, therefore, known as n-type, semiconductors. For n-type semiconductors, we have,, (14.3), ne >> nh, (ii) p-type semiconductor, This is obtained when Si or Ge is doped with a trivalent impurity, like Al, B, In, etc. The dopant has one valence electron less than, Si or Ge and, therefore, this atom can form covalent bonds with, neighbouring three Si atoms but does not have any electron to, offer to the fourth Si atom. So the bond between the fourth, neighbour and the trivalent atom has a vacancy or hole as shown, in Fig. 14.8. Since the neighbouring Si atom in the lattice wants, an electron in place of a hole, an electron in the outer orbit of, an atom in the neighbourhood may jump to fill this vacancy,, leaving a vacancy or hole at its own site. Thus the hole is, available for conduction. Note that the trivalent foreign atom, becomes effectively negatively charged when it shares fourth, electron with neighbouring Si atom. Therefore, the dopant atom, of p-type material can be treated as core of one negative charge, along with its associated hole as shown in Fig. 14.8(b). It is, obvious that one acceptor atom gives one hole. These holes are, in addition to the intrinsically generated holes while the source, of conduction electrons is only intrinsic generation. Thus, for, such a material, the holes are the majority carriers and electrons, FIGURE 14.8 (a) Trivalent, are minority carriers. Therefore, extrinsic semiconductors doped, acceptor atom (In, Al, B etc.), with trivalent impurity are called p-type semiconductors. For, doped in tetravalent Si or Ge, p-type semiconductors, the recombination process will reduce, lattice giving p-type semiconthe number (ni )of intrinsically generated electrons to ne. We, ductor. (b) Commonly used, have,, for p-type semiconductors, schematic representation of, >> ne, (14.4), n, p-type material which shows, h, only the fixed core of the, Note that the crystal maintains an overall charge neutrality, substituent acceptor with, as the charge of additional charge carriers is just equal and, one effective additional, opposite to that of the ionised cores in the lattice., negative charge and its, In extrinsic semiconductors, because of the abundance of, associated hole., majority current carriers, the minority carriers produced, thermally have more chance of meeting majority carriers and, thus getting destroyed. Hence, the dopant, by adding a large number of, current carriers of one type, which become the majority carriers, indirectly, helps to reduce the intrinsic concentration of minority carriers., The semiconductor’s energy band structure is affected by doping. In, the case of extrinsic semiconductors, additional energy states due to donor, impurities (ED ) and acceptor impurities (EA ) also exist. In the energy band, diagram of n-type Si semiconductor, the donor energy level ED is slightly, below the bottom EC of the conduction band and electrons from this level, move into the conduction band with very small supply of energy. At room, temperature, most of the donor atoms get ionised but very few (~10–12), atoms of Si get ionised. So the conduction band will have most electrons, 476, coming from the donor impurities, as shown in Fig. 14.9(a). Similarly,

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, for p-type semiconductor, the acceptor energy level EA is slightly above, the top EV of the valence band as shown in Fig. 14.9(b). With very small, supply of energy an electron from the valence band can jump to the level, EA and ionise the acceptor negatively. (Alternately, we can also say that, with very small supply of energy the hole from level EA sinks down into, the valence band. Electrons rise up and holes fall down when they gain, external energy.) At room temperature, most of the acceptor atoms get, ionised leaving holes in the valence band. Thus at room temperature the, density of holes in the valence band is predominantly due to impurity in, the extrinsic semiconductor. The electron and hole concentration in a, semiconductor in thermal equilibrium is given by, nenh = ni2, (14.5), Though the above description is grossly approximate and, hypothetical, it helps in understanding the difference between metals,, insulators and semiconductors (extrinsic and intrinsic) in a simple, manner. The difference in the resistivity of C, Si and Ge depends upon, the energy gap between their conduction and valence bands. For C, (diamond), Si and Ge, the energy gaps are 5.4 eV, 1.1 eV and 0.7 eV,, respectively. Sn also is a group IV element but it is a metal because the, energy gap in its case is 0 eV., , FIGURE 14.9 Energy bands of (a) n-type semiconductor at T > 0K, (b) p-type, semiconductor at T > 0K., Example 14.2 Suppose a pure Si crystal has 5 × 1028 atoms m–3. It is, doped by 1 ppm concentration of pentavalent As. Calculate the, number of electrons and holes. Given that ni =1.5 × 1016 m–3., , ~ 4.5 × 109 m–3, , EXAMPLE 14.2, , Solution Note that thermally generated electrons (ni ~1016 m–3 ) are, negligibly small as compared to those produced by doping., Therefore, ne | ND., Since nenh = ni2, The number of holes, nh = (2.25 × 1032 )/(5 ×1022 ), , 477

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Physics, , http://www.mtmi.vu.lt/pfk/funkc_dariniai/diod/p n_devices.htm, , 14.5 p-n JUNCTION, A p-n junction is the basic building block of many semiconductor devices, like diodes, transistor, etc. A clear understanding of the junction behaviour, is important to analyse the working of other semiconductor devices., We will now try to understand how a junction is formed and how the, junction behaves under the influence of external applied voltage (also, called bias)., , 14.5.1 p-n junction formation, , Formation and working of p-n junction diode, , Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding, precisely a small quantity of pentavelent impurity, part of the p-Si wafer, can be converted into n-Si. There are several processes by which a, semiconductor can be formed. The wafer now contains p-region and, n-region and a metallurgical junction between p-, and n- region., Two important processes occur during the formation of a p-n junction:, diffusion and drift. We know that in an n-type semiconductor, the, concentration of electrons (number of electrons per unit volume) is more, compared to the concentration of holes. Similarly, in a p-type, semiconductor, the concentration of holes is more than the concentration, of electrons. During the formation of p-n junction, and due to the, concentration gradient across p-, and n- sides, holes diffuse from p-side, to n-side (p o n) and electrons diffuse from n-side to p-side (n o p). This, motion of charge carries gives rise to diffusion current across the junction., When an electron diffuses from n o p, it leaves behind an ionised, donor on n-side. This ionised donor (positive charge) is immobile as it is, bonded to the surrounding atoms. As the electrons continue to diffuse, from n o p, a layer of positive charge (or positive space-charge region) on, n-side of the junction is developed., Similarly, when a hole diffuses from p o n due to the concentration, gradient, it leaves behind an ionised acceptor (negative charge) which is, immobile. As the holes continue to diffuse, a layer of negative charge (or, negative space-charge region) on the p-side of the junction is developed., This space-charge region on either side of the junction together is known, as depletion region as the electrons and holes taking part in the initial, movement across the junction depleted the region of its, free charges (Fig. 14.10). The thickness of depletion region, is of the order of one-tenth of a micrometre. Due to the, positive space-charge region on n-side of the junction and, negative space charge region on p-side of the junction,, an electric field directed from positive charge towards, negative charge develops. Due to this field, an electron on, p-side of the junction moves to n-side and a hole on nside of the junction moves to p-side. The motion of charge, carriers due to the electric field is called drift. Thus a, FIGURE 14.10 p-n junction, drift current, which is opposite in direction to the diffusion, formation process., 478, current (Fig. 14.10) starts.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, Initially, diffusion current is large and drift current is small., As the diffusion process continues, the space-charge regions, on either side of the junction extend, thus increasing the electric, field strength and hence drift current. This process continues, until the diffusion current equals the drift current. Thus a p-n, junction is formed. In a p-n junction under equilibrium there, is no net current., The loss of electrons from the n-region and the gain of, electron by the p-region causes a difference of potential across, the junction of the two regions. The polarity of this potential is, such as to oppose further flow of carriers so that a condition of, equilibrium exists. Figure 14.11 shows the p-n junction at, equilibrium and the potential across the junction. The, n-material has lost electrons, and p material has acquired, electrons. The n material is thus positive relative to the p, material. Since this potential tends to prevent the movement of, electron from the n region into the p region, it is often called a, barrier potential., , FIGURE 14.11 (a) Diode under, equilibrium (V = 0), (b) Barrier, potential under no bias., , Solution No! Any slab, howsoever flat, will have roughness much, larger than the inter-atomic crystal spacing (~2 to 3 Å) and hence, continuous contact at the atomic level will not be possible. The junction, will behave as a discontinuity for the flowing charge carriers., , EXAMPLE 14.3, , Example 14.3 Can we take one slab of p-type semiconductor and, physically join it to another n-type semiconductor to get p-n junction?, , 14.6 SEMICONDUCTOR DIODE, A semiconductor diode [Fig. 14.12(a)] is basically a, p-n junction with metallic contacts provided at the, ends for the application of an external voltage. It is a, two terminal device. A p-n junction diode is, symbolically represented as shown in Fig. 14.12(b)., The direction of arrow indicates the conventional, direction of current (when the diode is under forward, bias). The equilibrium barrier potential can be altered, by applying an external voltage V across the diode., The situation of p-n junction diode under equilibrium, (without bias) is shown in Fig. 14.11(a) and (b)., , FIGURE 14.12 (a) Semiconductor diode,, (b) Symbol for p-n junction diode., , 14.6.1 p-n junction diode under forward bias, When an external voltage V is applied across a semiconductor diode such, that p-side is connected to the positive terminal of the battery and n-side, to the negative terminal [Fig. 14.13(a)], it is said to be forward biased., The applied voltage mostly drops across the depletion region and the, voltage drop across the p-side and n-side of the junction is negligible., (This is because the resistance of the depletion region – a region where, there are no charges – is very high compared to the resistance of n-side, and p-side.) The direction of the applied voltage (V ) is opposite to the, , 479

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Physics, , FIGURE 14.13 (a) p-n, junction diode under forward, bias, (b) Barrier potential, (1) without battery, (2) Low, battery voltage, and (3) High, voltage battery., , built-in potential V0. As a result, the depletion layer width, decreases and the barrier height is reduced [Fig. 14.13(b)]. The, effective barrier height under forward bias is (V0 – V )., If the applied voltage is small, the barrier potential will be, reduced only slightly below the equilibrium value, and only a, small number of carriers in the material—those that happen to, be in the uppermost energy levels—will possess enough energy, to cross the junction. So the current will be small. If we increase, the applied voltage significantly, the barrier height will be reduced, and more number of carriers will have the required energy. Thus, the current increases., Due to the applied voltage, electrons from n-side cross the, depletion region and reach p-side (where they are minority, carries). Similarly, holes from p-side cross the junction and reach, the n-side (where they are minority carries). This process under, forward bias is known as minority carrier injection. At the, junction boundary, on each side, the minority carrier, concentration increases significantly compared to the locations, far from the junction., Due to this concentration gradient, the injected electrons on, p-side diffuse from the junction edge of p-side to the other end, of p-side. Likewise, the injected holes on n-side diffuse from the, junction edge of n-side to the other end of n-side, (Fig. 14.14). This motion of charged carriers on either side, gives rise to current. The total diode forward current is sum, of hole diffusion current and conventional current due to, electron diffusion. The magnitude of this current is usually, in mA., , 14.6.2 p-n junction diode under reverse bias, FIGURE 14.14 Forward bias, minority carrier injection., , 480, , When an external voltage (V ) is applied across the diode such, that n-side is positive and p-side is negative, it is said to be, reverse biased [Fig.14.15(a)]. The applied voltage mostly, drops across the depletion region. The direction of applied voltage is same, as the direction of barrier potential. As a result, the barrier height increases, and the depletion region widens due to the change in the electric field., The effective barrier height under reverse bias is (V0 + V ), [Fig. 14.15(b)]., This suppresses the flow of electrons from n o p and holes from p o n., Thus, diffusion current, decreases enormously compared to the diode, under forward bias., The electric field direction of the junction is such that if electrons on, p-side or holes on n-side in their random motion come close to the, junction, they will be swept to its majority zone. This drift of carriers, gives rise to current. The drift current is of the order of a few PA. This is, quite low because it is due to the motion of carriers from their minority, side to their majority side across the junction. The drift current is also, there under forward bias but it is negligible (PA) when compared with, current due to injected carriers which is usually in mA., The diode reverse current is not very much dependent on the applied, voltage. Even a small voltage is sufficient to sweep the minority carriers, from one side of the junction to the other side of the junction. The current

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, is not limited by the magnitude of the applied voltage but is, limited due to the concentration of the minority carrier on either, side of the junction., The current under reverse bias is essentially voltage, independent upto a critical reverse bias voltage, known as, breakdown voltage (Vbr ). When V = Vbr, the diode reverse current, increases sharply. Even a slight increase in the bias voltage causes, large change in the current. If the reverse current is not limited by, an external circuit below the rated value (specified by the, manufacturer) the p-n junction will get destroyed. Once it exceeds, the rated value, the diode gets destroyed due to overheating. This, can happen even for the diode under forward bias, if the forward, current exceeds the rated value., The circuit arrangement for studying the V-I characteristics, of a diode, (i.e., the variation of current as a function of applied, FIGURE 14.15 (a) Diode, voltage) are shown in Fig. 14.16(a) and (b). The battery is connected, under reverse bias,, to the diode through a potentiometer (or reheostat) so that the, (b) Barrier potential under, applied voltage to the diode can be changed. For different values, reverse bias., of voltages, the value of the current is noted. A graph between V, and I is obtained as in Fig. 14.16(c). Note that in forward bias, measurement, we use a milliammeter since the expected current is large, (as explained in the earlier section) while a micrometer is used in reverse, bias to measure the current. You can see in Fig. 14.16(c) that in forward, , FIGURE 14.16 Experimental circuit arrangement for studying V-I characteristics of, a p-n junction diode (a) in forward bias , (b) in reverse bias. (c) Typical V-I, 481, characteristics of a silicon diode.

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Physics, bias, the current first increases very slowly, almost negligibly, till the, voltage across the diode crosses a certain value. After the characteristic, voltage, the diode current increases significantly (exponentially), even for, a very small increase in the diode bias voltage. This voltage is called the, threshold voltage or cut-in voltage (~0.2V for germanium diode and, ~0.7 V for silicon diode)., For the diode in reverse bias, the current is very small (~PA) and almost, remains constant with change in bias. It is called reverse saturation, current. However, for special cases, at very high reverse bias (break down, voltage), the current suddenly increases. This special action of the diode, is discussed later in Section 14.8. The general purpose diode are not, used beyond the reverse saturation current region., The above discussion shows that the p-n junction diode primerly, allows the flow of current only in one direction (forward bias). The forward, bias resistance is low as compared to the reverse bias resistance. This, property is used for rectification of ac voltages as discussed in the next, section. For diodes, we define a quantity called dynamic resistance as, the ratio of small change in voltage 'V to a small change in current 'I:, rd, , 'V, 'I, , (14.6), , Example 14.4 The V-I characteristic of a silicon diode is shown in, the Fig. 14.17. Calculate the resistance of the diode at (a) ID = 15 mA, and (b) VD = –10 V., , 482, , EXAMPLE 14.4, , FIGURE 14.17, , Solution Considering the diode characteristics as a straight line, between I = 10 mA to I = 20 mA passing through the origin, we can, calculate the resistance using Ohm’s law., (a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V, rf b ='V/'I = 0.1V/10 mA = 10 :, (b) From the curve at V = –10 V, I = –1 PA,, Therefore,, rrb = 10 V/1PA= 1.0 × 107 :

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, , 14.7 APPLICATION OF JUNCTION DIODE AS A RECTIFIER, From the V-I characteristic of a junction diode we see that it allows current, to pass only when it is forward biased. So if an alternating voltage is, applied across a diode the current flows only in that part of the cycle, when the diode is forward biased. This property, is used to rectify alternating voltages and the, circuit used for this purpose is called a rectifier., If an alternating voltage is applied across a, diode in series with a load, a pulsating voltage will, appear across the load only during the half cycles, of the ac input during which the diode is forward, biased. Such rectifier circuit, as shown in, Fig. 14.18, is called a half-wave rectifier. The, secondary of a transformer supplies the desired, ac voltage across terminals A and B. When the, voltage at A is positive, the diode is forward biased, and it conducts. When A is negative, the diode is, reverse-biased and it does not conduct. The reverse, saturation current of a diode is negligible and can, be considered equal to zero for practical purposes., (The reverse breakdown voltage of the diode must, be sufficiently higher than the peak ac voltage at, the secondary of the transformer to protect the, diode from reverse breakdown.), Therefore, in the positive half-cycle of ac there, FIGURE 14.18 (a) Half-wave rectifier, is a current through the load resistor RL and we, circuit, (b) Input ac voltage and output, get an output voltage, as shown in Fig. 14.18(b),, voltage waveforms from the rectifier circuit., whereas there is no current in the negative halfcycle. In the next positive half-cycle, again we get, the output voltage. Thus, the output voltage, though still varying, is, restricted to only one direction and is said to be rectified. Since the, rectified output of this circuit is only for half of the input ac wave it is, called as half-wave rectifier., The circuit using two diodes, shown in Fig. 14.19(a), gives output, rectified voltage corresponding to both the positive as well as negative, half of the ac cycle. Hence, it is known as full-wave rectifier. Here the, p-side of the two diodes are connected to the ends of the secondary of the, transformer. The n-side of the diodes are connected together and the, output is taken between this common point of diodes and the midpoint, of the secondary of the transformer. So for a full-wave rectifier the, secondary of the transformer is provided with a centre tapping and so it, is called centre-tap transformer. As can be seen from Fig.14.19(c) the, voltage rectified by each diode is only half the total secondary voltage., Each diode rectifies only for half the cycle, but the two do so for alternate, cycles. Thus, the output between their common terminals and the centretap of the transformer becomes a full-wave rectifier output. (Note that, there is another circuit of full wave rectifier which does not need a centre483, tap transformer but needs four diodes.) Suppose the input voltage to A

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Physics, with respect to the centre tap at any instant, is positive. It is clear that, at that instant,, voltage at B being out of phase will be, negative as shown in Fig.14.19(b). So, diode, D1 gets forward biased and conducts (while, D2 being reverse biased is not conducting)., Hence, during this positive half cycle we get, an output current (and a output voltage, across the load resistor RL) as shown in, Fig.14.19(c). In the course of the ac cycle, when the voltage at A becomes negative with, respect to centre tap, the voltage at B would, be positive. In this part of the cycle diode, D1 would not conduct but diode D2 would,, giving an output current and output, voltage (across RL ) during the negative half, cycle of the input ac. Thus, we get output, voltage during both the positive as well as, the negative half of the cycle. Obviously,, this is a more efficient circuit for getting, rectified voltage or current than the halfwave rectifier, The rectified voltage is in the form of, pulses of the shape of half sinusoids., Though it is unidirectional it does not have, a steady value. To get steady dc output, from the pulsating voltage normally a, capacitor is connected across the output, terminals (parallel to the load RL). One can, also use an inductor in series with RL for, FIGURE 14.19 (a) A Full-wave rectifier, the same purpose. Since these additional, circuit; (b) Input wave forms given to the, circuits appear to filter out the ac ripple, diode D1 at A and to the diode D2 at B;, and give a pure dc voltage, so they are, (c) Output waveform across the, called, filters., load RL connected in the full-wave, Now, we shall discuss the role of, rectifier circuit., capacitor in filtering. When the voltage, across the capacitor is rising, it gets, charged. If there is no external load, it remains charged to the peak voltage, of the rectified output. When there is a load, it gets discharged through, the load and the voltage across it begins to fall. In the next half-cycle of, rectified output it again gets charged to the peak value (Fig. 14.20). The, rate of fall of the voltage across the capacitor depends upon the inverse, product of capacitor C and the effective resistance RL used in the circuit, and is called the time constant. To make the time constant large value of, C should be large. So capacitor input filters use large capacitors. The, output voltage obtained by using capacitor input filter is nearer to the, peak voltage of the rectified voltage. This type of filter is most widely, 484, used in power supplies.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, , FIGURE 14.20 (a) A full-wave rectifier with capacitor filter, (b) Input and output, voltage of rectifier in (a)., , 14.8 SPECIAL PURPOSE p-n JUNCTION DIODES, In the section, we shall discuss some devices which are basically junction, diodes but are developed for different applications., , 14.8.1 Zener diode, It is a special purpose semiconductor diode, named after its inventor, C. Zener. It is designed to operate under reverse bias in the breakdown, region and used as a voltage regulator. The symbol for Zener diode is, shown in Fig. 14.21(a)., Zener diode is fabricated by heavily doping both p-, and, n- sides of the junction. Due to this, depletion region formed, is very thin (<10–6 m) and the electric field of the junction is, extremely high (~5×106 V/m) even for a small reverse bias, voltage of about 5V. The I-V characteristics of a Zener diode is, shown in Fig. 14.21(b). It is seen that when the applied reverse, bias voltage(V ) reaches the breakdown voltage (Vz ) of the Zener, diode, there is a large change in the current. Note that after, the breakdown voltage Vz, a large change in the current can, be produced by almost insignificant change in the reverse bias, voltage. In other words, Zener voltage remains constant, even, though current through the Zener diode varies over a wide, range. This property of the Zener diode is used for regulating, supply voltages so that they are constant., Let us understand how reverse current suddenly increases, at the breakdown voltage. We know that reverse current is, due to the flow of electrons (minority carriers) from p on and, holes from n op. As the reverse bias voltage is increased, the, electric field at the junction becomes significant. When the, reverse bias voltage V = Vz, then the electric field strength is, high enough to pull valence electrons from the host atoms on, the p-side which are accelerated to n-side. These electrons, account for high current observed at the breakdown. The, FIGURE 14.21 Zener diode,, emission of electrons from the host atoms due to the high, (a) symbol, (b) I-V, electric field is known as internal field emission or field, characteristics., ionisation. The electric field required for field ionisation is of, 485, the order of 106 V/m.

