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LIGHT : REFLECTION, , 3, CHAPTER, , CONTENTS, , , , , , , , , , The Nature of light, Reflection of light, Laws of reflection of light, Nature of image, Reflection from the plane mirror, Reflection from spherical mirrors, , Rules for image formation by ray, diagram method,  Image formation by spherical, mirrors in different cases,  Numerical method in spherical mirror,  Summary of images by spherical, mirror, THE NATURE OF LIGHT, Light is a form of energy (optical energy), which helps us in seeing objects by its, presence., (A) THEORIES ABOUT NATURE OF LIGHT :, (i) Particle nature of light (Newton's, corpuscular theory) :, According to Newton light travels in space, with a great speed as a stream of very small, particles called corpuscles., This theory was failed to explain interference, of light and diffraction of light. So wave, theory of light was discovered., (ii) Wave nature of light :, Light waves are electromagnetic waves so, there is no need of medium for the propagation, , of these waves. They can travel in vacuum, also. The speed of these waves in air or in, vacuum is maximum i.e., 3 × 108 m/s., Photoelectric effect was not explained with the, help of wave theory, so Plank gave a new theory, which was known as quantum theory of light., (iii) Quantum theory of light :, When light falls on the surface of metals like, caesium, potassium etc., electrons are given, out. These electrons are called 'photoelectrons' and phenomenon is called 'photoelectric effect'., This was explained by Einstein. According to, plank light consisted of packets or quanta's of, energy called photons. The rest mass of, photon is zero. Each quanta carries energy, E = h., h  Planck's constant = 6.6 × 10–34 J-s.,   Frequency of light, Some phenomenons like interference of light,, diffraction of light are explained with the help, of wave theory but wave theory was failed to, explain the photo electric effect of light. It, was explained with the help of quantum, theory. So, light has dual nature., (i) Wave nature, (ii) Particle nature,  (B) SOURCES OF LIGHT., The objects which emit (give) light are called, luminous objects. It may be natural or manmade. Sun is a natural source of light and, electric lamp, and oil lamp, etc. are manmade source of light., The Non-luminous objects do not emit light., However, such objects become visible due to the, reflection of the light falling on them. Moon does, not emit light. It becomes visible due to the, reflection of the sunlight falling on it.

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too. Scientists have assigned a value of 299,, 792, 458 m/s to the speed of light in vacuum., , (C), PROPAGATION OF LIGHT, Light travels along straight lines in a medium, or in vacuum. The path of light changes only, when there is an object in its path or where, the medium changes. Apart from vacuum and, gases, light can travel through some liquids, and solids as well.,  Transparent medium : A medium in which, light can travel freely over large distances is, called a transparent medium., , , , REFLECTION OF LIGHT, , , Examples: Water, glycerin, glass and clear, plastics are transparent., , , Opaque : A medium in which light cannot, travel is called opaque., Examples : Wood, metals, bricks, etc., are, opaque., , , , Translucent : A medium in which light can, travel some distance, but its intensity reduces, rapidly., Such, materials, are, called, translucent., , According to current scientific theories, no, material particle can travel at a speed greater, than that of light in vacuum., , Definition. When light rays are incident on, an opaque polished surface (medium), these, are returned back in the same medium., This phenomenon of returning of ray of light, in the same medium, is called reflection of, light., , Definition of some associated terms :, P, , X, , , , , , Light is a transverse wave, and does not need, any medium to travel. Light can travel, through vaccum. Its speed through vaccum is, 3 × 108 m/s., The velocity of light changes when it travels, from one medium to another., , , , The wavelength () of light changes when it, goes from one medium to another., , , , The frequency (f) of the light wave remains, the same in all media., , , , Point of incidence : The point on the, reflecting surface at which a ray of light, strikes, is called the point of incidence. In, diagram, O is the point of incidence., , , , Normal : A perpendicular drawn on the, reflecting surface at the point of incidence, is, called the normal. In diagram, ON is the, normal., , , , Incident ray : The ray of light which strikes, the reflecting surface at the point of incidence, is called the incident ray. In diagram, PO is, the incident ray., , , , Reflected ray : The ray of light reflected, from the reflecting surface from the point of, incidence, is called the reflected ray. In, diagram, OQ is the reflected ray., , surfaces, such as mirrors, polished metal, surfaces, etc., travels from one transparent medium to, another., , , Light does not need a material medium to, travel, that is, it can travel through a vacuum, , Y, , Reflecting surface : The surface from which, the light is reflected, is called the reflecting, surface. In diagram, XY is the reflecting, surface., , Light gets reflected back from polished, , Light undergoes refraction (bending) when it, , O, , , , Light is an electromagnetic wave., , Light travels in a straight line., , , Q, , i r, , Examples : Oil, (D) THE CHARACTERISTICS OF LIGHT, , N