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Physics, Zener diode as a voltage regulator, We know that when the ac input voltage of a rectifier fluctuates, its rectified, output also fluctuates. To get a constant dc voltage from the dc, unregulated output of a rectifier, we use a Zener diode. The circuit diagram, of a voltage regulator using a Zener diode is shown in Fig. 14.22., The unregulated dc voltage (filtered output of a, rectifier) is connected to the Zener diode through a series, resistance Rs such that the Zener diode is reverse biased., If the input voltage increases, the current through Rs, and Zener diode also increases. This increases the, voltage drop across Rs without any change in the, voltage across the Zener diode. This is because in the, breakdown region, Zener voltage remains constant even, though the current through the Zener diode changes., Similarly, if the input voltage decreases, the current, through Rs and Zener diode also decreases. The voltage, drop across Rs decreases without any change in the, FIGURE 14.22 Zener diode as DC, voltage across the Zener diode. Thus any increase/, voltage regulator (to be corrected)., decrease in the input voltage results in, increase/, decrease of the voltage drop across Rs without any, change in voltage across the Zener diode. Thus the Zener diode acts as a, voltage regulator. We have to select the Zener diode according to the, required output voltage and accordingly the series resistance Rs., , EXAMPLE 14.5, , Example 14.5 In a Zener regulated power supply a Zener diode with, VZ = 6.0 V is used for regulation. The load current is to be 4.0 mA and, the unregulated input is 10.0 V. What should be the value of series, resistor RS?, Solution, The value of RS should be such that the current through the Zener, diode is much larger than the load current. This is to have good load, regulation. Choose Zener current as five times the load current, i.e.,, IZ = 20 mA. The total current through RS is, therefore, 24 mA. The, voltage drop across R S is 10.0 – 6.0 = 4.0 V. This gives, RS = 4.0V/(24 × 10–3) A = 167 :. The nearest value of carbon resistor, is 150 :. So, a series resistor of 150 : is appropriate. Note that slight, variation in the value of the resistor does not matter, what is important, is that the current IZ should be sufficiently larger than IL., , 14.8.2 Optoelectronic junction devices, , 486, , We have seen so far, how a semiconductor diode behaves under applied, electrical inputs. In this section, we learn about semiconductor diodes in, which carriers are generated by photons (photo-excitation). All these, devices are called optoelectronic devices. We shall study the functioning, of the following optoelectronic devices:, (i) Photodiodes used for detecting optical signal (photodetectors)., (ii) Light emitting diodes (LED) which convert electrical energy into light., (iii) Photovoltaic devices which convert optical radiation into electricity, (solar cells).

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, (i) Photodiode, A Photodiode is again a special purpose p-n, junction diode fabricated with a transparent, window to allow light to fall on the diode. It is, operated under reverse bias. When the photodiode, is illuminated with light (photons) with energy (hQ), greater than the energy gap (E g ) of the, semiconductor, then electron-hole pairs are, generated due to the absorption of photons. The, diode is fabricated such that the generation of, e-h pairs takes place in or near the depletion region, of the diode. Due to electric field of the junction,, electrons and holes are separated before they, recombine. The direction of the electric field is such, that electrons reach n-side and holes reach p-side., Electrons are collected on n-side and holes are, collected on p-side giving rise to an emf. When an, external load is connected, current flows. The, magnitude of the photocurrent depends on the, intensity of incident light (photocurrent is, proportional to incident light intensity)., It is easier to observe the change in the current, with change in the light intensity, if a reverse bias, is applied. Thus photodiode can be used as a, photodetector to detect optical signals. The circuit, diagram used for the measurement of I-V, characteristics of a photodiode is shown in, Fig. 14.23(a) and a typical I-V characteristics in, Fig. 14.23(b)., , FIGURE 14.23 (a) An illuminated, photodiode under reverse bias , (b) I-V, characteristics of a photodiode for different, illumination intensity I4 > I3 > I2 > I1., , Example 14.6 The current in the forward bias is known to be more, (~mA) than the current in the reverse bias (~PA). What is the reason, then to operate the photodiodes in reverse bias?, Solution Consider the case of an n-type semiconductor. Obviously,, the majority carrier density (n ) is considerably larger than the, minority hole density p (i.e., n >> p). On illumination, let the excess, electrons and holes generated be 'n and 'p, respectively:, pc = p +'p, Here nc and p c are the electron and hole concentrations* at any, particular illumination and n and p are carriers concentration when, there is no illumination. Remember 'n = 'p and n >> p. Hence, the, , * Note that, to create an e-h pair, we spend some energy (photoexcitation, thermal, excitation, etc.). Therefore when an electron and hole recombine the energy is, released in the form of light (radiative recombination) or heat (non-radiative, recombination). It depends on semiconductor and the method of fabrication of, the p-n junction. For the fabrication of LEDs, semiconductors like GaAs, GaAsGaP are used in which radiative recombination dominates., , EXAMPLE 14.6, , nc = n +'n, , 487

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EXAMPLE 14.6, , Physics, fractional change in the majority carriers (i.e., 'n/n ) would be much, less than that in the minority carriers (i.e., 'p/p). In general, we can, state that the fractional change due to the photo-effects on the, minority carrier dominated reverse bias current is more easily, measurable than the fractional change in the forward bias current., Hence, photodiodes are preferably used in the reverse bias condition, for measuring light intensity., , (ii) Light emitting diode, , 488, , It is a heavily doped p-n junction which under forward bias emits, spontaneous radiation. The diode is encapsulated with a transparent, cover so that emitted light can come out., When the diode is forward biased, electrons are sent from n op (where, they are minority carriers) and holes are sent from p on (where they are, minority carriers). At the junction boundary the concentration of minority, carriers increases compared to the equilibrium concentration (i.e., when, there is no bias). Thus at the junction boundary on either side of the, junction, excess minority carriers are there which recombine with majority, carriers near the junction. On recombination, the energy is released in, the form of photons. Photons with energy equal to or slightly less than, the band gap are emitted. When the forward current of the diode is small,, the intensity of light emitted is small. As the forward current increases,, intensity of light increases and reaches a maximum. Further increase in, the forward current results in decrease of light intensity. LEDs are biased, such that the light emitting efficiency is maximum., The V-I characteristics of a LED is similar to that of a Si junction, diode. But the threshold voltages are much higher and slightly different, for each colour. The reverse breakdown voltages of LEDs are very low,, typically around 5V. So care should be taken that high reverse voltages, do not appear across them., LEDs that can emit red, yellow, orange, green and blue light are, commercially available. The semiconductor used for fabrication of visible, LEDs must at least have a band gap of 1.8 eV (spectral range of visible, light is from about 0.4 Pm to 0.7 Pm, i.e., from about 3 eV to 1.8 eV). The, compound semiconductor Gallium Arsenide – Phosphide (GaAs1–x Px ) is, used for making LEDs of different colours. GaAs0.6 P0.4 (Eg ~ 1.9 eV) is, used for red LED. GaAs (Eg ~ 1.4 eV) is used for making infrared LED., These LEDs find extensive use in remote controls, burglar alarm systems,, optical communication, etc. Extensive research is being done for, developing white LEDs which can replace incandescent lamps., LEDs have the following advantages over conventional incandescent, low power lamps:, (i) Low operational voltage and less power., (ii) Fast action and no warm-up time required., (iii) The bandwidth of emitted light is 100 Å to 500 Å or in other words it, is nearly (but not exactly) monochromatic., (iv) Long life and ruggedness., (v) Fast on-off switching capability.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, (iii) Solar cell, A solar cell is basically a p-n junction which, generates emf when solar radiation falls on the, p-n junction. It works on the same principle, (photovoltaic effect) as the photodiode, except that, no external bias is applied and the junction area, is kept much larger for solar radiation to be, incident because we are interested in more power., A simple p-n junction solar cell is shown in, Fig. 14.24., A p-Si wafer of about 300 Pm is taken over, which a thin layer (~0.3 Pm) of n-Si is grown on, one-side by diffusion process. The other side of, p-Si is coated with a metal (back contact). On the, top of n-Si layer, metal finger electrode (or metallic, FIGURE 14.24 (a) Typical p-n junction, grid) is deposited. This acts as a front contact. The, solar cell; (b) Cross-sectional view., metallic grid occupies only a very small fraction, of the cell area (<15%) so that light can be incident, on the cell from the top., The generation of emf by a solar cell, when light falls on, it is due to, the following three basic processes: generation, separation and collection—, (i) generation of e-h pairs due to light (with hQ > Eg ), close to the junction; (ii) separation of electrons and, holes due to electric field of the depletion region., Electrons are swept to n-side and holes to p-side;, (iii) the electrons reaching the n-side are collected by, the front contact and holes reaching p-side are collected, by the back contact. Thus p-side becomes positive and, n-side becomes negative giving rise to photovoltage., When an external load is connected as shown in, the Fig. 14.25(a) a photocurrent IL flows through the, load. A typical I-V characteristics of a solar cell is shown, in the Fig. 14.25(b)., Note that the I – V characteristics of solar cell is, drawn in the fourth quadrant of the coordinate axes., This is because a solar cell does not draw current but, supplies the same to the load., Semiconductors with band gap close to 1.5 eV are, ideal materials for solar cell fabrication. Solar cells are, made with semiconductors like Si (Eg = 1.1 eV), GaAs, (Eg = 1.43 eV), CdTe (Eg = 1.45 eV), CuInSe2 (Eg = 1.04, eV), etc. The important criteria for the selection of a, material for solar cell fabrication are (i) band gap (~1.0, to 1.8 eV), (ii) high optical absorption (~104 cm–1), (iii), electrical conductivity, (iv) availability of the raw, FIGURE 14.25 (a) A typical, material, and (v) cost. Note that sunlight is not always, illuminated, p-n junction solar cell;, required for a solar cell. Any light with photon energies, (b) I-V characteristics of a solar cell., greater than the bandgap will do. Solar cells are used, to power electronic devices in satellites and space, vehicles and also as power supply to some calculators. Production of, low-cost photovoltaic cells for large-scale solar energy is a topic, 489, for research.

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Physics, Example 14.7 Why are Si and GaAs are preferred materials for, solar cells?, Solution The solar radiation spectrum received by us is shown in, Fig. 14.26., , EXAMPLE 14.7, , FIGURE 14.26, , The maxima is near 1.5 eV. For photo-excitation, hQ > Eg. Hence,, semiconductor with band gap ~1.5 eV or lower is likely to give better, solar conversion efficiency. Silicon has Eg ~ 1.1 eV while for GaAs it is, ~1.53 eV. In fact, GaAs is better (in spite of its higher band gap) than, Si because of its relatively higher absorption coefficient. If we choose, materials like CdS or CdSe (Eg ~ 2.4 eV), we can use only the high, energy component of the solar energy for photo-conversion and a, significant part of energy will be of no use., The question arises: why we do not use material like PbS (Eg ~ 0.4 eV), which satisfy the condition hQ > Eg for Q maxima corresponding to the, solar radiation spectra? If we do so, most of the solar radiation will be, absorbed on the top-layer of solar cell and will not reach in or near, the depletion region. For effective electron-hole separation, due to, the junction field, we want the photo-generation to occur in the, junction region only., , 14.9 JUNCTION TRANSISTOR, , 490, , The credit of inventing the transistor in the year 1947 goes to J. Bardeen, and W.H. Brattain of Bell Telephone Laboratories, U.S.A. That transistor, was a point-contact transistor. The first junction transistor consisting of, two back-to-back p-n junctions was invented by William Schockley, in 1951., As long as only the junction transistor was known, it was known, simply as transistor. But over the years new types of transistors were, invented and to differentiate it from the new ones it is now called the, Bipolar Junction Transistor (BJT). Even now, often the word transistor

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, is used to mean BJT when there is no confusion. Since our study is, limited to only BJT, we shall use the word transistor for BJT without, any ambiguity., , 14.9.1 Transistor: structure and action, A transistor has three doped regions forming two p-n junctions, between them. Obviously, there are two types of transistors, as shown, in Fig. 14.27., (i) n-p-n transistor : Here two segments of n-type semiconductor, (emitter and collector) are separated by a segment of p-type, semiconductor (base)., (ii) p-n-p transistor : Here two segments of p-type semiconductor, (termed as emitter and collector) are separated by a segment of, n-type semiconductor (termed as base)., The schematic representations of an n-p-n and a p-n-p, configuration are shown in Fig. 14.27(a). All the three segments of a, transistor have different thickness and their doping levels are also, different. In the schematic symbols used for representing p-n-p and, n-p-n transistors [Fig. 14.27(b)] the arrowhead shows the direction of, conventional current in the transistor. A brief description of the three, segments of a transistor is given below:, x Emitter : This is the segment on one side of the transistor shown in, Fig. 14.27(a). It is of moderate size and heavily doped. It supplies, a large number of majority carriers for the current flow through, the transistor., x Base: This is the central segment. It is very thin and lightly doped., x Collector: This segment collects a major portion of the majority, carriers supplied by the emitter. The collector side is moderately, doped and larger in size as compared to the emitter., We have seen earlier in the case of a p-n junction, that there is a, formation of depletion region acorss the junction. In case of a transistor, depletion regions are formed at the emitter base-junction and the basecollector junction. For understanding the action of a transistor, we, have to consider the nature of depletion regions formed at these, junctions. The charge carriers move across different regions of the, transistor when proper voltages are applied across its terminals., The biasing of the transistor is done differently for different uses., The transistor can be used in two distinct ways. Basically, it was, invented to function as an amplifier, a device which produces a enlarged, copy of a signal. But later its use as a switch acquired equal, importance. We shall study both these functions and the ways the, transistor is biased to achieve these mutually exclusive functions., First we shall see what gives the transistor its amplifying capabilities., The transistor works as an amplifier, with its emitter-base junction, forward biased and the base-collector junction reverse biased. This, situation is shown in Fig. 14.28, where VCC and VEE are used for creating, the respective biasing. When the transistor is biased in this way it is, said to be in active state.We represent the voltage between emitter and, base as VEB and that between the collector and the base as VCB. In, , FIGURE 14.27, (a) Schematic, representations of a, n-p-n transistor and, p-n-p transistor, and, (b) Symbols for n-p-n, and p-n-p transistors., , 491

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Physics, Fig. 14.28, base is a common terminal for the two, power supplies whose other terminals are, connected to emitter and collector, respectively. So, the two power supplies are represented as VEE, and, VCC, respectively. In circuits, where emitter is the, common terminal, the power supply between the, base and the emitter is represented as VBB and that, between collector and emitter as VCC., Let us see now the paths of current carriers in, the transistor with emitter-base junction forward, biased and base-collector junction reverse biased., The heavily doped emitter has a high concentration, of majority carriers, which will be holes in a p-n-p, transistor and electrons in an n-p-n transistor., These majority carriers enter the base region in, large numbers. The base is thin and lightly doped., So the majority carriers there would be few. In a, p-n-p transistor the majority carriers in the base, are electrons since base is of n-type semiconductor., The large number of holes entering the base from, the emitter swamps the small number of electrons, there. As the base collector-junction is reversebiased, these holes, which appear as minority, carriers at the junction, can easily cross the, junction and enter the collector. The holes in the, base could move either towards the base terminal, to combine with the electrons entering from outside, or cross the junction to enter into the collector and, reach the collector terminal. The base is made thin, so that most of the holes find themselves near, the reverse-biased base-collector junction and so, cross the junction instead of moving to the base, FIGURE 14.28 Bias Voltage applied on: (a), terminal., p-n-p transistor and (b) n-p-n transistor., It is interesting to note that due to forward, bias a large current enters the emitter -base, junction, but most of it is diverted to adjacent reverse-biased base-collector, junction and the current coming out of the base becomes a very small, fraction of the current that entered the junction. If we represent the hole, current and the electron current crossing the forward biased junction by, Ih and Ie respectively then the total current in a forward biased diode is, the sum Ih + Ie. We see that the emitter current IE = Ih + Ie but the base, current IB << Ih + Ie, because a major part of IE goes to collector instead of, coming out of the base terminal. The base current is thus a small fraction, of the emitter current., The current entering into the emitter from outside is equal to the, emitter current IE. Similarly the current emerging from the base terminal, is IB and that from collector terminal is IC. It is obvious from the above, description and also from a straight forward application of Kirchhoff’s, law to Fig. 14.28(a) that the emitter current is the sum of collector current, 492, and base current:

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, IE = IC + IB, , (14.7), , We also see that IC | IE., Our description of the direction of motion of the holes is identical, with the direction of the conventional current. But the direction of motion, of electrons is just opposite to that of the current. Thus in a p-n-p, transistor the current enters from emitter into base whereas in a n-p-n, transistor it enters from the base into the emitter. The arrowhead in the, emitter shows the direction of the conventional current., The description about the paths followed by the majority and minority, carriers in a n-p-n is exactly the same as that for the p-n-p transistor., But the current paths are exactly opposite, as shown in Fig. 14.28. In, Fig. 14.28(b) the electrons are the majority carriers supplied by the, n-type emitter region. They cross the thin p-base region and are able to, reach the collector to give the collector current, IC . From the above, description we can conclude that in the active state of the transistor the, emitter-base junction acts as a low resistance while the base collector, acts as a high resistance., , 14.9.2 Basic transistor circuit configurations and transistor, characteristics, In a transistor, only three terminals are available, viz., Emitter (E), Base, (B) and Collector (C). Therefore, in a circuit the input/output connections, have to be such that one of these (E, B or C) is common to both the input, and the output. Accordingly, the transistor can be connected in either of, the following three configurations:, Common Emitter (CE), Common Base (CB), Common Collector (CC), The transistor is most widely used in the CE configuration and we, shall restrict our discussion to only this configuration. Since more, commonly used transistors are n-p-n Si transistors, we shall confine, our discussion to such transistors only. With p-n-p transistors the, polarities of the external power supplies are to be inverted., Common emitter transistor characteristics, When a transistor is used in CE configuration, the input, is between the base and the emitter and the output is, between the collector and the emitter. The variation of, the base current IB with the base-emitter voltage VBE is, called the input characteristic. Similarly, the variation, of the collector current IC with the collector-emitter, voltage VCE is called the output characteristic. You will, see that the output characteristics are controlled by, the input characteristics. This implies that the collector, current changes with the base current., The input and the output characteristics of an, n-p-n transistors can be studied by using the circuit, shown in Fig. 14.29., To study the input characteristics of the transistor, in CE configuration, a curve is plotted between the base, current IB against the base-emitter voltage VBE. The, , FIGURE 14.29 Circuit arrangement, for studying the input and output, characteristics of n-p-n transistor in, CE configuration., , 493

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Physics, collector-emitter voltage VCE is kept fixed while, studying the dependence of IB on VBE. We are, interested to obtain the input characteristic, when the transistor is in active state. So the, collector -emitter voltage V CE is kept large, enough to make the base collector junction, reverse biased. Since VCE = VCB + VBE and for Si, transistor VBE is 0.6 to 0.7 V, VCE must be, sufficiently larger than 0.7 V. Since the, transistor is operated as an amplifier over large, range of VCE, the reverse bias across the basecollector junction is high most of the time., Therefore, the input characteristics may be, obtained for VCE somewhere in the range of 3 V, to 20 V. Since the increase in VCE appears as, increase in VCB, its effect on IB is negligible. As, a consequence, input characteristics for various, values of VCE will give almost identical curves., Hence, it is enough to determine only one input, characteristics. The input characteristics of a, transistor is as shown in Fig. 14.30(a)., The output characteristic is obtained by, observing the variation of IC as VCE is varied, keeping IB constant. It is obvious that if VBE is, increased by a small amount, both hole current, from the emitter region and the electron current, from the base region will increase. As a, consequence both I B and I C will increase, FIGURE 14.30 (a) Typical input, proportionately. This shows that when I B, characteristics, and (b) Typical output, increases IC also increases. The plot of IC versus, characteristics., VCE for different fixed values of IB gives one, output characteristic. So there will be different output characteristics, corresponding to different values of IB as shown in Fig. 14.30(b)., The linear segments of both the input and output characteristics can, be used to calculate some important ac parameters of transistors as, shown below., (i) Input resistance (ri ): This is defined as the ratio of change in baseemitter voltage ('VBE) to the resulting change in base current ('IB ) at, constant collector-emitter voltage (VCE ). This is dynamic (ac resistance), and as can be seen from the input characteristic, its value varies with, the operating current in the transistor:, ri, , § 'VBE ·, ¨© 'I ¸¹, B, V, , (14.8), , CE, , 494, , The value of ri can be anything from a few hundreds to a few thousand, ohms.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, (ii) Output resistance (ro): This is defined as the ratio of change in, collector-emitter voltage ('VCE) to the change in collector current ('IC), at a constant base current IB., ro, , § 'VCE ·, ¨© 'I ¸¹, C, , (14.9), IB, , The output characteristics show that initially for very small values of, VCE , IC increases almost linearly. This happens because the base-collector, junction is not reverse biased and the transistor is not in active state. In, fact, the transistor is in the saturation state and the current is controlled, by the supply voltage VCC (=VCE) in this part of the characteristic. When, VCE is more than that required to reverse bias the base-collector junction,, IC increases very little with VCE. The reciprocal of the slope of the linear, part of the output characteristic gives the values of ro. The output, resistance of the transistor is mainly controlled by the bias of the basecollector junction. The high magnitude of the output resistance (of the, order of 100 k:) is due to the reverse-biased state of this diode. This, also explains why the resistance at the initial part of the characteristic,, when the transistor is in saturation state, is very low., (iii) Current amplification factor (E ): This is defined as the ratio of, the change in collector current to the change in base current at a, constant collector-emitter voltage (VCE) when the transistor is in, active state., , Eac, , § 'I C ·, ¨© 'I ¸¹, B, , (14.10), VCE, , This is also known as small signal current gain and its value is very, large., If we simply find the ratio of IC and IB we get what is called dc Eof the, transistor. Hence,, IC, (14.11), IB, Since IC increases with IB almost linearly and IC = 0 when IB = 0, the values, of both Edc and Eac are nearly equal. So, for most calculations Edc can be, used. Both Eac and Edc vary with VCE and IB (or IC) slightly., , Edc, , Example 14.8 From the output characteristics shown in Fig. 14.30(b),, calculate the values of E ac and E dc of the transistor when VCE is, 10 V and IC = 4.0 mA., Solution, , Eac, , B, , , Edc, VCE, , IC, IB, , For determining Eac and Edc at the stated values of VCE and IC one can, proceed as follows. Consider any two characteristics for two values, of IB which lie above and below the given value of IC . Here IC = 4.0 mA., (Choose characteristics for IB= 30 and 20 PA.) At VCE = 10 V we read, the two values of IC from the graph. Then, , EXAMPLE 14.8, , § 'I C ·, ¨© 'I ¸¹, , 495