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, , , , Angle of incidence : The angle that the, incident ray makes with the normal, is called, the angle of incidence. It is represented by the, symbol i. In diagram, angle PON is the angle, of incidence., Angle of reflection : The angle that the, reflected ray makes with the normal, is called, the angle of reflection. It is represented by the, symbol r. In diagram, QON is the angle of, reflection., , Plane of incidence : The plane in which the, , normal and the incident ray lie, is called the, plane of incidence. In diagram, the plane of, the bookpage, is the plane of incidence., Plane of reflection : The plane in which the, , normal and the reflected ray lie, is called the, plane of reflection. In diagram, the plane of, the book page, is the plane of reflection., , LAWS OF REFLECTION OF LIGHT, , , First law : The incident ray, the reflected ray, and the normal at the point of incidence, all, lie in the same plane., , , , Second law : The angle of reflection (r) is, always equal to the angle of incidence (i)., i.e.,, , Incident ray, Reflected ray, , Normal Incidence, (ii) Laws of reflection are also obeyed when light, is reflected from the spherical or curved, surfaces as shown in figure (a) and (b), , I, , N, , R, , I, , N, , i r, , i r, , (a), , (b), , R, , Reflection from curved surface, (iii) Regular and Irregular Reflection :, Regular Reflection – The phenomenon due, , to which a parallel beam of light travelling, through a certain medium, on striking some, smooth polished surface, bounces off from it,, as parallel beam, in some other fixed, direction is called Regular reflection., , r = i, , (For normal incidence, i = 0, r = 0. The ray is, reflected back along normal)., (i) A ray of light striking the surface normally, retraces its path., When a ray of light strikes a surface, normally, then angle of incidence is zero i.e.,, i = 0. According to the law of reflection,, r = i, r = 0 i.e. the reflected ray is also, perpendicular to the surface. Thus, an, incident ray normal to the surface (i.e., perpendicular to the surface) retraces its path, as shown in figure., , Regular reflection, Regular reflection takes place from the, objects like looking glass, still water, oil,, highly polished metals, etc., Regular reflection is useful in the formation of, images, e.g., we can see our face in a mirror, only on account of regular reflection. However,, it causes a very strong glare in our eyes.