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Physics, , EXAMPLE 14.8, , 'IB, , (30 – 20) PA = 10 PA, 'IC, , (4.5 – 3.0) mA = 1.5 mA, , Therefore, Eac = 1.5 mA/ 10 PA = 150, For determining Edc, either estimate the value of IB corresponding to, IC = 4.0 mA at VCE = 10 V or calculate the two values of Edc for the two, characteristics chosen and find their mean., Therefore, for IC = 4.5 mA and IB = 30 PA,, Edc = 4.5 mA/ 30 PA = 150, and for IC = 3.0 mA and IB = 20 PA, Edc =3.0 mA / 20 PA = 150, Hence, Edc =(150 + 150) /2 = 150, , 14.9.3 Transistor as a device, The transistor can be used as a device application depending on the, configuration used (namely CB, CC and CE), the biasing of the E-B and, B-C junction and the operation region namely cutoff, active region and, saturation. As mentioned earlier we have confined only to the CE, configuration and will be concentrating on the biasing and the operation, region to understand the working of a device., When the transistor is used in the cutoff, or saturation state it acts as a switch. On, the other hand for using the transistor as, an amplifier, it has to operate in the active, region., (i) Transistor as a switch, We shall try to understand the operation of, the transistor as a switch by analysing the, behaviour of the base-biased transistor in, CE configuration as shown in Fig. 14.31(a)., Applying Kirchhoff’s voltage rule to the, input and output sides of this circuit, we, get, VBB = IBRB + VBE, (14.12), and, VCE = VCC – ICRC., (14.13), We shall treat VBB as the dc input, voltage Vi and VCE as the dc output voltage, VO. So, we have, Vi = IBRB + VBE and, Vo = VCC – ICRC., Let us see how V o changes as V i, increases from zero onwards. In the case, of Si transistor, as long as input Vi is less, than 0.6 V, the transistor will be in cut off, FIGURE 14.31 (a) Base-biased transistor in CE, state and current IC will be zero., configuration, (b) Transfer characteristic., Hence Vo = VCC, When Vi becomes greater than 0.6 V the transistor is in active state with, 496, some current IC in the output path and the output Vo decrease as the

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, term ICRC increases. With increase of Vi , IC increases almost linearly, and so V o decreases linearly till its value becomes less than, about 1.0 V., Beyond this, the change becomes non linear and transistor goes into, saturation state. With further increase in Vi the output voltage is found to, decrease further towards zero though it may never become zero. If we plot, the Vo vs Vi curve, [also called the transfer characteristics of the base-biased, transistor (Fig. 14.31(b)], we see that between cut off state and active state, and also between active state and saturation state there are regions of, non-linearity showing that the transition from cutoff state to active state, and from active state to saturation state are not sharply defined., Let us see now how the transistor is operated as a switch. As long as, Vi is low and unable to forward-bias the transistor, Vo is high (at VCC ). If, Vi is high enough to drive the transistor into saturation, then Vo is low,, very near to zero. When the transistor is not conducting it is said to be, switched off and when it is driven into saturation it is said to be switched, on. This shows that if we define low and high states as below and above, certain voltage levels corresponding to cutoff and saturation of the, transistor, then we can say that a low input switches the transistor off, and a high input switches it on. Alternatively, we can say that a low, input to the transistor gives a high output and a high input gives a low, output. The switching circuits are designed in such a way that the, transistor does not remain in active state., (ii) Transistor as an amplifier, For using the transistor as an amplifier we will use the active region of, the Vo versus Vi curve. The slope of the linear part of the curve represents, the rate of change of the output with the input. It is negative because the, output is VCC – ICRC and not ICRC. That is why as input voltage of the CE, amplifier increases its output voltage decreases and the output is said to, be out of phase with the input. If we consider 'Vo and'Vi as small, changes in the output and input voltages then 'Vo/'Vi is called the small, signal voltage gain AV of the amplifier., If the VBB voltage has a fixed value corresponding to the mid point of, the active region, the circuit will behave as a CE amplifier with voltage, gain 'Vo/'Vi . We can express the voltage gain AV in terms of the resistors, in the circuit and the current gain of the transistor as follows., We have, Vo = VCC – ICRC, Therefore, 'Vo = 0 – RC ' IC, Similarly, from Vi = IBRB + VBE, 'Vi = RB'IB + 'VBE, But 'VBE is negligibly small in comparison to 'IBRB in this circuit., So, the voltage gain of this CE amplifier (Fig. 14.32) is given by, AV = –RC ' IC /RB 'IB, = –Eac(RC /RB ), (14.14), where Eac is equal to ' IC/'IB from Eq. (14.10). Thus the linear, portion of the active region of the transistor can be exploited for the use, in amplifiers. Transistor as an amplifier (CE configuration) is discussed, in detail in the next section., , 497

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Physics, 14.9.4 Transistor as an Amplifier (CE-Configuration), To operate the transistor as an amplifier it is necessary to fix its operating, point somewhere in the middle of its active region. If we fix the value of, VBB corresponding to a point in the middle of the linear part of the transfer, curve then the dc base current IB would be constant and corresponding, collector current IC will also be constant. The dc voltage VCE = VCC - ICRC, would also remain constant. The operating values of VCE and IB determine, the operating point, of the amplifier., If a small sinusoidal voltage with amplitude vs is superposed on, the dc base bias by connecting the source of that signal in series with the, VBB supply, then the base current will have sinusoidal variations, superimposed on the value of IB. As a consequence the collector current, also will have sinusoidal variations, superimposed on the value of IC, producing, in turn corresponding change in the value, of VO. We can measure the ac variations, across the input and output terminals by, blocking the dc voltages by large capacitors., In the discription of the amplifier given, above we have not considered any ac signal., In general, amplifiers are used to amplify, alternating signals. Now let us superimpose, an ac input signal vi (to be amplified) on the, FIGURE 14.32 A simple circuit of a, bias VBB (dc) as shown in Fig. 14.32. The, CE-transistor amplifier., output is taken between the collector and, the ground., The working of an amplifier can be easily understood, if we first, assume that vi = 0. Then applying Kirchhoff’s law to the output loop,, we get, Vcc = VCE + IcRL, Likewise, the input loop gives, , (14.15), , VBB = VBE + IB RB, , (14.16), , When vi is not zero, we get, VBE + vi = VBE + IB RB + 'IB (RB + ri ), The change in VBE can be related to the input resistance ri [see, Eq. (14.8)] and the change in IB. Hence, vi = 'IB (RB + ri ), , 498, , = r 'IB, The change in IB causes a change in Ic. We define a parameter Eac,, which is similar to the Edc defined in Eq. (14.11), as, 'I c i c, Eac, (14.17), 'I B i b, which is also known as the ac current gain Ai . Usually Eac is close to Edc, in the linear region of the output characteristics., The change in Ic due to a change in IB causes a change in VCE and the, voltage drop across the resistor RL because VCC is fixed.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, These changes can be given by Eq. (14.15) as, 'VCC = 'VCE + RL 'IC = 0, or 'VCE = –RL 'IC, The change in VCE is the output voltage v0. From Eq. (14.10), we get, v0 = 'VCE = –Eac RL 'IB, The voltage gain of the amplifier is, Av, , v0, vi, –, , 'VCE, r 'I B, , Eac R L, , (14.18), r, The negative sign represents that output voltage is opposite with phase, with the input voltage., From the discussion of the transistor characteristics you have seen, that there is a current gain Eac in the CE configuration. Here we have also, seen the voltage gain Av. Therefore the power gain Ap can be expressed, as the product of the current gain and voltage gain. Mathematically, (14.19), Ap = Eac × Av, Since Eac and Av are greater than 1, we get ac power gain. However it, should be realised that transistor is not a power generating device. The, energy for the higher ac power at the output is supplied by the battery., Example 14.9 In Fig. 14.31(a), the VBB supply can be varied from 0V, to 5.0 V. The Si transistor has Edc = 250 and RB = 100 k:, RC = 1 K:,, VCC = 5.0V. Assume that when the transistor is saturated, VCE = 0V, and VBE = 0.8V. Calculate (a) the minimum base current, for which, the transistor will reach saturation. Hence, (b) determine V1 when, the transistor is ‘switched on’. (c) find the ranges of V1 for which the, transistor is ‘switched off’ and ‘switched on’., , Note that the transistor is in active state when IB varies from 0.0mA, to 20mA. In this range, IC = EIB is valid. In the saturation range,, IC d EIB., , EXAMPLE 14.9, , Solution, Given at saturation VCE = 0V, VBE = 0.8V, VCE = VCC – ICRC, IC = VCC/RC = 5.0V/1.0k: = 5.0 mA, Therefore IB = IC/E = 5.0 mA/250 = 20PA, The input voltage at which the transistor will go into saturation is, given by, VIH = VBB = IBRB +VBE, = 20PA × 100 k: + 0.8V = 2.8V, The value of input voltage below which the transistor remains cutoff, is given by, VIL = 0.6V, VIH = 2.8V, Between 0.0V and 0.6V, the transistor will be in the ‘switched off’, state. Between 2.8V and 5.0V, it will be in ‘switched on’ state., , 499

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Physics, , EXAMPLE 14.10, , Example 14.10 For a CE transistor amplifier, the audio signal voltage, across the collector resistance of 2.0 k: is 2.0 V. Suppose the current, amplification factor of the transistor is 100, What should be the value, of RB in series with VBB supply of 2.0 V if the dc base current has to be, 10 times the signal current. Also calculate the dc drop across the, collector resistance. (Refer to Fig. 14.33)., Solution The output ac voltage is 2.0 V. So, the ac collector current, iC = 2.0/2000 = 1.0 mA. The signal current through the base is,, therefore given by iB = iC /E = 1.0 mA/100 = 0.010 mA. The dc base, current has to be 10× 0.010 = 0.10 mA., From Eq.14.16, RB = (VBB - VBE ) /IB. Assuming VBE = 0.6 V,, RB = (2.0 – 0.6 )/0.10 = 14 k:, The dc collector current IC = 100×0.10 = 10 mA., , 14.9.5 Feedback amplifier and transistor oscillator, In an amplifier, we have seen that a sinusoidal input is given which appears, as an amplified signal in the output. This means that an external input is, necessary to sustain ac signal in the, output for an amplifier. In an oscillator, we, get ac output without any external input, signal. In other words, the output in an, oscillator is self-sustained. To attain this,, an amplifier is taken. A portion of the, output power is returned back (feedback), to the input in phase with the starting, power (this process is termed positive, feedback) as shown in Fig. 14.33(a). The, feedback can be achieved by inductive, coupling (through mutual inductance) or, LC or RC networks. Different types of, oscillators essentially use different methods, of coupling the output to the input, (feedback network), apart from the resonant, circuit for obtaining oscillation at a, particular frequency. For understanding, the oscillator action, we consider the circuit, shown in Fig. 14.33(b) in which the, feedback is accomplished by inductive, coupling from one coil winding (T 1) to, another coil winding (T2). Note that the coils, T2 and T1 are wound on the same core and, hence are inductively coupled through their, mutual inductance. As in an amplifier, the, base-emitter junction is forward biased, FIGURE 14.33 (a) Principle of a transistor, while the base-collector junction is reverse, amplifier with positive feedback working as an, biased. Detailed biasing circuits actually, oscillator and (b) Tuned collector oscillator, (c) Rise, used have been omitted for simplicity., and fall (or built up) of current Ic and Ie due to the, Let us try to understand how oscillations, inductive coupling., 500, are built. Suppose switch S1 is put on to

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, apply proper bias for the first time. Obviously, a surge of collector current, flows in the transistor. This current flows through the coil T2 where, terminals are numbered 3 and 4 [Fig. 14.33(b)]. This current does not, reach full amplitude instantaneously but increases from X to Y, as shown, in Fig. [14.33(c)(i)]. The inductive coupling between coil T2 and coil T1, now causes a current to flow in the emitter circuit (note that this actually, is the ‘feedback’ from input to output). As a result of this positive feedback,, this current (in T1; emitter current) also increases from X´ to Y´ [Fig., 14.33(c)(ii)]. The current in T2 (collector current) connected in the collector, circuit acquires the value Y when the transistor becomes saturated. This, means that maximum collector current is flowing and can increase no, further. Since there is no further change in collector current, the magnetic, field around T2 ceases to grow. As soon as the field becomes static, there, will be no further feedback from T2 to T1. Without continued feedback,, the emitter current begins to fall. Consequently, collector current decreases, from Y towards Z [Fig. 14.33(c)(i)]. However, a decrease of collector current, causes the magnetic field to decay around the coil T2. Thus, T1 is now, seeing a decaying field in T2 (opposite from what it saw when the field was, growing at the initial start operation). This causes a further decrease in, the emitter current till it reaches Zcwhen the transistor is cut-off. This, means that both IE and IC cease to flow. Therefore, the transistor has, reverted back to its original state (when the power was first switched on)., The whole process now repeats itself. That is, the transistor is driven to, saturation, then to cut-off, and then back to saturation. The time for, change from saturation to cut-off and back is determined by the constants, of the tank circuit or tuned circuit (inductance L of coil T2 and C connected, in parallel to it). The resonance frequency (Q ) of this tuned circuit, determines the frequency at which the oscillator will oscillate., , § 1 ·, (14.20), ¨©, ¸, 2ʌ LC ¹, In the circuit of Fig. 14.33(b), the tank or tuned circuit is connected, in the collector side. Hence, it is known as tuned collector oscillator. If the, tuned circuit is on the base side, it will be known as tuned base oscillator., There are many other types of tank circuits (say RC) or feedback circuits, giving different types of oscillators like Colpitt’s oscillator, Hartley, oscillator, RC-oscillator., , Q, , 14.10 DIGITAL ELECTRONICS, , AND, , LOGIC GATES, , In electronics circuits like amplifiers, oscillators, introduced to you in, earlier sections, the signal (current or voltage) has been in the form of, continuous, time-varying voltage or current. Such signals are called, continuous or analogue signals. A typical analogue signal is shown in, Figure. 14.34(a). Fig. 14.34(b) shows a pulse waveform in which only, discrete values of voltages are possible. It is convenient to use binary, numbers to represent such signals. A binary number has only two digits, ‘0’ (say, 0V) and ‘1’ (say, 5V). In digital electronics we use only these two, levels of voltage as shown in Fig. 14.34(b). Such signals are called Digital, Signals. In digital circuits only two values (represented by 0 or 1) of the, input and output voltage are permissible., , 501

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Physics, This section is intended to provide the first step in our understanding, of digital electronics. We shall restrict our study to some basic building, blocks of digital electronics (called Logic Gates) which process the digital, signals in a specific manner. Logic gates are used in calculators, digital, watches, computers, robots, industrial control systems, and in, telecommunications., A light switch in your house can be used as an example of a digital, circuit. The light is either ON or OFF depending on the switch position., When the light is ON, the output value is ‘1’. When the light is OFF the, output value is ‘0’. The inputs are the position of the light switch. The, switch is placed either in the ON or OFF position to activate the light., , FIGURE 14.34 (a) Analogue signal, (b) Digital signal., , 14.10.1 Logic gates, , Input, , Output, , A, , Y, , 0, , 1, , 1, , 0, , (b), FIGURE 14.35, (a) Logic symbol,, (b) Truth table of, NOT gate., , 502, , A gate is a digital circuit that follows curtain logical relationship, between the input and output voltages. Therefore, they are generally, known as logic gates — gates because they control the flow of, information. The five common logic gates used are NOT, AND, OR,, NAND, NOR. Each logic gate is indicated by a symbol and its function, is defined by a truth table that shows all the possible input logic level, combinations with their respective output logic levels. Truth tables, help understand the behaviour of logic gates. These logic gates can, be realised using semiconductor devices., (i) NOT gate, This is the most basic gate, with one input and one output. It produces, a ‘1’ output if the input is ‘0’ and vice-versa. That is, it produces an, inverted version of the input at its output. This is why it is also known, as an inverter. The commonly used symbol together with the truth, table for this gate is given in Fig. 14.35., (ii) OR Gate, An OR gate has two or more inputs with one output. The logic symbol, and truth table are shown in Fig. 14.36. The output Y is 1 when either, input A or input B or both are 1s, that is, if any of the input is high, the, output is high.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, Input, , Output, , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 1, (b), , FIGURE 14.36 (a) Logic symbol (b) Truth table of OR gate., , Apart from carrying out the above mathematical logic operation, this, gate can be used for modifying the pulse waveform as explained in the, following example., Example 14.11 Justify the output waveform (Y) of the OR gate for, the following inputs A and B given in Fig. 14.37., Solution Note the following:, A = 0, B = 0;, Hence Y = 0, x At t < t1;, x For t1 to t2; A = 1, B = 0;, Hence Y = 1, x For t2 to t3; A = 1, B = 1;, Hence Y = 1, Hence Y = 1, x For t3 to t4; A = 0, B = 1;, x For t4 to t5; A = 0, B = 0;, Hence Y = 0, x For t5 to t6; A = 1, B = 0;, Hence Y = 1, Hence Y = 1, x For t > t6; A = 0, B = 1;, Therefore the waveform Y will be as shown in the Fig. 14.37., , (iii) AND Gate, An AND gate has two or more inputs and one output. The output Y of, AND gate is 1 only when input A and input B are both 1. The logic, symbol and truth table for this gate are given in Fig. 14.38, , FIGURE 14.38 (a) Logic symbol, (b) Truth table of AND gate., , EXAMPLE 14.11, , FIGURE 14.37, , Input, , Output, , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , (b), , 503

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Physics, Example 14.12 Take A and B input waveforms similar to that in, Example 14.11. Sketch the output waveform obtained from AND gate., , EXAMPLE 14.12, , Solution, A = 0, B = 0;, x For t d t1;, A = 1, B = 0;, x For t1 to t2;, x For t2 to t3;, A = 1, B = 1;, x For t3 to t4;, A = 0, B = 1;, A = 0, B = 0;, x For t4 to t5;, x For t5 to t6;, A = 1, B = 0;, x For t > t6;, A = 0, B = 1;, Based on the above, the output waveform for AND gate, as given below., , Hence Y = 0, Hence Y = 0, Hence Y = 1, Hence Y = 0, Hence Y = 0, Hence Y = 0, Hence Y = 0, can be drawn, , FIGURE 14.39, , (iv) NAND Gate, This is an AND gate followed by a NOT gate. If inputs A and B are both, ‘1’, the output Y is not ‘1’. The gate gets its name from this NOT AND, behaviour. Figure 14.40 shows the symbol and truth table of NAND gate., NAND gates are also called Universal Gates since by using these, gates you can realise other basic gates like OR, AND and NOT (Exercises, 14.16 and 14.17)., Input, Output, A, , B, , Y, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 0, , (b), , 504, , EXAMPLE 14.13, , FIGURE 14.40 (a) Logic symbol, (b) Truth table of NAND gate., Example 14.13 Sketch the output Y from a NAND gate having inputs, A and B given below:, Solution, x For t < t1;, x For t1 to t2;, x For t2 to t3;, x For t3 to t4;, , A, A, A, A, , =, =, =, =, , 1,, 0,, 0,, 1,, , B, B, B, B, , =, =, =, =, , 1;, 0;, 1;, 0;, , Hence, Hence, Hence, Hence, , Y, Y, Y, Y, , =, =, =, =, , 0, 1, 1, 1

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, x For t4 to t5;, x For t5 to t6;, x For t > t6;, , A = 1, B = 1;, A = 0, B = 0;, A = 0, B = 1;, , Hence Y = 0, Hence Y = 1, Hence Y = 1, , EXAMPLE 14.13, , FIGURE 14.41, , (v) NOR Gate, It has two or more inputs and one output. A NOT- operation applied, after OR gate gives a NOT-OR gate (or simply NOR gate). Its output Y is, ‘1’ only when both inputs A and B are ‘0’, i.e., neither one input nor the, other is ‘1’. The symbol and truth table for NOR gate is given in, Fig. 14.42., Input, Output, A, , B, , Y, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 0, , (b), , FIGURE 14.42 (a) Logic symbol, (b) Truth table of NOR gate., , NOR gates are considered as universal gates because you can obtain, all the gates like AND, OR, NOT by using only NOR gates (Exercises 14.18, and 14.19)., , 14.11 INTEGRATED CIRCUITS, The conventional method of making circuits is to choose components, like diodes, transistor, R, L, C etc., and connect them by soldering wires, in the desired manner. Inspite of the miniaturisation introduced by the, discovery of transistors, such circuits were still bulky. Apart from this,, such circuits were less reliable and less shock proof. The concept of, fabricating an entire circuit (consisting of many passive components like, R and C and active devices like diode and transistor) on a small single, block (or chip) of a semiconductor has revolutionised the electronics, technology. Such a circuit is known as Integrated Circuit (IC). The most, widely used technology is the Monolithic Integrated Circuit. The word, , 505

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Physics, monolithic is a combination of two greek words, monos, means single and lithos means stone. This, in effect,, means that the entire circuit is formed on a single, silicon crystal (or chip). The chip dimensions are as, small as 1mm × 1mm or it could even be smaller. Figure, 14.43 shows a chip in its protective plastic case, partly, removed to reveal the connections coming out from the, ‘chip’ to the pins that enable it to make external, connections., FIGURE 14.43 The casing and, Depending on nature of input signals, IC’s can be, connection of a ‘chip’., grouped in two categories: (a) linear or analogue IC’s, and (b) digital IC’s. The linear IC’s process analogue, signals which change smoothly and continuously over a range of values, between a maximum and a minimum. The output is more or less directly, proportional to the input, i.e., it varies linearly with the input. One of the, most useful linear IC’s is the operational amplifier., The digital IC’s process signals that have only two values. They, contain circuits such as logic gates. Depending upon the level of, integration (i.e., the number of circuit components or logic gates), the ICs, are termed as Small Scale Integration, SSI (logic gates < 10); Medium, Scale Integration, MSI (logic gates < 100); Large Scale Integration, LSI, (logic gates < 1000); and Very Large Scale Integration, VLSI (logic gates >, 1000). The technology of fabrication is very involved but large scale, industrial production has made them very inexpensive., , FASTER, , AND SMALLER: THE FUTURE OF COMPUTER TECHNOLOGY, , The Integrated Chip (IC) is at the heart of all computer systems. In fact ICs are found in, almost all electrical devices like cars, televisions, CD players, cell phones etc. The, miniaturisation that made the modern personal computer possible could never have, happened without the IC. ICs are electronic devices that contain many transistors, resistors,, capacitors, connecting wires – all in one package. You must have heard of the, microprocessor. The microprocessor is an IC that processes all information in a computer,, like keeping track of what keys are pressed, running programmes, games etc. The IC was, first invented by Jack Kilky at Texas Instruments in 1958 and he was awarded Nobel Prize, for this in 2000. ICs are produced on a piece of semiconductor crystal (or chip) by a process, called photolithography. Thus, the entire Information Technology (IT) industry hinges on, semiconductors. Over the years, the complexity of ICs has increased while the size of its, features continued to shrink. In the past five decades, a dramatic miniaturisation in, computer technology has made modern day computers faster and smaller. In the 1970s,, Gordon Moore, co-founder of INTEL, pointed out that the memory capacity of a chip (IC), approximately doubled every one and a half years. This is popularly known as Moore’s, law. The number of transistors per chip has risen exponentially and each year computers, are becoming more powerful, yet cheaper than the year before. It is intimated from current, trends that the computers available in 2020 will operate at 40 GHz (40,000 MHz) and, would be much smaller, more efficient and less expensive than present day computers., The explosive growth in the semiconductor industry and computer technology is best, expressed by a famous quote from Gordon Moore: “If the auto industry advanced as rapidly, as the semiconductor industry, a Rolls Royce would get half a million miles per gallon, and, it would be cheaper to throw it away than to park it”., , 506

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, SUMMARY, 1., 2., , 3., , 4., 5., , 6., , 7., 8., , 9., 10., , 11., 12., , 13., , 14., , 15., , Semiconductors are the basic materials used in the present solid state, electronic devices like diode, transistor, ICs, etc., Lattice structure and the atomic structure of constituent elements, decide whether a particular material will be insulator, metal or, semiconductor., Metals have low resistivity (10–2 to 10–8 : m), insulators have very high, resistivity (>108 : m–1), while semiconductors have intermediate values, of resistivity., Semiconductors are elemental (Si, Ge) as well as compound (GaAs,, CdS, etc.)., Pure semiconductors are called ‘intrinsic semiconductors’. The presence, of charge carriers (electrons and holes) is an ‘intrinsic’ property of the, material and these are obtained as a result of thermal excitation. The, number of electrons (ne ) is equal to the number of holes (nh ) in intrinsic, conductors. Holes are essentially electron vacancies with an effective, positive charge., The number of charge carriers can be changed by ‘doping’ of a suitable, impurity in pure semiconductors. Such semiconductors are known as, extrinsic semiconductors. These are of two types (n-type and p-type)., In n-type semiconductors, ne >> nh while in p-type semiconductors nh >> ne., n-type semiconducting Si or Ge is obtained by doping with pentavalent, atoms (donors) like As, Sb, P, etc., while p-type Si or Ge can be obtained, by doping with trivalent atom (acceptors) like B, Al, In etc., nenh = ni2 in all cases. Further, the material possesses an overall charge, neutrality., There are two distinct band of energies (called valence band and, conduction band) in which the electrons in a material lie. Valence, band energies are low as compared to conduction band energies. All, energy levels in the valence band are filled while energy levels in the, conduction band may be fully empty or partially filled. The electrons in, the conduction band are free to move in a solid and are responsible for, the conductivity. The extent of conductivity depends upon the energy, gap (Eg ) between the top of valence band (EV ) and the bottom of the, conduction band EC. The electrons from valence band can be excited by, heat, light or electrical energy to the conduction band and thus, produce, a change in the current flowing in a semiconductor., For insulators Eg > 3 eV, for semiconductors Eg is 0.2 eV to 3 eV, while, for metals Eg | 0., p-n junction is the ‘key’ to all semiconductor devices. When such a, junction is made, a ‘depletion layer’ is formed consisting of immobile, ion-cores devoid of their electrons or holes. This is responsible for a, junction potential barrier., By changing the external applied voltage, junction barriers can be, changed. In forward bias (n-side is connected to negative terminal of the, battery and p-side is connected to the positive), the barrier is decreased, while the barrier increases in reverse bias. Hence, forward bias current, is more (mA) while it is very small (PA) in a p-n junction diode., Diodes can be used for rectifying an ac voltage (restricting the ac voltage, to one direction). With the help of a capacitor or a suitable filter, a dc, voltage can be obtained., There are some special purpose diodes., , 507

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Physics, 16. Zener diode is one such special purpose diode. In reverse bias, after a, certain voltage, the current suddenly increases (breakdown voltage) in, a Zener diode. This property has been used to obtain voltage regulation., 17. p-n junctions have also been used to obtain many photonic or, optoelectronic devices where one of the participating entity is ‘photon’:, (a) Photodiodes in which photon excitation results in a change of reverse, saturation current which helps us to measure light intensity; (b) Solar, cells which convert photon energy into electricity; (c) Light Emitting, Diode and Diode Laser in which electron excitation by a bias voltage, results in the generation of light., 18. Transistor is an n-p-n or p-n-p junction device. The central block, (thin and lightly doped) is called ‘Base’ while the other electrodes are, ‘Emitter’ and ‘Collectors’. The emitter-base junction is forward biased, while collector-base junction is reverse biased., 19. The transistors can be connected in such a manner that either C or E, or B is common to both the input and output. This gives the three, configurations in which a transistor is used: Common Emitter (CE),, Common Collector (CC) and Common Base (CB). The plot between IC, and VCE for fixed IB is called output characteristics while the plot between, IB and VBE with fixed VCE is called input characteristics. The important, transistor parameters for CE-configuration are:, , § 'VBE ·, ¨© 'I ¸¹, B, VCE, § 'VCE ·, output resistance, ro ¨, © 'I C ¸¹ I, input resistance, ri, , current amplification factor, E, , B, , § 'I C ·, ¨© 'I ¸¹, B, , VCE, , 20. Transistor can be used as an amplifier and oscillator. In fact, an, oscillator can also be considered as a self-sustained amplifier in which, a part of output is fed-back to the input in the same phase (positive, feed back). The voltage gain of a transistor amplifier in common emitter, , 21., 22., 23., 24., , § vo ·, ¨© v ¸¹, , RC, , where RC and RB are respectively, RB, i, the resistances in collector and base sides of the circuit., When the transistor is used in the cutoff or saturation state, it acts as, a switch., There are some special circuits which handle the digital data consisting, of 0 and 1 levels. This forms the subject of Digital Electronics., The important digital circuits performing special logic operations are, called logic gates. These are: OR, AND, NOT, NAND, and NOR gates., In modern day circuit, many logical gates or circuits are integrated in, one single ‘Chip’. These are known as Intgrated circuits (IC)., configuration is: Av, , E, , POINTS TO PONDER, 1., , 508, , The energy bands (EC or EV ) in the semiconductors are space delocalised, which means that these are not located in any specific place inside the, solid. The energies are the overall averages. When you see a picture in, which EC or EV are drawn as straight lines, then they should be, respectively taken simply as the bottom of conduction band energy levels, and top of valence band energy levels.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, 2., , 3., , 4., , 5., , In elemental semiconductors (Si or Ge), the n-type or p-type, semiconductors are obtained by introducing ‘dopants’ as defects. In, compound semiconductors, the change in relative stoichiometric ratio, can also change the type of semiconductor. For example, in ideal GaAs, the ratio of Ga:As is 1:1 but in Ga-rich or As-rich GaAs it could, respectively be Ga1.1 As0.9 or Ga0.9 As1.1. In general, the presence of, defects control the properties of semiconductors in many ways., In transistors, the base region is both narrow and lightly doped,, otherwise the electrons or holes coming from the input side (say, emitter, in CE-configuration) will not be able to reach the collector., We have described an oscillator as a positive feedback amplifier. For, stable oscillations, the voltage feedback (Vfb) from the output voltage, (Vo) should be such that after amplification (A) it should again become, Vo. If a fraction Ec is feedback, then Vfb = Vo. Ec and after amplification, its value A(vo.Ec) should be equal to Vo. This means that the criteria for, stable oscillations to be sustained is A Ec = 1. This is known as, Barkhausen's Criteria., In an oscillator, the feedback is in the same phase (positive feedback)., If the feedback voltage is in opposite phase (negative feedback), the, gain is less than 1 and it can never work as oscillator. It will be an, amplifier with reduced gain. However, the negative feedback also reduces, noise and distortion in an amplifier which is an advantageous feature., , EXERCISES, 14.1, , 14.2, 14.3, , 14.4, , In an n-type silicon, which of the following statement is true:, (a) Electrons are majority carriers and trivalent atoms are the, dopants., (b) Electrons are minority carriers and pentavalent atoms are the, dopants., (c) Holes are minority carriers and pentavalent atoms are the, dopants., (d) Holes are majority carriers and trivalent atoms are the dopants., Which of the statements given in Exercise 14.1 is true for p-type, semiconductos., Carbon, silicon and germanium have four valence electrons each., These are characterised by valence and conduction bands separated, by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which, of the following statements is true?, (a) (Eg)Si < (Eg)Ge < (Eg)C, (b) (Eg)C < (Eg)Ge > (Eg)Si, (c) (Eg)C > (Eg)Si > (Eg)Ge, (d) (Eg)C = (Eg)Si = (Eg)Ge, In an unbiased p-n junction, holes diffuse from the p-region to, n-region because, (a) free electrons in the n-region attract them., (b) they move across the junction by the potential difference., (c) hole concentration in p-region is more as compared to n-region., (d) All the above., , 509

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Physics, When a forward bias is applied to a p-n junction, it, (a) raises the potential barrier., (b) reduces the majority carrier current to zero., (c) lowers the potential barrier., (d) None of the above., 14.6 For transistor action, which of the following statements are correct:, (a) Base, emitter and collector regions should have similar size and, doping concentrations., (b) The base region must be very thin and lightly doped., (c) The emitter junction is forward biased and collector junction is, reverse biased., (d) Both the emitter junction as well as the collector junction are, forward biased., 14.7 For a transistor amplifier, the voltage gain, (a) remains constant for all frequencies., (b) is high at high and low frequencies and constant in the middle, frequency range., (c) is low at high and low frequencies and constant at mid, frequencies., (d) None of the above., 14.8 In half-wave rectification, what is the output frequency if the input, frequency is 50 Hz. What is the output frequency of a full-wave rectifier, for the same input frequency., 14.9 For a CE-transistor amplifier, the audio signal voltage across the, collected resistance of 2 k: is 2 V. Suppose the current amplification, factor of the transistor is 100, find the input signal voltage and base, current, if the base resistance is 1 k:., 14.10 Two amplifiers are connected one after the other in series (cascaded)., The first amplifier has a voltage gain of 10 and the second has a, voltage gain of 20. If the input signal is 0.01 volt, calculate the output, ac signal., 14.11 A p-n photodiode is fabricated from a semiconductor with band gap, of 2.8 eV. Can it detect a wavelength of 6000 nm?, 14.5, , ADDITIONAL EXERCISES, 14.12 The number of silicon atoms per m3 is 5 u 10 28. This is doped, simultaneously with 5 u 1022 atoms per m3 of Arsenic and 5 u 1020, per m3 atoms of Indium. Calculate the number of electrons and holes., Given that ni = 1.5 u 1016 m–3. Is the material n-type or p-type?, 14.13 In an intrinsic semiconductor the energy gap Eg is 1.2eV. Its hole, mobility is much smaller than electron mobility and independent of, temperature. What is the ratio between conductivity at 600K and, that at 300K? Assume that the temperature dependence of intrinsic, carrier concentration ni is given by, ni, , 510, , Eg ·, §, n 0 exp ¨ –, © 2k B T ¸¹, , where n0 is a constant.

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Semiconductor Electronics:, Materials, Devices and, Simple Circuits, 14.14 In a p-n junction diode, the current I can be expressed as, , I, , § eV, ·, I 0 exp ¨, – 1¸, k, T, 2, © B, ¹, , where I0 is called the reverse saturation current, V is the voltage, across the diode and is positive for forward bias and negative for, reverse bias, and I is the current through the diode, k B is the, Boltzmann constant (8.6×10 –5 eV/K) and T is the absolute, temperature. If for a given diode I0 = 5 × 10–12 A and T = 300 K, then, (a) What will be the forward current at a forward voltage of 0.6 V?, (b) What will be the increase in the current if the voltage across the, diode is increased to 0.7 V?, (c) What is the dynamic resistance?, (d) What will be the current if reverse bias voltage changes from 1 V, to 2 V?, 14.15 You are given the two circuits as shown in Fig. 14.44. Show that, circuit (a) acts as OR gate while the circuit (b) acts as AND gate., , FIGURE 14.44, , 14.16 Write the truth table for a NAND gate connected as given in, Fig. 14.45., , FIGURE 14.45, , Hence identify the exact logic operation carried out by this circuit., 14.17 You are given two circuits as shown in Fig. 14.46, which consist, of NAND gates. Identify the logic operation carried out by the two, circuits., , FIGURE 14.46, , 14.18 Write the truth table for circuit given in Fig. 14.47 below consisting, of NOR gates and identify the logic operation (OR, AND, NOT) which, this circuit is performing., , 511

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Physics, , FIGURE 14.47, , (Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0, and hence Y=1. Similarly work out the values of Y for other, combinations of A and B. Compare with the truth table of OR, AND,, NOT gates and find the correct one.), 14.19 Write the truth table for the circuits given in Fig. 14.48 consisting of, NOR gates only. Identify the logic operations (OR, AND, NOT) performed, by the two circuits., , FIGURE 14.48, , 512

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Chapter Fifteen, , COMMUNICATION, SYSTEMS, , 15.1 INTRODUCTION, Communication is the act of transmission of information. Every living, creature in the world experiences the need to impart or receive information, almost continuously with others in the surrounding world. For, communication to be successful, it is essential that the sender and the, receiver understand a common language. Man has constantly made, endeavors to improve the quality of communication with other human, beings. Languages and methods used in communication have kept, evolving from prehistoric to modern times, to meet the growing demands, in terms of speed and complexity of information. It would be worthwhile, to look at the major milestones in events that promoted developments in, communications, as presented in Table 15.1., Modern communication has its roots in the 19th and 20th century in, the work of scientists like J.C. Bose, F.B. Morse, G. Marconi and Alexander, Graham Bell. The pace of development seems to have increased, dramatically after the first half of the 20th century. We can hope to see, many more accomplishments in the coming decades. The aim of this, chapter is to introduce the concepts of communication, namely the mode, of communication, the need for modulation, production and deduction, of amplitude modulation., , 15.2 ELEMENTS, , OF A, , COMMUNICATION SYSTEM, , Communication pervades all stages of life of all living creatures. Irrespective, of its nature, every communication system has three essential elements-

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Physics, TABLE 15.1 SOME MAJOR MILESTONES IN THE HISTORY OF COMMUNICATION, Year, , Event, , Remarks, , Around, 1565 A.D., , The reporting of the delivery of, a child by queen using drum, beats from a distant place to, King Akbar., , It is believed that minister Birbal, experimented with the arrangement to, decide the number of drummers posted, between the place where the queen, stayed and the place where the king, stayed., , 1835, , Invention of telegraph by, Samuel F.B. Morse and Sir, Charles Wheatstone, , It resulted in tremendous growth of, messages through post offices and, reduced physical travel of messengers, considerably., , 1876, , Telephone invented by, Alexander Graham Bell and, Antonio Meucci, , Perhaps the most widely used means of, communication in the history of, mankind., , 1895, , Jagadis Chandra Bose and, Guglielmo, Marconi, demonstrated, wireless, telegraphy., , It meant a giant leap – from an era of, communication using wires, to, communicating without using wires., (wireless), , 1936, , Television broadcast(John, Logi Baird), , First television broadcast by BBC, , 1955, , First radio FAX transmitted, across continent.(Alexander, Bain), , The idea of FAX transmission was, patented by Alexander Bain in 1843., , 1968, , ARPANET - the first internet, came into existence(J.C.R., Licklider), , ARPANET was a project undertaken by, the U.S. defence department. It allowed, file transfer from one computer to, another connected to the network., , 1975, , Fiber optics developed at Bell, Laboratories, , Fiber optical systems are superior and, more economical compared to, traditional communication systems., , 1989-91, , Tim Berners-Lee invented the, World Wide Web., , WWW may be regarded as the mammoth, encyclopedia of knowledge accessible to, everyone round the clock throughout the, year., , 514

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Communication System, transmitter, medium/channel and receiver. The block diagram shown in, Fig. 15.1 depicts the general form of a communication system., , FIGURE 15.1 Block diagram of a generalised communication system., , In a communication system, the transmitter is located at one place,, the receiver is located at some other place (far or near) separate from the, transmitter and the channel is the physical medium that connects them., Depending upon the type of communication system, a channel may be in, the form of wires or cables connecting the transmitter and the receiver or, it may be wireless. The purpose of the transmitter is to convert the message, signal produced by the source of information into a form suitable for, transmission through the channel. If the output of the information source, is a non-electrical signal like a voice signal, a transducer converts it to, electrical form before giving it as an input to the transmitter. When a, transmitted signal propagates along the channel it may get distorted due, to channel imperfection. Moreover, noise adds to the transmitted signal, and the receiver receives a corrupted version of the transmitted signal., The receiver has the task of operating on the received signal. It reconstructs, a recognisable form of the original message signal for delivering it to the, user of information., There are two basic modes of communication: point-to-point and, broadcast., In point-to-point communication mode, communication takes place, over a link between a single transmitter and a receiver. Telephony is an, example of such a mode of communication. In contrast, in the broadcast, mode, there are a large number of receivers corresponding to a single, transmitter. Radio and television are examples of broadcast mode of, communication., , 15.3 B ASIC T ERMINOLOGY U SED, COMMUNICATION SYSTEMS, , IN, , ELECTRONIC, , By now, we have become familiar with some terms like information source,, transmitter, receiver, channel, noise, etc. It would be easy to understand, the principles underlying any communication, if we get ourselves, acquainted with the following basic terminology., , 515

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JAGADIS CHANDRA BOSE (1858 – 1937), , Physics, , 516, , Jagadis Chandra Bose, (1858 – 1937) He, developed an apparatus, for generating ultrashort, electro-magnetic waves, and studied their quasioptical properties. He, was said to be the first to, employ a semiconductor, like galena as a selfrecovering detector of, electromagnetic waves., Bose published three, papers in the British, magazine,, ‘The, Electrician’ of 27 Dec., 1895. His invention was, published, in, the, ‘Proceedings of The Royal, Society’ on 27 April 1899, over two years before, Marconi’s first wireless, communication on 13, December 1901. Bose, also invented highly, sensitive instruments for, the detection of minute, responses by living, organisms to external, stimulii and established, parallelism, between, animal, and, plant, tissues., , (i) Transducer: Any device that converts one form of, energy into another can be termed as a transducer., In electronic communication systems, we usually, come across devices that have either their inputs, or outputs in the electrical form. An electrical, transducer may be defined as a device that converts, some physical variable (pressure, displacement,, force, temperature, etc) into corresponding, variations in the electrical signal at its output., (ii) Signal: Information converted in electrical form, and suitable for transmission is called a signal., Signals can be either analog or digital. Analog, signals are continuous variations of voltage or, current. They are essentially single-valued, functions of time. Sine wave is a fundamental, analog signal. All other analog signals can be fully, understood in terms of their sine wave components., Sound and picture signals in TV are analog in, nature. Digital signals are those which can take, only discrete stepwise values. Binary system that, is extensively used in digital electronics employs, just two levels of a signal. ‘0’ corresponds to a low, level and ‘1’ corresponds to a high level of voltage/, current. There are several coding schemes useful, for digital communication. They employ suitable, combinations of number systems such as the, binary coded decimal (BCD)*. American Standard, Code for Information Interchange (ASCII)** is a, universally popular digital code to represent, numbers, letters and certain characters., (iii) Noise: Noise refers to the unwanted signals that, tend to disturb the transmission and processing, of message signals in a communication system., The source generating the noise may be located, inside or outside the system., (iv) Transmitter: A transmitter processes the incoming, message signal so as to make it suitable for, transmission through a channel and subsequent, reception., (v) Receiver: A receiver extracts the desired message, signals from the received signals at the channel, output., (vi) Attenuation: The loss of strength of a signal while, propagating through a medium is known as, attenuation., , * In BCD, a digit is usually represented by four binary (0 or 1) bits. For example, the numbers 0, 1, 2, 3, 4 in the decimal system are written as 0000, 0001, 0010,, 0011 and 0100. 1000 would represent eight., ** It is a character encoding in terms of numbers based on English alphabet since, the computer can only understand numbers.

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Communication System, (vii), , (viii), (ix), (x), , (xi), (xii), , Amplification: It is the process of increasing the amplitude (and, consequently the strength) of a signal using an electronic circuit, called the amplifier (reference Chapter 14). Amplification is, necessary to compensate for the attenuation of the signal in, communication systems. The energy needed for additional signal, strength is obtained from a DC power source. Amplification is, done at a place between the source and the destination wherever, signal strength becomes weaker than the required strength., Range: It is the largest distance between a source and a destination, up to which the signal is received with sufficient strength., Bandwidth: Bandwidth refers to the frequency range over which, an equipment operates or the portion of the spectrum occupied, by the signal., Modulation: The original low frequency message/information, signal cannot be transmitted to long distances because of, reasons given in Section 15.7. Therefore, at the transmitter,, information contained in the low frequency message signal is, superimposed on a high frequency wave, which acts as a carrier, of the information. This process is known as modulation. As, will be explained later, there are several types of modulation,, abbreviated as AM, FM and PM., Demodulation: The process of retrieval of information from the, carrier wave at the receiver is termed demodulation. This is the, reverse process of modulation., Repeater: A repeater is a combination of a receiver and a, transmitter. A repeater, picks up the signal from the transmitter,, amplifies and retransmits it to the receiver sometimes with a, change in carrier frequency. Repeaters are used to extend the, range of a communication system as shown in Fig. 15.2. A, communication satellite is essentially a repeater station in space., , FIGURE 15.2 Use of repeater station to increase the range of communication., , 15.4 BANDWIDTH, , OF, , SIGNALS, , In a communication system, the message signal can be voice, music,, picture or computer data. Each of these signals has different ranges of, frequencies. The type of communication system needed for a given signal, depends on the band of frequencies which is considered essential for the, communication process., For speech signals, frequency range 300 Hz to 3100 Hz is considered, adequate. Therefore speech signal requires a bandwidth of 2800 Hz (3100 Hz, – 300 Hz) for commercial telephonic communication. To transmit music,, , 517

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Physics, an approximate bandwidth of 20 kHz is required because of the high, frequencies produced by the musical instruments. The audible range of, frequencies extends from 20 Hz to 20 kHz., Video signals for transmission of pictures require about 4.2 MHz of, bandwidth. A TV signal contains both voice and picture and is usually, allocated 6 MHz of bandwidth for transmission., In the preceeding paragraph, we have considered only analog signals., Digital signals are in the form of rectangular waves as shown in Fig. 15.3., One can show that this rectangular wave can be decomposed into a, superposition of sinusoidal waves of frequencies Q0, 2Q0, 3Q0, 4Q0 ... nQ0, where n is an integer extending to infinity and Q0 = 1/T0. The fundamental, (Q0 ), fundamental (Q0 ) + second harmonic (2Q0 ), and fundamental (Q0 ) +, second harmonic (2Q0 ) +, third harmonic (3Q0 ), are, shown in the same figure to, illustrate this fact. It is clear, that to reproduce the, rectangular wave shape, exactly we need to, superimpose all the, harmonics Q 0 , 2 Q 0 , 3 Q 0 ,, 4 Q0..., which implies an, infinite, bandwidth., However, for practical, purposes, the contribution, from higher harmonics can, be neglected, thus limiting, FIGURE 15.3 Approximation of a rectangular wave in terms of a, the bandwidth. As a result,, fundamental sine wave and its harmonics., received waves are a, distorted version of the, transmitted one. If the bandwidth is large enough to accommodate a few, harmonics, the information is not lost and the rectangular signal is more, or less recovered. This is so because the higher the harmonic, less is its, contribution to the wave form., , 15.5 BANDWIDTH, , 518, , OF, , TRANSMISSION MEDIUM, , Similar to message signals, different types of transmission media offer, different bandwidths. The commonly used transmission media are wire,, free space and fiber optic cable. Coaxial cable is a widely used wire, medium, which offers a bandwidth of approximately 750 MHz. Such cables, are normally operated below 18 GHz. Communication through free space, using radio waves takes place over a very wide range of frequencies: from, a few hundreds of kHz to a few GHz. This range of frequencies is further, subdivided and allocated for various services as indicated in Table 15.2., Optical communication using fibers is performed in the frequency range, of 1 THz to 1000 THz (microwaves to ultraviolet). An optical fiber can, offer a transmission bandwidth in excess of 100 GHz., Spectrum allocations are arrived at by an international agreement., The International Telecommunication Union (ITU) administers the present, system of frequency allocations.

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Communication System, TABLE 15.2 SOME IMPORTANT WIRELESS COMMUNICATION FREQUENCY BANDS, Service, , Frequency bands, , Standard AM broadcast, , 540-1600 kHz, , FM broadcast, , 88-108 MHz, , Television, , 54-72 MHz, , VHF (very high frequencies), , 76-88 MHz, , TV, , 174-216 MHz, 420-890 MHz, , UHF (ultra high frequencies), TV, , Cellular Mobile Radio, , 896-901 MHz, 840-935 MHz, , Mobile to base station, Base station to mobile, , Satellite Communication, , 5.925-6.425 GHz, 3.7-4.2 GHz, , Uplink, Downlink, , 15.6 PROPAGATION, , OF, , Comments, , ELECTROMAGNETIC WAVES, , In communication using radio waves, an antenna at the transmitter, radiates the Electromagnetic waves (em waves), which travel through the, space and reach the receiving antenna at the other end. As the em wave, travels away from the transmitter, the strength of the wave keeps on, decreasing. Several factors influence the propagation of em waves and, the path they follow. At this point, it is also important to understand the, composition of the earth’s atmosphere as it plays a vital role in the, propagation of em waves. A brief discussion on some useful layers of the, atmosphere is given in Table 15.3., , 15.6.1 Ground wave, To radiate signals with high efficiency, the antennas should have a size, comparable to the wavelength O of the signal (at least ~ O/4). At longer, wavelengths (i.e., at lower frequencies), the antennas have large physical, size and they are located on or very near to the ground. In standard AM, broadcast, ground based vertical towers are generally used as transmitting, antennas. For such antennas, ground has a strong influence on the, propagation of the signal. The mode of propagation is called surface wave, propagation and the wave glides over the surface of the earth. A wave, induces current in the ground over which it passes and it is attenuated, as a result of absorption of energy by the earth. The attenuation of surface, waves increases very rapidly with increase in frequency. The maximum, range of coverage depends on the transmitted power and frequency (less, than a few MHz)., , 519

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Physics, TABLE 15.3 DIFFERENT LAYERS OF ATMOSPHERE AND THEIR INTERACTION WITH THE, PROPAGATING ELECTROMAGNETIC WAVES, , Name of the, stratum (layer), , Approximate height, over earth’s surface, , Troposphere, , 10 km, , Day and, night, , VHF (up to several GHz), , 65-75 km, , Day only, , Reflects LF, absorbs MF, and HF to some degree, , 100 km, , Day only, , Helps surface waves,, reflects HF, , 170-190 km, , Daytime,, merges with, F2 at night, , Partially absorbs HF, waves yet allowing them, to reach F2, , 300 km at night,, 250-400 km, during daytime, , Day and, night, , Efficiently reflects HF, waves, particularly at, night, , D (part of, stratosphere), E (part of, Stratosphere), F1 (Part of, Mesosphere), F2, (Thermosphere), , P, A, R, T, S, , Exists during, , O, F, I, O, N, O, S, P, H, E, R, E, , Frequencies most, affected, , 15.6.2 Sky waves, In the frequency range from a few MHz up to 30 to 40 MHz, long distance, communication can be achieved by ionospheric reflection of radio waves, back towards the earth. This mode of propagation is called sky wave, propagation and is used by short wave broadcast services. The ionosphere, is so called because of the presence of a large number of ions or charged, particles. It extends from a height of ~ 65 Km to about 400 Km above the, earth’s surface. Ionisation occurs due to the absorption of the ultraviolet, and other high-energy radiation coming from the sun by air molecules., The ionosphere is further subdivided into several layers, the details of, which are given in Table 15.3. The degree of ionisation varies with the, height. The density of atmosphere decreases with height. At great heights, the solar radiation is intense but there are few molecules to be ionised., Close to the earth, even though the molecular concentration is very high,, the radiation intensity is low so that the ionisation is again low. However,, at some intermediate heights, there occurs a peak of ionisation density., The ionospheric layer acts as a reflector for a certain range of frequencies, (3 to 30 MHz). Electromagnetic waves of frequencies higher than 30 MHz, penetrate the ionosphere and escape. These phenomena are shown in the, Fig. 15.4. The phenomenon of bending of em waves so that they are, diverted towards the earth is similar to total internal reflection in optics*., , 520, , * Compare this with the phenomenon of mirage.

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Communication System, , FIGURE 15.4 Sky wave propagation. The layer nomenclature, is given in Table 15.3., , 15.6.3 Space wave, Another mode of radio wave propagation is by space waves. A space, wave travels in a straight line from transmitting antenna to the receiving, antenna. Space waves are used for line-of-sight (LOS) communication as, well as satellite communication. At frequencies above 40 MHz,, communication is essentially limited to line-of-sight paths. At these, frequencies, the antennas are relatively smaller and can be placed at, heights of many wavelengths above the ground. Because of line-of-sight, nature of propagation, direct waves get blocked at some point by the, curvature of the earth as illustrated in Fig. 15.5. If the signal is to be, received beyond the horizon then the receiving antenna must be high, enough to intercept the line-of-sight waves., , FIGURE 15.5 Line of sight communication by space waves., , If the transmitting antenna is at a height hT, then you can show that, the distance to the horizon dT is given as d T, , 2Rh T , where R is the, , radius of the earth (approximately 6400 km). dT is also called the radio, horizon of the transmitting antenna. With reference to Fig. 15.5 the, maximum line-of-sight distance dM between the two antennas having, heights hT and hR above the earth is given by, dM, , 2Rh T  2Rh R, , where hR is the height of receiving antenna., , (15.1), , 521

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Physics, Television broadcast, microwave links and satellite communication, are some examples of communication systems that use space wave mode, of propagation. Figure 15.6 summarises the various modes of wave, propagation discussed so far., , FIGURE 15.6 Various propagation modes for em waves., , EXAMPLE 15.1, , Example 15.1 A transmitting antenna at the top of a tower has a height, 32 m and the height of the receiving antenna is 50 m. What is the, maximum distance between them for satisfactory communication in, LOS mode? Given radius of earth 6.4 × 106 m., Solution, , dm, , 2 u 64 u 105 u 32  2 u 64 u 105 u 50 m, 2, , 144 u 102 u 10 m, , 15.7 MODULATION, , 522, , 3, , 64 u 10 u 10  8 u 10 u 10 m, 45.5 km, , AND ITS, , NECESSITY, , As already mentioned, the purpose of a communication system is to, transmit information or message signals. Message signals are also called, baseband signals, which essentially designate the band of frequencies, representing the original signal, as delivered by the source of information., No signal, in general, is a single frequency sinusoid, but it spreads over a, range of frequencies called the signal bandwidth. Suppose we wish to, transmit an electronic signal in the audio frequency (AF) range (baseband, signal frequency less than 20 kHz) over a long distance directly. Let us, find what factors prevent us from doing so and how we overcome these, factors,

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Communication System, 15.7.1 Size of the antenna or aerial, For transmitting a signal, we need an antenna or an aerial. This antenna, should have a size comparable to the wavelength of the signal (at least, O/4 in dimension) so that the antenna properly senses the time variation, of the signal. For an electromagnetic wave of frequency 20 kHz, the, wavelength O is 15 km. Obviously, such a long antenna is not possible to, construct and operate. Hence direct transmission of such baseband signals, is not practical. We can obtain transmission with reasonable antenna, lengths if transmission frequency is high (for example, if Q is 1 MHz, then, O is 300 m). Therefore, there is a need of translating the information, contained in our original low frequency baseband signal into high or, radio frequencies before transmission., , 15.7.2 Effective power radiated by an antenna, A theoretical study of radiation from a linear antenna (length l ) shows, that the power radiated is proportional to (l/O)2 . This implies that for the, same antenna length, the power radiated increases with decreasing O,, i.e., increasing frequency.Hence, the effective power radiated by a long, wavelength baseband signal would be small. For a good transmission,, we need high powers and hence this also points out to the need of using, high frequency transmission., , 15.7.3 Mixing up of signals from different transmitters, Another important argument against transmitting baseband signals, directly is more practical in nature. Suppose many people are talking at, the same time or many transmitters are transmitting baseband information, signals simultaneously. All these signals will get, mixed up and there is no simple way to distinguish, between them. This points out towards a possible, solution by using communication at high, frequencies and allotting a band of frequencies to, each message signal for its transmission., The above arguments suggest that there is a, need for translating the original low frequency, baseband message or information signal into high, frequency wave before transmission such that the, translated signal continues to possess the, information contained in the original signal. In, doing so, we take the help of a high frequency signal,, known as the carrier wave, and a process known, as modulation which attaches information to it. The, carrier wave may be continuous (sinusoidal) or in, the form of pulses as shown in Fig. 15.7., FIGURE 15.7 (a) Sinusoidal, and, (b) pulse shaped signals., A sinusoidal carrier wave can be represented as, c(t ) = Ac sin (Zct + I), (15.2), where c(t) is the signal strength (voltage or current), Ac is the amplitude,, Zc ( = 2SQc ) is the angular frequency and I is the initial phase of the carrier, wave. During the process of modulation, any of the three parameters, viz, Ac, Zc and I, of the carrier wave can be controlled by the message or, , 523

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Physics, , http://www.williamson labs.com/480_mod.htm, , Modulation and Demodulation, , information signal. This results in three types of modulation: (i) Amplitude, modulation (AM), (ii) Frequency modulation (FM) and, (iii) Phase modulation (PM), as shown in Fig. 15.8., , FIGURE 15.8 Modulation of a carrier wave: (a) a sinusoidal carrier wave;, (b) a modulating signal; (c) amplitude modulation; (d) frequency, modulation; and (e) phase modulation., , Similarly, the significant characteristics of a pulse are: pulse amplitude,, pulse duration or pulse Width, and pulse position (denoting the time of, rise or fall of the pulse amplitude) as shown in Fig. 15.7(b). Hence, different, types of pulse modulation are: (a) pulse amplitude modulation (PAM),, (b) pulse duration modulation (PDM) or pulse width modulation (PWM),, and (c) pulse position modulation (PPM). In this chapter, we shall confine, to amplitude modulation on ly., , 15.8 AMPLITUDE MODULATION, In amplitude modulation the amplitude of the carrier is varied in, accordance with the information signal. Here we explain amplitude, modulation process using a sinusoidal signal as the modulating signal., Let c(t) = Ac sin Zct represent carrier wave and m(t) = Am sin Zmt represent, the message or the modulating signal where Zm = 2Sfm is the angular, frequency of the message signal. The modulated signal cm (t ) can be, written as, cm (t) = (Ac + Am sin Zmt) sin Zct, §, ·, A, Ac ¨1  m sin Zm t ¸ sin Zc t, (15.3), Ac, ©, ¹, Note that the modulated signal now contains the message signal. This, can also be seen from Fig. 15.8(c). From Eq. (15.3), we can write,, , 524, , cm (t ), , Ac sin Zct  P Ac sin Zm t sin Zct, , (15.4)

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Communication System, Here P= Am/Ac is the modulation index; in practice,P is kept d 1 to avoid, distortion., Using the trignomatric relation sinA sinB = ½ (cos(A – B) – cos (A + B),, we can write cm (t) of Eq. (15.4) as, , P Ac, cos(Z c  Zm ) t , cos(Z c  Zm )t, (15.5), 2, 2, Here Zc –Zm and Zc +Zm are respectively called the lower side and upper, side frequencies. The modulated signal now consists of the carrier wave, of frequency Zc plus two sinusoidal waves each with a frequency slightly, different from, known as side bands. The frequency spectrum of the, amplitude modulated signal is shown in Fig. 15.9., cm (t ), , Ac sin Z c t , , P Ac, , FIGURE 15.9 A plot of amplitude versus Zfor, an amplitude modulated signal., , As long as the broadcast frequencies (carrier waves) are sufficiently, spaced out so that sidebands do not overlap, different stations can operate, without interfering with each other., , Solution, (a) Modulation index =10/20 = 0.5, (b) The side bands are at (1000+10 kHz)=1010 kHz and, (1000 –10 kHz) = 990 kHz., , 15.9 PRODUCTION, , OF, , EXAMPLE 15.2, , Example 15.2 A message signal of frequency 10 kHz and peak voltage, of 10 volts is used to modulate a carrier of frequency 1 MHz and peak, voltage of 20 volts. Determine (a) modulation index, (b) the side bands, produced., , AMPLITUDE MODULATED WAVE, , Amplitude modulation can be produced by a variety of methods. A, conceptually simple method is shown in the block diagram of Fig. 15.10., , FIGURE 15.10 Block diagram of a simple modulator, for obtaining an AM signal., , 525

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Physics, Here the modulating signal Am sin Zmt is added to the carrier signal, Ac sin Zct to produce the signal x (t). This signal x (t) = Am sinZmt, + Ac sinZct is passed through a square law device which is a non-linear, device which produces an output, y (t) = B x (t ) + Cx 2 (t ), (15.6), where B and C are constants. Thus,, y (t)= BAm sin Zmt + BAc sin Zct, , C ª¬ Am2 sin2 Zm t  Ac2 sin2 Zc t  2 Am Ac sin Zm t sin Zc t º¼, , (15.7), , = BAm sin Zmt + BAc sin Zct, C Am2, C Am2, C Ac2,  Ac2 –, cos 2Zm t –, cos 2Z c t, 2, 2, 2, + CAm Ac cos (Zc – Zm ) t – CAm Ac cos (Zc+ Zm ) t, , , (15.8), , 2, , where the trigonometric relations sin A = (1 – cos2A)/2 and the relation, for sin A sin B mentioned earlier are used., In Eq. (15.8), there is a dc term C/2 Am2  Ac2, , and sinusoids of, , frequencies Zm, 2Zm, Zc, 2Zc, Zc –Zm and Zc +Zm. As shown in Fig. 15.10, this signal is passed through a band pass filter* which rejects dc and the, sinusoids of frequencies Zm , 2Zm and 2 Zc and retains the frequencies Zc,, Zc –Zm and Zc +Zm. The output of the band pass filter therefore is of the, same form as Eq. (15.5) and is therefore an AM wave., It is to be mentioned that the modulated signal cannot be transmitted, as such. The modulator is to be followed by a power amplifier which, provides the necessary power and then the modulated signal is fed to an, antenna of appropriate size for radiation as shown in Fig. 15.11., , FIGURE 15.11 Block diagram of a transmitter., , 15.10 DETECTION, , OF, , AMPLITUDE MODULATED WAVE, , The transmitted message gets attenuated in propagating through the, channel. The receiving antenna is therefore to be followed by an amplifier, and a detector. In addition, to facilitate further processing, the carrier, frequency is usually changed to a lower frequency by what is called an, intermediate frequency (IF) stage preceding the detection. The detected, signal may not be strong enough to be made use of and hence is required, , 526, , * A band pass filter rejects low and high frequencies and allows a band of frequencies, to pass through.

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Communication System, to be amplified. A block diagram of a typical receiver is shown in, Fig. 15.12, , FIGURE 15.12 Block diagram of a receiver., , Detection is the process of recovering the modulating signal from the, modulated carrier wave. We just saw that the modulated carrier wave, contains the frequencies Zc and Zc r Zm. In order to obtain the original, message signal m(t ) of angular frequency Zm , a simple method is shown, in the form of a block diagram in Fig. 15.13., , FIGURE 15.13 Block diagram of a detector for AM signal. The quantity, on y-axis can be current or voltage., , The modulated signal of the form given in (a) of fig. 15.13 is passed, through a rectifier to produce the output shown in (b). This envelope of, signal (b) is the message signal. In order to retrieve m (t ), the signal is, passed through an envelope detector (which may consist of a simple RC, circuit)., In the present chapter we have discussed some basic concepts of, communication and communication systems. We have also discussed, one specific type of analog modulation namely Amplitude Modulation, (AM). Other forms of modulation and digital communication systems play, an important role in modern communication. These and other exciting, developments are taking place everyday., So far we have restricted our discussion to some basic communication, systems. Before we conclude this chapter, it is worth taking a glance at, some of the communication systems (see the box) that in recent times, have brought major changes in the way we exchange information even in, our day-to-day life:, , 527

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Physics, ADDITIONAL INFORMATION, The Internet, It is a system with billions of users worldwide. It permits communication and sharing of all, types of information between any two or more computers connected through a large and, complex network. It was started in 1960’s and opened for public use in 1990’s. With the, passage of time it has witnessed tremendous growth and it is still expanding its reach. Its, applications include, (i) E mail – It permits exchange of text/graphic material using email software. We can write a, letter and send it to the recipient through ISP’s (Internet Service Providers) who work like, the dispatching and receiving post offices., (ii) File transfer – A FTP (File Transfer Programmes) allows transfer of files/software from one, computer to another connected to the Internet., (iii) World Wide Web (WWW) – Computers that store specific information for sharing with others, provide websites either directly or through web service providers. Government, departments, companies, NGO’s (Non-Government Organisations) and individuals can post, information about their activities for restricted or free use on their websites. This information, becomes accessible to the users. Several search engines like Google, Yahoo! etc. help us in, finding information by listing the related websites. Hypertext is a powerful feature of the, web that automatically links relevant information from one page on the web to another, using HTML (hypertext markup language)., (iv) E-commerce – Use of the Internet to promote business using electronic means such as, using credit cards is called E-commerce. Customers view images and receive all the, information about various products or services of companies through their websites. They, can do on-line shopping from home/office. Goods are dispatched or services are provided, by the company through mail/courier., (v) Chat – Real time conversation among people with common interests through typed, messages is called chat. Everyone belonging to the chat group gets the message, instantaneously and can respond rapidly., Facsimile (FAX), It scans the contents of a document (as an image, not text) to create electronic signals. These, signals are then sent to the destination (another FAX machine) in an orderly manner using, telephone lines. At the destination, the signals are reconverted into a replica of the original, document. Note that FAX provides image of a static document unlike the image provided by, television of objects that might be dynamic., Mobile telephony, The concept of mobile telephony was developed first in 1970’s and it was fully implemented in, the following decade. The central concept of this system is to divide the service area into a, suitable number of cells centred on an office called MTSO (Mobile Telephone Switching Office)., Each cell contains a low-power transmitter called a base station and caters to a large number, of mobile receivers (popularly called cell phones). Each cell could have a service area of a few, square kilometers or even less depending upon the number of customers. When a mobile, receiver crosses the coverage area of one base station, it is necessary for the mobile user to be, transferred to another base station. This procedure is called handover or handoff. This process, is carried out very rapidly, to the extent that the consumer does not even notice it. Mobile, telephones operate typically in the UHF range of frequencies (about 800-950 MHz)., , 528

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Communication System, SUMMARY, 1., , 2., , Electronic communication refers to the faithful transfer of information, or message (available in the form of electrical voltage and current), from one point to another point., Transmitter, transmission channel and receiver are three basic units, of a communication system., , 3., , Two important forms of communication system are: Analog and Digital., The information to be transmitted is generally in continuous waveform, for the former while for the latter it has only discrete or quantised, levels., , 4., , Every message signal occupies a range of frequencies. The bandwidth, of a message signal refers to the band of frequencies, which are necessary, for satisfactory transmission of the information contained in the signal., Similarly, any practical communication system permits transmission, of a range of frequencies only, which is referred to as the bandwidth of, the system., , 5., , Low frequencies cannot be transmitted to long distances. Therefore,, they are superimposed on a high frequency carrier signal by a process, known as modulation., , 6., , In modulation, some characteristic of the carrier signal like amplitude,, frequency or phase varies in accordance with the modulating or message, signal. Correspondingly, they are called Amplitude Modulated (AM),, Frequency Modulated (FM) or Phase Modulated (PM) waves., , 7., , Pulse modulation could be classified as: Pulse Amplitude Modulation, (PAM), Pulse Duration Modulation (PDM) or Pulse Width Modulation, (PWM) and Pulse Position Modulation (PPM)., , 8., , For transmission over long distances, signals are radiated into space, using devices called antennas. The radiated signals propagate as, electromagnetic waves and the mode of propagation is influenced by, the presence of the earth and its atmosphere. Near the surface of the, earth, electromagnetic waves propagate as surface waves. Surface wave, propagation is useful up to a few MHz frequencies., , 9., , Long distance communication between two points on the earth is, achieved through reflection of electromagnetic waves by ionosphere., Such waves are called sky waves. Sky wave propagation takes place up, to frequency of about 30 MHz. Above this frequency, electromagnetic, waves essentially propagate as space waves. Space waves are used for, line-of-sight communication and satellite communication., , 10. If an antenna radiates electromagnetic waves from a height hT, then, the range dT is given by, , 2Rh T where R is the radius of the earth., , 11. Amplitude modulated signal contains frequencies (Z c, , Z m ), Zc, , and, , (Z c  Zm )., 12. Amplitude modulated waves can be produced by application of the, message signal and the carrier wave to a non-linear device, followed by, a band pass filter., 13. AM detection, which is the process of recovering the modulating signal, from an AM waveform, is carried out using a rectifier and an envelope, detector., , 529

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Physics, POINTS TO PONDER, 1., , 2., , 3., , In the process of transmission of message/ information signal, noise gets, added to the signal anywhere between the information source and the, receiving end. Can you think of some sources of noise?, In the process of modulation, new frequencies called sidebands are generated, on either side (higher and lower than the carrier frequency) of the carrier by, an amount equal to the highest modulating frequency. Is it possible to, retrieve the message by transmitting (a) only the side bands, (b) only one, side band?, In amplitude modulation, modulation index P d 1 is used. What will happen, if P ! 1?, , EXERCISES, 15.1, , Which of the following frequencies will be suitable for beyond-thehorizon communication using sky waves?, (a) 10 kHz, (b) 10 MHz, (c) 1 GHz, (d) 1000 GHz, , 530, , 15.2, , Frequencies in the UHF range normally propagate by means of:, (a) Ground waves., (b) Sky waves., (c) Surface waves., (d) Space waves., , 15.3, , Digital signals, (i) do not provide a continuous set of values,, (ii) represent values as discrete steps,, (iii) can utilize binary system, and, (iv) can utilize decimal as well as binary systems., Which of the above statements are true?, (a) (i) and (ii) only, (b) (ii) and (iii) only, (c) (i), (ii) and (iii) but not (iv), (d) All of (i), (ii), (iii) and (iv)., , 15.4, , Is it necessary for a transmitting antenna to be at the same height, as that of the receiving antenna for line-of-sight communication? A, TV transmitting antenna is 81m tall. How much service area can it, cover if the receiving antenna is at the ground level?, , 15.5, , A carrier wave of peak voltage 12V is used to transmit a message, signal. What should be the peak voltage of the modulating signal in, order to have a modulation index of 75%?, , 15.6, , A modulating signal is a square wave, as shown in Fig. 15.14.

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Communication System, , FIGURE 15.14, , 15.7, , 15.8, , The carrier wave is given by c (t ) 2 sin(8St ) volts., (i) Sketch the amplitude modulated waveform, (ii) What is the modulation index?, For an amplitude modulated wave, the maximum amplitude is found, to be 10V while the minimum amplitude is found to be 2V. Determine, the modulation index, P., What would be the value of P if the minimum amplitude is zero volt?, Due to economic reasons, only the upper sideband of an AM wave is, transmitted, but at the receiving station, there is a facility for, generating the carrier. Show that if a device is available which can, multiply two signals, then it is possible to recover the modulating, signal at the receiver station., , 531

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Physics, , INDEX, , Absorption spectra, AC current, AC Generator, AC voltage, applied to a capacitor, applied to a resistor, applied to an inductor, applied to a series LCR circuit, Accelerators in India, Accommodation of eye, Activity of radioactive substances, Additivity of charges, Alpha decay, Alpha particle scattering, Ammeter, Ampere, Amperes circuital law, Amplification, Amplitude modulation, Analog signal, AND gate, Andre, Ampere, Angle, of deviation, of incidence, of reflection, of refraction, Angular magnification, Apparent depth, Area element vector, Astigmatism, Atomic, mass unit, number, spectra, Attenuation, Aurora Boriolis, Band gap, Bandwidth of signal, , 421, 233, 224, 233, 241, 234, 237, 244, 142, 336, 447, 8, 449, 415, 165, 155, 147, 517, 524, 502, 503, 148, 330, 355, 355, 355, 341, 318, 26, 337, 439, 440, 420, 516, 139, 470, 517, , Bandwidth of transmission medium, Bar magnet, as solenoid, Barrier potential, Base, Becquerel, Beta decay, Binding energy per nucleon, Biot-Savart law, Bohr magneton, Bohr radius, Bohr’s model of atom, Bohr’s postulates, Brewster’s angle, Brewster’s law, C A. Volta, Capacitance, Capacitive reactance, Capacitive circuit, Capacitor, parallel plate, in parallel, in series, Cartesian sign convention, Cassegrain telescope, Cells, in parallel, in series, Chain reaction, Channel, Charging by induction, Charles August de Coulomb, Chromatic aberration, Ciliary muscles, Coercivity, Coherent source, Collector, Colour code of resistors, Combination of lenses, , 518, 174, 176, 479, 491, 448, 450, 444, 143, 163, 425, 422, 424, 380, 381, 53, 73, 241, 252, 74, 79, 78, 311, 342, 110, 114, 113, 452, 515, 6, 11, 333, 337, 195, 360, 491, 103, 328

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Index, Combination of resistors, series, parallel, Composition of nucleus, Concave mirror, Conduction band, Conductivity, Conductors, Conservation of charge, Conservative force, Continuous charge distribution, Control rods, Convex mirror, Coulomb, Coulomb’s law, Critical angle, Curie temperature, Curie, Current, amplification factor, density, loop as a magnetic dipole, sensitivity of galvanometer, Cut-off voltage/Stopping potential, Cyclotron, frequency, Davisson & Germer Experiment, de Broglie, relation, wavelength, explanation, Decay constant, Detection of amplitude modulated wave, Diamagnetism, Dielectrics, Dielectric, constant, strength, Diffraction, single slit, Digital, electronics, signal, Dioptre, Dipole, moment, moment vector, in uniform electric field, physical significance, Dispersion by a prism, Displacement current, Doppler effect, Drift velocity, Earth’s magnetism, Earthing, , 107, 108, 438, 312, 469, 97, 468, 5, 8, 51, 32, 454, 312, 11, 10, 320, 194, 448, 94, 495, 97, 160, 165, 391, 140, 141, 403, 398, 398, 430, 446, 526, 192, 71, 76, 74, 367, 368, 501, 502, 328, 28, 28, 31, 29, 332, 270, 358, 98, 185, 6, , Eddy currents, Einstein’s photoelectric equation, Electric, charge, current, dipole, displacement, field, field, physical significance, field due to a system of charges, field lines, flux, susceptibility, Electrical energy, Electromagnetic, waves, sources, waves, nature, damping, spectrum, Electron emission, Electrostatic, analog, potential, shielding, Electrostatics, of conductors, Electromotive force (emf ), Emission spectra, Emitter, Energy, bands, generation in stars, levels, stored in a capacitor, Equipotential surfaces, Excited state, Experiments of Faraday & Henry, Extrinsic semiconductor, Eye, Farad, Faraday’s law of Induction, Fast breeder reactor, Ferromagnetism, Field, due to infinite plane sheet, due to uniformly charged thin spherical, shell, Field emission, Flemings left hand rule, Flux leakage, Focal length, Force between two parallel currents, Forward bias, Franck-Hertz experiment, , 218, 394, 1, 93, 27, 77, 18, 20, 19, 23, 25, 72, 105, 274, 275, 218, 280, 387, 180, 53, 69, 1, 67, 110, 421, 491, 469, 455, 427, 80, 60, 427, 205, 474, 336, 75, 207, 453, 193, 38, 39, 388, 261, 311, 154, 479, 428, , 555

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Physics, Fringe width, Full-wave rectifier, G.S. Ohm, Gamma, rays, decay, Gauss’s law, its applications, in magnetism, Gaussian surface, Geographic meridian, Gold leaf electroscope, Ground, state, wave, H.A. Lorentz, Half life, Half-wave rectifier, Hallwachs’ and Lenard’s observations, Henry, Hertz Experiment, Holes, Horizontal component of earth’s, magnetic field, Huygen’s Principle, Hypermetropia, Impact parameter, Impedence diagram, Inductance, mutual, self, Induction, of charge, Inductive, circuit, reactance, Input resistance of a transistor, Insulators, Integrated circuits (IC), Interference, constructive, destructive, fringes, Internal resistance, Intrinsic semiconductor, Ionisation energy, Iris, Isobars, Isotones, Isotopes, J.C. Maxwell, Junction transistor, K.F. Gauss, Kirchhoff’s rules, , 556, , 364, 483, 96, 283, 451, 33, 37, 181, 35, 186, 4, 427, 519, 134, 448, 483, 388, 220, 274, 472, 187, 353, 337, 418, 246, 219, 220, 222, 6, 6, 252, 238, 494, 5, 505, 361, 361, 363, 110, 472, 427, 337, 441, 441, 439, 270, 490, 182, 115, , Lateral shift, 317, Law, of radioactive decay, 447, of reflection, 357, of refraction, 356, LC oscillations, 255, Least distance of distinct vision, 336, Lenz’s law, 210, Lens maker’s formula, 326, Light emitting diode, 488, Limitations of Ohm’s law, 101, Linear, charge density, 32, magnification/Magnifying power, 339, Logic gates, 502, Lorentz force, 134, Magnetic, declination, 186, dipole, 177, dipole moment of a revolving electron, 162, field, 132, field lines, 175, field on the axis of a circular current loop 145, flux, 182, 206, force on a current carrying conductor, 135, force, 133, hysteresis, 195, inclination, 187, intensity, 190, meridian, 186, moment of a current loop, 158, moment, 178, permeability, 190, potential energy, 178, susceptibility, 190, torque, 178, Magnetisation, 189, Majority carriers, 476, Mass, defect, 443, number, 440, energy relation, 442, Maxwell’s equations, 273, Mean life, 448, Meter bridge, 120, Michael Faraday, 208, Microscope, 339, compound, 340, Microwaves, 281, Minority carriers, 476, Mirage, 321, Mirror equation, 314, Mobility, 100, Moderator, 454

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Index, Modulation, index, Motion in a magnetic field, Motional emf, Moving coil galvanometer, Multiplication factor (fission), Myopia, NAND gate, Near point, Neutrons, Noise, Non-polar molecules, NOR gate, North pole, NOT gate, n-p-n transistor, n-type semi conductor, Nuclear, binding energy, density, energy, fission, force, fusion, holocaust, reactor, size, winter, Numerical aperture, Ohm, Ohm’s law, Optical fibers, OR gate, Orbital magnetic moment, Output resistance of a transistor, Paramagnetism, Permanent magnets, Permeability of free space, Permittivity, of free space, of medium, Phasors, diagram, Photodiode, Photoelectric effect, Photocell, Photoelectric emission, Photoelectrons, Photon, Pith ball, Plane polarised wave, p-n Junction, p-n-p transistor, Point charge, Polar molecules, , 517, 522, 525, 137, 212, 163, 454, 336, 504, 336, 440, 516, 72, 505, 174, 502, 491, 475, 442, 442, 451, 452, 445, 455, 457, 452, 441, 457, 375, 95, 95, 322, 502, 163, 495, 192, 195, 143, 11, 76, 76, 237, 237, 487, 388, 399, 388, 389, 395, 2, 377, 478, 491, 10, 72, , Polarisation, 71, 376, by reflection, 380, by scattering, 379, Polarity of charge, 2, Polaroid, 378, Potential, 53, due to an electric dipole, 55, due to a point charge, 54, due to a system of charges, 57, energy difference, 53, energy for a system of charges, 61, energy of a dipole, 66, energy of a single charge, 64, energy of a system of two charges, 65, energy, 52, Potentiometer, 122, Power (electrical), 106, factor, 252, in ac circuit, 252, of lens, 327, Pressurised heavy water reactors, 453, Primary coil, Principal focus, 311, Principle of superposition, 15, Principle quantum number, 425, Prism formula, 331, Production of amplitude modulated wave, 525, Properties of electric charge, 8, p-type semi conductor, 476, Q factor/quality factor, 250, Quanta of energy, 393, Quantisation of charge, 8, Radio waves, 281, Radioactivity, 446, Rainbow, 333, Ray optics, validity of, 375, Rayleigh scattering, 335, Rectifier, 483, Red shift, 358, Reflection of light, 310, Refraction, 318, of a plane wave, 355, Refractive index, 317, 356, Relation between field and potential, 61, Relaxation time, 98, Rententivity, 195, Repeater, 517, Resistance, 95, Resistivity, 96, 468, of some materials, 102, Resolving power, 373, of eye, 374, Resonance, 248, Sharpness, 249, Resonant frequency, 248, Reverse bias, 480, , 557

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Physics, Right hand rule, Root mean square (rms) or effective, current, voltage, Roget’s spiral, Rutherford’s model of atom, Saturation current, Scattering of light, Secondary wavelet, Semiconductors, diode, elemental, compound, Shunt resistance, Signal, Sky wave, Snell’s law, Solar cell, Solenoid, South pole, Space wave, Spectral series, Brackett, Fund, Lyman, Paschen, Spectrum of light, Spherical mirror, Spin magnetic moment, Surface charge density, Telescope, Temperature dependence of, resistivity, Tesla, Thermionic emission, Thermonuclear fusion, Thin lens formula, Threshold frequency, , 558, , 149, 235, 235, 156, 415, 390, 335, 354, 469, 479, 468, 468, 164, 516, 520, 317, 356, 489, 151, 174, 521, 421, 422, 422, 422, 422, 332, 310, 311, 163, 32, 341, 103, 135, 388, 456, 326, 392, , Tokamak, Toroid, Torque, on a current loop, on a dipole, Total internal reflection, Transducer, Transformer, Step-down, Step-up, Transistor, as a switch, as an amplifier, oscillator, common emitter configuration, Truth table, Uncertainty Principle, Unpolarised wave, Ultraviolet rays, Valence band, Van de Graaff Generator, Velocity selector, Visible rays, Voltage Regulator, Voltage sensitivity of a galvanometer, Voltmeter, Volume charge density, Wattless current, Wavefront, plane, spherical, Wheatstone bridge, Work function, X rays, Young’s experiment, Zener, diode, breakdown, , 153, 152, 157, 31, 319, 516, 259, 261, 261, 496, 497, 500, 493, 502, 400, 377, 282, 469, 83, 140, 282, 486, 165, 165, 32, 252, 353, 354, 354, 118, 394, 283, 362, 485, 485

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Physics, , ANSWERS, , CHAPTER 9, 9.1, 9.2, 9.3, , v = –54 cm. The image is real, inverted and magnified. The size of the, image is 5.0 cm. As u o f, v of; for u < f, image is virtual., v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As, u of; v of (but never beyond) while m o0., 1.33; 1.7 cm, , 9.4, , nga = 1.51; n wa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e.,, r ~ 38º., , 9.5, , r = 0.8 × tan ic and sin i c 1/1.33 # 0.75 , where r is the radius (in m), of the largest circle from which light comes out and ic is the critical, angle for water-air interface, Area = 2.6 m2, n # 1.53 and Dm for prism in water # 10º, R = 22 cm, Here the object is virtual and the image is real. u = +12 cm (object on, right; virtual), (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right, side., (b) f = –16 cm. Image is real and at 48 cm from the lens on its right, side., v = 8.4 cm, image is erect and virtual. It is diminished to a size, 1.8 cm. As u of, v o f (but never beyond f while m o 0)., Note that when the object is placed at the focus of the concave lens, (21 cm), the image is located at 10.5 cm (not at infinity as one might, wrongly think)., A diverging lens of focal length 60 cm, (a) ve = –25 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm =, 10 cm,, fO = u O = – 2.5 cm; Magnifying power = 20, (b) u O = – 2.59 cm., Magnifying power = 13.5., Angular magnification of the eye-piece for image at 25 cm, , 9.6, 9.7, 9.8, , 9.9, , 9.10, 9.11, , 9.12, , 25, , 25, cm 2.27cm ; v = 7.2 cm, O, 11, Separation = 9.47 cm; Magnifying power = 88, 2.5, ,  1 11; | u e |

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Answers, 9.13, 9.14, 9.15, , 9.16, , 9.17, , 24; 150 cm, (a) Angular magnification = 1500, (b) Diameter of the image = 13.7 cm., Apply mirror equation and the condition:, (a) f < 0 (concave mirror); u < 0 (object on left), (b) f > 0; u < 0, (c) f > 0 (convex mirror) and u < 0, (d) f < 0 (concave mirror); f < u < 0, to deduce the desired result., The pin appears raised by 5.0 cm. It can be seen with an explicit ray, diagram that the answer is independent of the location of the slab, (for small angles of incidence)., (a) sin icc = 1.44/1.68 which gives icc = 59°. Total internal reflection, takes place when i  ! 59q or when r < r max = 31° Now,, (sin i, , 9.18, , 9.19, , 9.20, , max, , /sin r, , max, , ) 1.68 , which gives i, max, , ~, , 60°. Thus , all, , incident rays of angles in the range 0 < i < 60° will suffer total, internal reflections in the pipe. (If the length of the pipe is, finite, which it is in practice, there will be a lower limit on i, determined by the ratio of the diameter to the length of the, pipe.), (b) If there is no outer coating, i cc = sin–1(1/1.68) = 36.5° Now,, i = 90° will have r = 36.5° and i c = 53.5° which is greater than, icc. Thus, all incident rays (in the range 53.5° < i < 90°) will, suffer total internal reflections., (a) Rays converging to a point ‘behind’ a plane or convex mirror are, reflected to a point in front of the mirror on a screen. In other, words, a plane or convex mirror can produce a real image if the, object is virtual. Convince yourself by drawing an appropriate, ray diagram., (b) When the reflected or refracted rays are divergentthe image is, virtual. The divergent rays can be converged on to a screen by, means of an appropriate converging lens. The convex lens of the, eye does just that. The virtual image here serves as an object for, the lens to produce a real image Note the screen here is not, located at the position of the virtual image. There is no, contradiction., (c) Taller, (d) The apparent depth for oblique viewing decreases from its value, for near-normal viewing. Convince yourself of this fact by drawing, ray diagrams for different positions of the observer., (e) Refractive index of a diamond is about 2.42, much larger than, that of ordinary glass (about 1.5). The critical angle of diamond, is about 24° much less than that of glass A skilled diamondcutter exploits the larger range of angles of incidence (in the, diamond), 24° to 90°, to ensure that light entering the diamond, is totally reflected from many faces before getting out–thus, producing a sparkling effect., For fixed distance s between object and screen the lens equation, does not give a real solution for u or v if f is greater than s/4., Therefore, fmax = 0.75 m., 21.4 cm, , 535

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Physics, 9.21, , (a) (i) Let a parallel beam be the incident from the left on the convex, lens first., f1 = 30 cm and u1= – f ,give v1 = + 30 cm. This image becomes, a virtual object for the second lens., f 2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives,, v2 = – 220 cm. The parallel incident beam appears to diverge, from a point 216 cm from the centre of the two-lens system, (ii) Let the parallel beam be incident from the left on the concave, lens first: f1 = – 20 cm, u1 = – f, give v1 = – 20 cm. This image, becomes a real object for the second lens: f2 = + 30 cmu2 =, – (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel, incident beam appears to diverge from a point 416 cm on the, left of the centre of the two-lens system., Clearly, the answer depends on which side of the lens system, the parallel beam is incident. Further we do not have a simple lens, equation true for all u (and v) in terms of a definite constant of the, system (the constant being determined by f1 and f2, and the separation, between the lenses). The notion of effective focal length, therefore,, does not seem to be meaningful for this system., (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm., Magnitude of magnification due to the first (convex) lens is 3., u 2 = + (120 – 8) cm = +112 cm (object virtual);, f2 = – 20 cm which gives v 2, , 9.22, , 9. 23, , , , 112 u 20, cm, 92, , Magnitude of magnification due to the second (concave), lens = 20/92., Net magnitude of magnification = 0.652, Size of the image = 0.98 cm, If the refracted ray in the prism is incident on the second face at the, critical angle ic, the angle of refraction r at the first face is (60°–ic)., Now, ic = sin–1 (1/1.524) ~ 41°, Therefore, r = 19°, sin i = 0.4962; i ~ 30°, Two identical prisms made of the same glass placed with their, bases on opposite sides (of the incident white light) and faces, touching (or parallel) will neither deviate nor disperse, but will mearly, produce a parallel displacement of the beam., (a) To deviate without dispersion, choose, say, the first prism, to be of crown glass, and take for the second prism a flint, glass prism of suitably chosen refracting angle (smaller than, that of crown glass prism because the flint glass prism, disperses more) so that dispersion due to the first is nullified, by the second., , 536, , (b) To disperse without deviation, increase the angle of flint, glass prism (i.e., try flint glass prisms of greater and greater, angle) so that deviations due to the two prisms are equal, and opposite. (The flint glass prism angle will still be smaller, than that of crown glass because flint glass has higher, refractive index than that of crown glass). Because of the, adjustments involved for so many colours, these are not, meant to be precise arrangements for the purpose required.

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Answers, 9.24, , To see objects at infinity the eye uses its least converging power, = (40+20) dioptres = 60 dioptres. This gives a rough idea of the distance, between the retina and cornea-eye lens: (5/3) cm. To focus an object, at the near point (u = –25 cm)on the retina (v = 5/3 cm), the focal, length should be, , ª 1 3º, «¬ 25  5 »¼, , 9.25, , 9.26, , 9.27, , 9.28, , 1, , 25, cm, 16, , corresponding to a converging power of 64 dioptresThe power of the, eye lens then is (64 – 40) dioptres = 24 dioptres. The range of, accommodation of the eye-lens is roughly 20 to 24 dioptres., No, a person may have normal ability of accommodation of the eyelens and yet may be myopic or hypermetropic. Myopia arises when, the eye-ball from front to back gets too elongated; hypermetropia, arises when it gets too shortened. In practice, in addition, the eye, lens may also lose some of its ability of accommodation When the, eyeball has the normal length but the eye lens loses partially its, ability of accommodation (as happens with increasing age for any, normal eye), the ‘defect’ is called presbyopia and is corrected in the, same manner as hypermetropia., The far point of the person is 100 cmwhile his near point may have, been normal (about 25 cm). Objects at infinity produce virtual image, at 100 cm (using spectacles). To view closer objects i.e.those which, are (or whose images using the spectacles are) between 100 cm and, 25 cm, the person uses the ability of accommodation of his eye-lens., This ability usually gets partially lost in old age (presbyopia). The, near point of the person recedes to 50 cm. To view objects at 25 cm, clearly, the person needs converging lens of power + 2 dioptres., The defect (called astigmatism) arises because the curvature of the, cornea plus eye-lens refracting system is not the same in different, planes. [The eye-lens is usually spherical i.e. has the same, curvature on different planes but the cornea is not spherical in, case of an astigmatic eye.]In the present casethe curvature in the, vertical plane is enough, so sharp images of vertical lines can be, formed on the retina. But the curvature is insufficient in the, horizontal plane, so horizontal lines appear blurred. The defect can, be corrected by using a cylindrical lens with its axis along the vertical., Clearly parallel rays in the vertical plane will suffer no extra, refraction, but those in the horizontal plane can get the required, extra convergence due to refraction by the curved surface of the, cylindrical lens if the curvature of the cylindrical surface is chosen, appropriately., (a) Closest distance = 4, , 1, cm | 4.2 cm, 6, , Farthest distance = 5 cm, (b) Maximum angular magnification = [25/(25/6)] = 6 Minimum, angular magnification = (25/5) = 5, 9.29, , (a), , 1 1, , v 9, , 1, 10, , i.e.,, v = – 90 cm,, Magnitude of magnification = 90/9 = 10., , 537

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Physics, , 9.30, , 9.31, , Each square in the virtual image has an area 10 × 10 × 1 mm2, = 100 mm2 = 1 cm2, (b) Magnifying power = 25/9 = 2.8, (c) No, magnification of an image by a lens and angular magnification, (or magnifying power) of an optical instrument are two separate, things. The latter is the ratio of the angular size of the object, (which is equal to the angular size of the image even if the image, is magnified) to the angular size of the object if placed at the near, point (25 cm). Thus, magnification magnitude is |(v/u )| and, magnifying power is (25/ |u|). Only when the image is located at, the near point |v| = 25 cm, are the two quantities equal., (a) Maximum magnifying power is obtained when the image is at, the near point (25 cm), u = – 7.14 cm., (b) Magnitude of magnification = (25/ |u|) = 3.5., (c) Magnifying power = 3.5, Yes, the magnifying power (when the image is produced at 25 cm), is equal to the magnitude of magnification., (6.25 / 1) = 2.5, , Magnification =, v = +2.5u, , , 9.32, , 538, , 1, 2.5u, , , , 1, , 1, , u, , 10, , i.e.,u = – 6 cm, |v| = 15 cm, The virtual image is closer than the normal near point (25 cm) and, cannot be seen by the eye distinctly., (a) Even though the absolute image size is bigger than the object, sizethe angular size of the image is equal to the angular size of, the object. The magnifier helps in the following way: without it, object would be placed no closer than 25 cm; with it the object, can be placed much closer. The closer object has larger angular, size than the same object at 25 cm. It is in this sense that angular, magnification is achieved., (b) Yes, it decreases a little because the angle subtended at the eye, is then slightly less than the angle subtended at the lens. The, effect is negligible if the image is at a very large distance away., [Note: When the eye is separated from the lens the angles, subtended at the eye by the first object and its image are not, equal.], (c) First, grinding lens of very small focal length is not easy. More, important, if you decrease focal length, aberrations ( both spherical, and chromatic ) become more pronounced. So, in practice, you, cannot get a magnifying power of more than 3 or so with a simple, convex lens. However, using an aberration corrected lens system,, one can increase this limit by a factor of 10 or so., (d) Angular magnification of eye-piece is [( 25/fe ) + 1] ( fe in cm) which, increases if fe is smaller. Further, magnification of the objective, is given by, , vO, , 1, , |u O |, , (|u O |/ f O )  1

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Answers, |u O | is slightly greater than fO. The microscope is used for viewing very close object. So |u O | is small, and, , which is large when, , 9.33, , so is fO., (e) The image of the objective in the eye-piece is known as ‘eye-ring’., All the rays from the object refracted by objective go through the, eye-ring. Therefore, it is an ideal position for our eyes for viewing., If we place our eyes too close to the eye-piecewe shall not collect, much of the light and also reduce our field of view. If we position, our eyes on the eye-ring and the area of the pupil of our eye is, greater or equal to the area of the eye-ring, our eyes will collect, all the light refracted by the objective. The precise location of, the eye-ring naturally depends on the separation between the, objective and the eye-piece. When you view through a microscope, by placing your eyes on one end,the ideal distance between the, eyes and eye-piece is usually built-in the design of the, instrument., Assume microscope in normal use i.e., image at 25 cm Angular, magnification of the eye-piece, =, , 25, , 1 6, 5, Magnification of the objective, , =, , 30, 6, , 5, , 1, 1, , 5u O u O, , 1, 1.25, , which gives uO= –1.5 cm; v0= 7.5 cm |u e| (25/6) cm = 4.17 cm. The, separation between the objective and the eye-piece should be (7.5 +, 4.17) cm = 11.67 cm. Further the object should be placed 1.5 cm from, the objective to obtain the desired magnification., 9.34, , (a) m = ( fO/fe ) = 28, fO º, ª, «¬1  25 »¼ = 33.6, (a) fO + fe = 145 cm, , (b) m =, 9.35, , fO, fe, , (b) Angle subtended by the tower = (100/3000) = (1/30) rad, Angle subtended by the image produced by the objective, =, , h, fO, , h, 140, , Equating the two, h = 4.7 cm., (c), 9.36, , Magnification (magnitude) of the eye-piece = 6. Height of the, final image (magnitude) = 28 cm., , The image formed by the larger (concave) mirror acts as virtual object, for the smaller (convex) mirror. Parallel rays coming from the object, at infinity will focus at a distance of 110 mm from the larger mirror., The distance of virtual object for the smaller mirror = (110 – 20) =, 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror, formula, image is formed at 315 mm from the smaller mirror., , 539

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Physics, 9.37, , The reflected rays get deflected by twice the angle of rotation of the, mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm., , 9.38, , n = 1.33, , CHAPTER 10, 10.1, , (a) Reflected light: (wavelength, frequency, speed same as incident, light), , O = 589 nm, Q = 5.09 u 1014 Hz, c = 3.00 u 108 m s–1, (b) Refracted light: (frequency same as the incident frequency), Q= 5.09u 1014Hz, v = (c/n) = 2.26 × 108 m s–1, O= (v/Q = 444 nm, 10.2, , (a) Spherical, (b) Plane, (c) Plane (a small area on the surface of a large sphere is nearly, planar)., , 10.3, , (a) 2.0 × 108 m s–1, (b) No. The refractive index, and hence the speed of light in a, medium, depends on wavelength. [When no particular, wavelength or colour of light is specified, we may take the given, refractive index to refer to yellow colour.] Now we know violet, colour deviates more than red in a glass prism, i.e. nv > nr., Therefore, the violet component of white light travels slower than, the red component., , 10.4, , O, , 10.5, , K/4, , 10.6, , (a) 1.17 mm, , 10.7, , 0.15°, , 10.8, , tan–1(1.5) ~ 56.3o, , 10.9, , 5000 Å, 6 × 1014 Hz; 45°, , –2, , 1.2 u10 u 0.28 u10, , –3, , 4 u1.4, , m = 600 nm, , (b) 1.56 mm, , 10.10 40 m, 10.11 Use the formula Oc – O =, i.e., v =, , c, , O, , ( O'  O =, , v, , O, , c, , 3 u 108 u 15, = 6.86 × 105 m s–1, 6563, , 10.12 In corpuscular (particle) picture of refraction, particles of light, incident from a rarer to a denser medium experience a force of, attraction normal to the surface. This results in an increase in the, normal component of the velocity but the component along the surface, is unchanged. This means, , 540, , c sin i = v sin r, , or, , v, , sin i, , c, , sin r, , n . Since n > 1, v > c.

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Answers, The prediction is opposite to the experimental results (v < c ). The, wave picture of light is consistent with the experiment., 10.13 With the point object at the centre, draw a circle touching the mirror., This is a plane section of the spherical wavefront from the object, that has just reached the mirror. Next draw the locations of this, same wavefront after a time t in the presence of the mirror, and in, the absence of the mirror. You will get two arcs symmetrically located, on either side of the mirror. Using simple geometry, the centre of the, reflected wavefront (the image of the object) is seen to be at the same, distance from the mirror as the object., 10.14 (a) The speed of light in vacuum is a universal constant independent, of all the factors listed and anything else. In particular, note the, surprising fact that it is independent of the relative motion, between the source and the observer. This fact is a basic axiom, of Einstein’s special theory of relativity., (b) Dependence of the speed of light in a medium:, (i) does not depend on the nature of the source (wave speed is, determined by the properties of the medium of propagation., This is also true for other waves, e.g., sound waves, water, waves, etc.)., (ii) independent of the direction of propagation for isotropic, media., (iii) independent of the motion of the source relative to the, medium but depends on the motion of the observer relative, to the medium., (iv) depends on wavelength., (v) independent of intensity. [For high intensity beams,, however, the situation is more complicated and need not, concern us here.], 10.15 Sound waves require a medium for propagation. Thus even though, the situations (i) and (ii) may correspond to the same relative motion, (between the source and the observer), they are not identical, physically since the motion of the observer relative to the medium, is different in the two situations. Therefore, we cannot expect, Doppler formulas for sound to be identical for (i) and (ii). For light, waves in vacuum, there is clearly nothing to distinguish between, (i) and (ii). Here only the relative motion between the source and the, observer counts and the relativistic Doppler formula is the same for, (i) and (ii). For light propagation in a medium, once again like for, sound waves, the two situations are not identical and we should, expect the Doppler formulas for this case to be different for the two, situations (i) and (ii)., 10.16 3.4 × 10 – 4 m., 10.17 (a) The size reduces by half according to the relation: size ~ O/d., Intensity increases four fold., (b) The intensity of interference fringes in a double-slit arrangement, is modulated by the diffraction pattern of each slit., (c) Waves diffracted from the edge of the circular obstacle interfere, constructively at the centre of the shadow producing a bright, spot., (d) For diffraction or bending of waves by obstacles/apertures by a, large angle, the size of the latter should be comparable to, wavelength. If the size of the obstacle/aperture is much too large, compared to wavelength, diffraction is by a small angle. Here, , 541

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Physics, , (e), , the size is of the order of a few metres. The wavelength of light is, about 5 × 10–7 m, while sound waves of, say, 1 kHz frequency, have wavelength of about 0.3 m. Thus, sound waves can bend, around the partition while light waves cannot., Justification based on what is explained in (d). Typical sizes of, apertures involved in ordinary optical instruments are much, larger than the wavelength of light., , 10.18 12.5 cm., 10.19 0.2 nm., 10.20 (a) Interference of the direct signal received by the antenna with, the (weak) signal reflected by the passing aircraft., (b) Superposition principle follows from the linear character of the, (differential) equation governing wave motion. If y 1 and y 2 are, solutions of the wave equation, so is any linear combination of, y1 and y2. When the amplitudes are large (e.g., high intensity laser, beams) and non-linear effects are important, the situation is far, more complicated and need not concern us here., 10.21 Divide the single slit into n smaller slits of width a c = a/n. The angle, T = n O/a = O/a c Each of the smaller slits sends zero intensity in the, direction T. The combination gives zero intensity as well., , CHAPTER 11, (a) 7.24 × 1018 Hz, (b) 0.041 nm, (b) 0.34 V, (c) 344 km/s, (a) 0.34 eV = 0.54 × 10–19J, 1.5 eV = 2.4 × 10–19 J, (b) 3 × 1016 photons/s, (a) 3.14 × 10–19J, 1.05 × 10–27 kg m/s, (c) 0.63 m/s, 11.5 4 × 1021 photons/m2 s, 11.6 6.59 × 10–34 J s, (b) 3.0 × 1020 photons/s, 11.7 (a) 3.38 × 10–19 J = 2.11 eV, 11.8 2.0 V, 11.9 No, because Q Qo, 11.10 4.73 × 1014 Hz, 11.11 2.16 eV = 3.46 × 10–19J, 11.1, 11.2, 11.3, 11.4, , 11.12 (a) 4.04 × 10–24 kg m s–1, –24, , 11.13 (a) 5.92 × 10, , 11.14 (a) 6.95 × 10, , –25, , –1, , kg m s, , (b) 0.164 nm, (b) 6.50 × 106 m s–1, , J = 4.34 PeV, , 11.15 (a) 1.7 × 10–35 m, , (b) 3.78 × 10, , (b) 1.1 × 10–32 m, , 11.16 (a) 6.63 × 10–25 kg m/s (for both), 11.17, 11.18, 11.19, 11.20, , 542, , –21, , –2, , –28, , (c) 0.112 nm, J = 0.236 neV, , (c) 3.0 × 10–23 m, (b) 1.24 keV, , (c) 1.51 eV, , (b) 0.145 nm, (a) 6.686 × 10 J = 4.174 × 10 eV, O h/p = h/(hQ/c) = c/Q, 0.028 nm, (a) Use eV = (m v 2/2) i.e., v = [(2eV/m)]1/2 ; v = 1.33 × 107 m s–1., (b) If we use the same formula with V = 107 V, we get v = 1.88 ×, 109 m s–1. This is clearly wrong, since nothing can move with a, speed greater than the speed of light (c = 3 × 108 m s–1). Actually,, the above formula for kinetic energy (m v 2/2) is valid only when

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Answers, (v/c) << 1. At very high speeds when (v/c ) is comparable to, (though always less than) 1, we come to the relativistic domain, where the following formulae are valid:, Relativistic momentum p = m v, Total energy E = m c 2, Kinetic energy K = m c 2 – m 0 c 2,, § v2 ·, where the relativistic mass m is given by m m 0 ¨1  2 ¸, © c ¹, , 1/ 2, , m0 is called the rest mass of the particle. These relations also, imply:, E = (p 2c 2 + m 02 c4)1/2, Note that in the relativisitc domain when v/c is comparable to 1, K, or energy t m 0c 2 (rest mass energy). The rest mass energy of electron, is about 0.51 MeV. Thus a kinetic energy of 10 MeV, being much, greater than electron’s rest mass energy, implies relativistic domain., Using relativistic formulas, v (for 10 MeV kinetic energy) = 0.999 c., 11.21 (a) 22.7 cm, (b) No. As explained above, a 20 MeV electron moves at relativistic, speed. Consequently, the non-relativistic formula R = (m 0v/e B ), is not valid. The relativistic formula is, R, , p / eB, , mv / eB, , or, , R = m 0v / eB 1 – v 2 / c 2, , 11.22 We have e V = (m v 2 /2) and R = (m v/e B) which gives (e/m) =, (2V/R 2 B 2 ); using the given data (e/m) = 1.73 × 1011 C kg –1., 11.23 (a) 27.6 keV, (b) of the order of 30 kV, 11.24 Use O= (hc/E) with E = 5.1 × 1.602 × 10–10J to get O= 2.43 × 10–16 m ., 11.25 (a) For O= 500 m, E = (h c/ O) = 3.98 × 10–28J. Number of photons, emitted per second, = 104J s–1/3.98 × 10–28J a 3 × 1031 s–1, We see that the energy of a radiophoton is exceedingly small,, and the number of photons emitted per second in a radio beam, is enormously large. There is, therefore, negligible error involved, in ignoring the existence of a minimum quantum of energy, (photon) and treating the total energy of a radio wave as, continuous., (b) For Q= 6 × 1014 Hz, E a 4 × 10–19J. Photon flux corresponding to, minimum intensity, = 10–10 W m–2/4×10–19J = 2.5 × 108 m–2 s–1, Number of photons entering the pupil per second = 2.5 × 108 ×, 0.4 × 10–4 s–1 = 104 s–1. Though this number is not as large as in, (a) above, it is large enough for us never to ‘sense’ or ‘count’, individual photons by our eye., , I0, = 1.0 × 1015 Hz; O= 6328Å, h, corresponds to Q = 4.7 × 1014 Hz < Q0. The photo-cell will not respond, howsoever high be the intensity of laser light., 11.27 Use e V0 = h Q – I0 for both sources. From the data on the first source,, I0 = 1.40 eV. Use this value to obtain for the second source V0 = 1.50 V., 11.26 I0 = h Q – e V0 = 6.7 × 10–19 J = 4.2 eV; Q0 =, , 543

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Physics, 11.28 Obtain V0 versus Q plot. The slope of the plot is (h/e) and its intercept, on the Q-axis is Q0. The first four points lie nearly on a straight line, which intercepts the Q-axis at Q0 = 5.0 × 1014 Hz (threshold frequency)., The fifth point corresponds to Q < Q 0; there is no photoelectric, emission and therefore no stopping voltage is required to stop the, current. Slope of the plot is found to be 4.15 × 10–15 V s. Using e = 1.6 ×, 10–19 C, h = 6.64 × 10–34 J s (standard value h = 6.626 × 10–34 J s), I0 =, h Q0 = 2.11 V., 11.29 It is found that the given incident frequency Q is greater than Q0, (Na), and Q0 (K); but less than Q0 (Mo), and Q0 (Ni). Therefore, Mo and, Ni will not give photoelectric emission. If the laser is brought closer,, intensity of radiation increases, but this does not affect the result, regarding Mo and Ni. However, photoelectric current from Na and K, will increase in proportion to intensity., 11.30 Assume one conduction electron per atom. Effective atomic area, ~10–20 m2, Number of electrons in 5 layers, =, , 5 u 2 u104 m2, = 1017, 1020 m2, , Incident power, = 10–5 W m–2 × 2 × 10–4 m2 = 2 × 10–9 W, In the wave picture, incident power is uniformly absorbed by all the, electrons continuously. Consequently, energy absorbed per second, per electron, = 2 × 10–9/1017 = 2 × 10–26 W, Time required for photoelectric emission, = 2 × 1.6 × 10–19J/2 × 10–26 W = 1.6 × 107 s, which is about 0.5 year., Implication: Experimentally, photoelectric emission is observed nearly, instantaneously (a10 –9 s): Thus, the wave picture is in gross, disagreement with experiment. In the photon-picture, energy of the, radiation is not continuously shared by all the electrons in the top, layers. Rather, energy comes in discontinuous ‘quanta’. and, absorption of energy does not take place gradually. A photon is either, not absorbed, or absorbed by an electron nearly instantly., 11.31 For O= 1 Å, electron’s energy = 150 eV; photon’s energy = 12.4 keV., Thus, for the same wavelength, a photon has much greater energy, than an electron., 11.32 (a), , h, Thus, for same K, O  decreases with m as, 2 mK, (1/ m ). Now (mn/me) = 1838.6; therefore for the same energy,, , O, , h, p, , (150 eV) as in Ex. 11.31, wavelength of neutron =  1838.6, , ×, , 10 –10 m = 2.33×10–12 m. The interatomic spacing is about a, hundred times greater. A neutron beam of 150 eV energy is, therefore not suitable for diffraction experiments., , 544, , (b) O= 1.45 × 10–10 m [Use O= (h / 3 m k T ) ] which is comparable, to interatomic spacing in a crystal.

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Answers, Clearly, from (a) and (b) above, thermal neutrons are a suitable, probe for diffraction experiments; so a high energy neutron beam, should be first thermalised before using it for diffraction., 11.33 O= 5.5 × 10–12 m, O( yellow light) = 5.9 × 10–7m, Resolving Power (RP) is inversely proportional to wavelength. Thus,, RP of an electron microscope is about 105 times that of an optical, microscope. In practice, differences in other (geometrical) factors, can change this comparison somewhat., 6.63 u 10 –34 Js, = 6.63 × 10–19 kg m s–1, O, 10 –15 m, Use the relativistic formula for energy:, E 2 = c 2p 2 + m 02 c 4 = 9 × (6.63)2 × 10–22 + (0.511 × 1.6)2 × 10–26, a9 × (6.63)2 × 10–22,, the second term (rest mass energy) being negligible., Therefore, E = 1.989 × 10–10 J = 1.24 BeV. Thus, electron energies, from the accelerator must have been of the order of a few BeV., , 11.34 p =, , h, , 4 u10 3, kg, 6 u1023, This gives O= 0.73 × 10–10 m. Mean separation, r = (V/N)1/3 = (kT/p)1/3, For T = 300 K, p = 1.01 × 105 Pa, r = 3.4 × 10–9 m. We find r >> O., 11.36 Using the same formula as in Exercise 11.35, O= 6.2 × 10–9 m which, is much greater than the given inter-electron separation., 11.37 (a) Quarks are thought to be confined within a proton or neutron, by forces which grow stronger if one tries to pull them apart. It,, therefore, seems that though fractional charges may exist in, nature, observable charges are still integral multiples of e., (b) Both the basic relations e V = (1/2 ) m v 2 or e E = m a and, e B v =m v 2/r, for electric and magnetic fields, respectively, show, that the dynamics of electrons is determined not by e, and m, separately but by the combination e/m ., (c) At low pressures, ions have a chance to reach their respective, electrodes and constitute a current. At ordinary pressures, ions, have no chance to do so because of collisions with gas molecules, and recombination., (d) Work function merely indicates the minimum energy required, for the electron in the highest level of the conduction band to, get out of the metal. Not all electrons in the metal belong to this, level. They occupy a continuous band of levels. Consequently,, for the same incident radiation, electrons knocked off from, different levels come out with different energies., (e) The absolute value of energy E (but not momentum p) of any, particle is arbitrary to within an additive constant. Hence, while, O is physically significant, absolute value of Q of a matter wave of, an electron has no direct physical meaning. The phase speed QO, is likewise not physically significant. The group speed given by, , 11.35 Use O, , h, 3m k T, , ; m =, He, , dQ, dE, d § p2 · p, =, ¨, ¸, d (1/ O ) dp, dp © 2m ¹ m, is physically meaningful., , 545

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Physics, CHAPTER 12, (a) No different from, (b) Thomson’s model; Rutherford’s model, (c) Rutherford’s model, (d) Thomson’s model; Rutherford’s model, (e) Both the models, 12.2 The nucleus of a hydrogen atom is a proton. The mass of it is, 1.67 × 10 –27 kg, whereas the mass of an incident D-particle is, 6.64 × 10–27 kg. Because the scattering particle is more massive than, the target nuclei (proton), the D-particle won’t bounce back in even, in a head-on collision. It is similar to a football colliding with a tenis, ball at rest. Thus, there would be no large-angle scattering., 12.3 820 nm., 12.4 5.6 u 1014 Hz, 12.5 13.6 eV; –27.2 eV, 12.6 9.7 × 10 – 8 m; 3.1 × 1015 Hz., 12.7 (a) 2.18 × 106 m/s; 1.09 × 106 m/s; 7.27 × 105 m/s, (b) 1.52 × 10–16 s; 1.22 × 10–15 s; 4.11 × 10–15 s., 12.8 2.12u10–10 m; 4.77 u 10–10 m, 12.9 Lyman series: 103 nm and 122 nm; Balmer series: 656 nm., 12.10 2.6 × 1074, 12.11 (a) About the same., (b) Much less., (c) It suggests that the scattering is predominantly due to a single, collision, because the chance of a single collision increases, linearly with the number of target atoms, and hence linearly, with thickness., (d) In Thomson’s model, a single collision causes very little, deflection. The observed average scattering angle can be, explained only by considering multiple scattering. So it is wrong, to ignore multiple scattering in Thomson’s model. In, Rutherford’s model, most of the scattering comes through a, single collision and multiple scattering effects can be ignored, as a first approximation., 12.12 The first orbit Bohr’s model has a radius a 0 given by, 12.1, , 4 S H 0 (h / 2S)2, . If we consider the atom bound by the gravitational, m ee 2, force (Gmpme/r 2 ), we should replace (e2/4 SH0) by Gm pm e. That is, the, , a0 =, , radius of the first Bohr orbit is given by a 0G, , (h / 2S)2, # 1.2 u 1029 m., Gm pm e2, , This is much greater than the estimated size of the whole universe!, 12.13, , 546, , Q, , 1º, me 4, ª 1,  2», 3 2, 3 «, 2, (4S) H 0 (h / 2S) ¬ (n  1) n ¼, , For large n, Q #, , me 4, 32 S3 H 02 (h / 2S)3 n 3, , me 4 (2n  1), (4S)3 H 02 (h / 2S)3 n 2 (n  1)2

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Answers, Orbital frequency Qc = (v/2 Sr ). In Bohr model v =, , r, , n (h / 2S), 4 S H 0 (h / 2S)2 2, n . This gives Qc =, 2, 2 S mr 2, me, , n (h / 2S), , and, mr, , me 4, 32 S H (h / 2S)3 n 3, 3, , 2, 0, , which is same as Q for large n., §, ·, e2, 12.14 (a) The quantity ¨, 2 ¸ has the dimensions of length. Its value, © 4 S H 0mc ¹, –15, is 2.82 u10 m – much smaller than the typical atomic size., , 4SH 0 (h / 2S)2, has the dimensions of length. Its, me 2, value is 0.53 u 10–10 m – of the order of atomic sizes.(Note that, the dimensional arguments cannot, of course, tell us that we, should use 4S and h/2S in place of h to arrive at the right size.), , (b) The quantity, , 12.15 In Bohr’s model, mvr = nh and, , mv 2, r, , Ze 2, 4 S H 0r 2, , which give, 4 S H 0 =2 2, 1, Ze 2, n, ;r =, mv 2, Ze 2 m, 2, 8 S H 0r, These relations have nothing to do with the choice of the zero of, potential energy. Now, choosing the zero of potential energy at infinity, we have V = – (Z e2/4 S H 0 r) which gives V = –2T and E = T + V = – T, (a) The quoted value of E = – 3.4 eV is based on the customary, choice of zero of potential energy at infinity. Using E = – T, the, kinetic energy of the electron in this state is + 3.4 eV., (b) Using V = – 2T, potential energy of the electron is = – 6.8 eV, (c) If the zero of potential energy is chosen differently, kinetic energy, does not change. Its value is + 3.4 eV independent of the choice, of the zero of potential energy. The potential energy, and the, total energy of the state, however, would alter if a different zero, of the potential energy is chosen., 12.16 Angular momenta associated with planetary motion are, incomparably large relative to h. For example, angular momentum, of the earth in its orbital motion is of the order of 1070h. In terms of, the Bohr’s quantisation postulate, this corresponds to a very large, value of n (of the order of 1070). For such large values of n, the, differences in the successive energies and angular momenta of the, quantised levels of the Bohr model are so small compared to the, energies and angular momenta respectively for the levels that one, can, for all practical purposes, consider the levels continuous., 12.17 All that is needed is to replace m e by mP in the formulas of the Bohr, model. We note that keeping other factors fixed, r v (1/m) and E v m., Therefore,, , T=, , rm, , E P=, , re m e, mm, Ee m m, me, , 0.53 u10, 207, , 13, , = 2.56 × 10–13 m, , – (13.6 × 207) eV # – 2.8 keV, , 547

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Physics, CHAPTER 13, 13.1, 13.2, 13.3, 13.4, 13.5, , (a) 6.941 u, (b) 19.9%, 80.1%, 20.18 u, 104.7 MeV, 8.79 MeV, 7.84 MeV, 1.584 × 1025 MeV or 2.535×1012J, , 13.6, , i), , 226, 88, , 32, 15, , iii), , 11, 6, , v), , vii), , 222, 86, , Ra o, , Rn + 24 He, , 32, P o 16, S + e– + Q, , C o 115 B + e+ +Q, , 120, 54, , ii), , 242, 94, , 4, Pu o 238, 92 U + 2 He, , iv), , 210, 83, , vi), , 97, 43, , –, B o 210, 84 Po + e +Q, , +, Tc o 97, 42 Mo + e +Q, , Xe + e+ o 120, 53 I + Q, , 13.7, 13.8, 13.9, 13.10, 13.11, 13.12, , (a) 5 T years, (b) 6.65 T years, 4224 years, 7.126 ×10– 6 g, 7.877 ×1010 Bq or 2.13 Ci, 1.23, (a) Q = 4.93 MeV, ED = 4.85 MeV, (b) Q = 6.41 MeV, ED = 6.29 MeV, , 13.13, , 11, 6, , +, +Q, C o 11, 6 B + e +Q, , Q, , ªm N, ¬, , 11, 6, , C – mN, , 11, 6, , B – m e º¼ c 2 ,, , where the masses used are those of nuclei and not of atoms. If we, use atomic masses, we have to add 6me in case of 11C and 5me in, case of 11B. Hence, Q, , ªm, ¬, , 11, 6, , C –m, , 11, 6, , B – 2m e º¼ c 2 (Note me has been doubled), , Using given masses, Q = 0.961 MeV., Q = Ed+ Ee+ EQ, The daughter nucleus is too heavy compared to e+ and Q, so it carries, negligible energy (Ed |0). If the kinetic energy (EQ) carried by the, neutrino is minimum (i.e., zero), the positron carries maximum, energy, and this is practically all energy Q; hence maximum Ee |Q)., 13.14, , 23, 23, 2, 23, Ne o 11, Na + e – +Q  Q ; Q = ª¬m N 10 Ne – m N 11 Na – m e º¼ c , where the, masses used are masses of nuclei and not of atoms as in Exercise, , 23, 10, , 23, 23, ªm 10, Ne – m 11, Na º¼ c 2 . Note me has, ¬, been cancelled. Using given masses, Q = 4.37 MeV. As in Exercise 13.13,, maximum kinetic energy of the electron (max Ee ) = Q = 4.37 MeV., 13.15 (i) Q = – 4.03 MeV; endothermic, , 13.13. Using atomic masses Q, , (ii) Q = 4.62 MeV; exothermic, 13.16 Q = m, , 56, 26, , Fe – 2m, , 28, 13, , Al = 26.90 MeV; not possible., , 26, , 13.17 4.536 × 10 MeV, , 548, , 13.18 Energy generated per gram of, , 235, 92, , U =, , 6 u 1023 u 200 u 1.6 u 10 13, J g 1, 235

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Answers, 235, 92, , The amount of, , U consumed in 5y with 80% on-time, , 5 u 0.8 u 3.154 u 1016 u 235, g, 1.2 u 1.6 u 1013, , =, , The initial amount of, , 235, 92, , = 1544 kg, , U = 3088 kg., , 4, , 13.19 About 4.9 × 10 y, 13.20 360 KeV, 13.22 Consider the competing processes:, A, Z, , X o ZA –1 Y + e+  Qe  Q1 (positron capture), , e– , , A, Z, , Q1, , ªm N, ¬, , A, Z, , X – mN, , ªm N, ¬, , A, Z, , X – Z me – m, , X o ZA –1 Y  Qe  Q2 (electron capture), , ªm, ¬, , A, Z, , X –m, , A, Z –1, , A, Z –1, , Y – m e º¼ c 2, A, Z –1, , Y – Z – 1 m e – m e º¼ c 2, , Y – 2m e º¼ c 2, , ªm N ZA X  m e – m N ZA –1 Y º c 2, ªm ZA X – m ZA –1 Y º c 2, ¬, ¼, ¬, ¼, This means Q1 > 0 implies Q2 > 0 but Q2 > 0 does not necessarily, mean Q1 > 0. Hence the result., , Q2, , 13.23, , 25, 12, , Mg : 9.3%,, , 26, 12, , Mg :11.7%, , 13.24 Neutron separation energy Sn of a nucleus, , A, Z, , X is, , A 1, A, 2, ¬ªm N ( Z X)  m n  m N ( Z X)¼º c, , Sn, , 41, From given data , Sn ( 20, Ca) 8.36MeV, Sn ( 27, 13 Al) 13.06MeV, , 13.25 209 d, 14, 6, , 13.26 For, , C emission, , 209, 14, 2, [m N ( 223, 88 Ra)  m N ( 82 Pb)  m N ( 6 C)]c, , Q, , 209, 14, 2, = 31.85 MeV, [m ( 223, 88 Ra)  m ( 82 Pb)  m ( 6 C)]c, , For, 13.27 Q, 13.28 (a), , 4, 2, , He emission, Q, , 219, 4, 2, = 5.98MeV, [m ( 223, 88 Ra)  m ( 86 Rn)  m ( 2 He)]c, , 140, 99, 2, = 231.1 MeV, [m ( 238, 92 U)  m n  m ( 58 Ce)  m ( 44 Ru)]c, , [m ( 21 H)  m ( 31 H)  m ( 42 He)  m n ]c 2 17.59 MeV, , Q, , (b) K.E. required to overcome Coulomb repulsion = 480.0 keV, 480.0 KeV = 7.68×10–14 J = 3kT, , ?T, , 7.68 u 10 –14, 3 u 1.381 u 10 –23, , (as k, , 1.381 u 10 –23 J K –1 ), , = 1.85 ×109 K (required temperature), 13.29 K max E1–, , Q J1, , 0.284 MeV, K max E2–, , 2.627 u 1020 Hz , Q J 2, , 0.960 MeV, 0.995 u 1020 Hz , Q J 3, , 1.632 u 1020 Hz, , 549

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Physics, 13.30 (a) Note that in the interior of Sun, four 11 H nuclei combine to form, one 42 He nucleus releasing about 26 MeV of energy per event., Energy released in fusion of 1kg of hydrogen = 39 ×1026 MeV, (b) Energy released in fission of 1kg of, , 235, 92, , U = 5.1×1026 MeV, , The energy released in fusion of 1 kg of hydrogen is about 8, times that of the energy released in the fusion of 1 kg of uranium., 13.31 3.076 × 104 kg, , CHAPTER 14, 14.1, 14.2, 14.3, 14.4, 14.5, 14.6, 14.7, 14.8, 14.9, 14.10, 14.11, 14.12, , (c), (d), (c), (c), (c), (b), (c), (c), 50 Hz for half-wave, 100 Hz for full-wave, vi = 0.01 V ; IB= 10 PA, 2V, No (hQhas to be greater than Eg )., ne | 4.95 u 1022; nh = 4.75 u 109 ; n-type since ne >> nh, For charge neutrality ND – NA = ne – nh ; ne.nh = n i2, Solving these equations, ne =, , 1ª, ND  NA , 2 ¬«, , ND  NA, , 2, ,  4n i2 º, ¼», , 14.13 About 1 × 105, 14.14 (a) 0.0629 A,, (b) 2.97 A,, (c) 0.336 :, (d) For both the voltages, the current I will be almost equal to I0,, showing almost infinite dynamic resistance in the reverse, bias., 14.16 NOT ;, A Y, 0 1, 1 0, 14.17 (a) AND, (b) OR, 14.18 OR gate, 14.19 (a) NOT,, (b) AND, , CHAPTER 15, 15.1, 15.2, 15.3, , 550, , (b) 10 kHz cannot be radiated (antenna size), 1 GHz and 1000 GHz, will penetrate., (d) Consult Table 15.2, (c) Decimal system implies continuous set of values

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Answers, 15.4, , No. Service area will be A = S d T2 =, , 15.5, , P, Am, , 15.6, , 0.75, , 22, u 162 u 6.4 u 106 = 3258 km2., 7, , Am, Ac, , 0.75 u 12, , 9 V., , (a), , (b) P= 0.5, 15.7, , Since the AM wave is given by (Ac + Am sinZm t) cos Zct, the, maximum amplitude is M1 = Ac + Am while the minimum amplitude, is M2 = Ac – Am. Hence the modulation index is, , m, , Am, Ac, , M1 – M 2, M1  M 2, , 8, 12, , 2, 3., , With M2 = 0, clearly, m =1, irrespective of M1., 15.8, , Let, for simplicity, the received signal be, A1 cos (Zc + Zm ) t, The carrier Ac cos Zct is available at the receiving station. By, multiplying the two signals, we get, A1Ac cos (Zc + Za ) t cos Zct, =, , A1 Ac, ªcos 2Z c  Zm t  cos Zm t º¼, 2 ¬, , If this signal is passed through a low-pass filter, we can record, the modulating signal, , A1 Ac, cos Zm t ., 2, , 551

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Physics, , APPENDICES, , APPENDIX A 1, THE GREEK ALPHABET, , APPENDIX A 2, COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES

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Appendices, APPENDIX A 3, SOME IMPORTANT CONSTANTS, , OTHER USEFUL CONSTANTS, , 533

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Physics, , BIBLIOGRAPHY, , TEXTBOOKS, For additional reading on the topics covered in this book, you may like to consult one or more of the following, books. Some of these books however are more advanced and contain many more topics than this book., 1, , Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984)., , 2, , Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987)., , 3, , Advanced Physics, Tom Duncan, John Murray (2000)., , 4, , Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition, John Wily (2004)., , 5, , University Physics (Sears and Zemansky’s), H.D. Young and R A. Freedman, 11th Edition,, Addison—Wesley (2004)., , 6, , Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and, V. Shalnov, MIR Publishers, (1971)., , 7, , Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965)., , 8, , Berkeley Physics Course (5 volumes) McGraw Hill (1965)., , 9, , a., , Vol. 1 – Mechanics: (Kittel, Knight and Ruderman), , b., , Vol. 2 – Electricity and Magnetism (E.M. Purcell), , c., , Vol. 3 – Waves and Oscillations (Frank S. Crawford), , d., , Vol. 4 – Quantum Physics (Wichmann), , e., , Vol. 5 – Statistical Physics (F. Reif ), , Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967)., , 10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw, Hill (1977)., 11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill, (1978)., 12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960)., 13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, NCERT (1967)., 14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000)., 15 Physics, Patrick Fullick, Heinemann (2000)., 16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998)., 17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999)., 18 University Physics, Harris Benson, John Wiley (1996)., 19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994)., 20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988)., 21 Physics, Hans C. Ohanian, W.W. Norton (1989).

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Bibliography, Advanced Physics, Keith Gibbs, Cambridge University Press (1996)., Understanding Basic Mechanics, F. Reif, John Wiley (1995)., College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997)., Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987)., Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988)., Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F., Redish, John Wiley (2005)., 28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005)., 22, 23, 24, 25, 26, 27, , GENERAL BOOKS, For instructive and entertaining general reading on science, you may like to read some of the following books., Remember however, that many of these books are written at a level far beyond the level of the present book., 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, , 12, 13, 14, 15, 16, , Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967)., The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962)., Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966)., Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986)., One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961)., The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965)., Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934)., The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press, (1930)., The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980)., The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977)., Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978)., Book 1: Physical Bodies, Book 2: Molecules, Book 3: Electrons, Book 4: Photons and Nuclei., Physics can be Fun, Y. Perelman, MIR Publishers (1986)., Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985)., Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985)., How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005)., Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M., Hazen, John Wiley (2004)., , 553