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Irregular reflection or Diffused reflection :, , REFLECTION FROM THE PLANE MIRROR, Relation between the distances of the object, and the image from the plane mirror is that, they are equal., To verify this, consider the geometrical, construction shown in figure. Rays OP and, OD, starting from the object O, fall on the, mirror. The ray OP is perpendicular to the, mirror and hence, reflects back along PO. The, incident ray OD and the reflected ray DE, make equal angles with the normal DG. The, two reflected rays when produced backwards, meet at I, producing a virtual image there., , Irregular or diffused reflection, The phenomenon due to which a parallel, beam of light, travelling through some, medium, gets reflected in various possible, directions, on striking some rough surface is, called irregular reflection or diffused, reflection., , E, , The reflection which takes places from, ground, walls, trees, suspended particles in, air, and a variety of other objects, which are, not very smooth, is irregular reflection., , G, O, , Irregular reflection helps in spreading light, energy over a vast region and also decreases, its intensity. Thus, it helps in the general, illumination of places and helps us to see, things around us., , 2. A real image can be, obtained on a screen., , 2. A virtual image cannot, be obtained on a screen., , 3. A real image is inverted 3. A virtual image is erect, with respect to the, with respect to the, object., object., , (DG || IO),, (law of reflection),, , and, GDO = DOI, , (DG || IO)., , Hence, DIO = DOI, , Now, , OD = DI, OP2 = OD2 – DP2, and, PI2 = DI2 – DP2, , Definition : Incident rays starting from a, point object, and reflected from a mirror,, either actually meet at or appear to come from, a point. The other point is called the image of, the point object., Virtual Image, 1. A virtual image is, formed when two or, more rays appear to, be coming from a point, behind the mirror., , I, , EDG = GDO, , NATURE OF IMAGE, , Real Image, 1. A real image is formed, when two or more, reflected rays meet at, a point in front of the, mirror., , P, , Now, EDG = DIO, ,  Note : Laws of reflection are always valid no, matter whether reflection is regular or, irregular., , , , D, , From (i), since OD = DI, OP2 = PI2 or OP = PI., So, in the case of a plane mirror, the image is, formed as far behind the mirror at the same, distance as the object is in front of it.,  SOME IMPORTANT RESULTS ABOUT, REFLECTION FROM PLANE SURFACES, , , Lateral inversion : When you see your, image in a vertical plane mirror such as that, fixed to an almirah, the head in the image is, up and the feet are down, the same way as, you actually stand on the floor. Such an, image is called an erect image. However, if, you move your right hand, it will appear as if

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the left hand of your image is moving. If you, keep a printed page in front of a plane mirror,, the image of the letters appear erect but, inverted laterally, or sideways. Such an, inversion is called lateral inversion., R, , L, , R, , , , Deviation :  is defined as the angle between, directions of incident ray and emergent ray., So if light is incident at an angle of incidence, i,,  = 180º – (i + r) = (180º – 2i), [as i = r], , L, r, i, , Image, , Object, , , , Plane mirror, , So if light is incident at angle of 30º, ,   Relative motion of object and image :, Case I :, If an object moves towards (or away from) a, plane mirror at speed v, , , , , ,  = (180º – 2 × 30º) = 120º and for normal, incidence i = 0º,  = 180º, , Characteristics of the image formed by a plane, mirror :, (i) The image formed by a plane mirror is, virtual., (ii) The image formed by a plane mirror is erect., , The image will also approach (or recede) at, speed v, The speed of image relative to object will be, v – (–v) = 2v., Case II :, If the mirror is moved towards or (away, from) the object with speed 'v', The image will move towards (or away from), the object with a speed '2v'., , , Multiple Reflection, Number of images formed by combination of plane, mirrors depends upon angle between mirrors., , 90°, ,  If there are two plane mirrors inclined to each, other at an angle 90° , the number of images of a, point object formed are 3., , (iii) The size of the image formed by a plane, mirror is same as that of the size of the object., If object is 10 cm high, then the image of this, object will also be 10 cm high., (iv) The image formed by a plane mirror is at the, same distance behind the mirror as the object is, in front of it. Suppose, an object is placed at 5 cm, in front of a plane mirror then its image will be at, 5 cm behind the plane mirror., (v) The image formed by a plane mirror is laterally, inverted, i.e., the right side of the object appears as, the left side of its image and vice-versa., , , , , , Lateral Inversion

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REFLECTION FROM SPHERICAL MIRROR, , Radius of curvature : The distance between, , the pole and the centre of curvature of the, mirror, is called the radius of curvature of the, mirror. It is equal to the radius of the, spherical shell of which the mirror is a, section. In diagram, PC is the radius of, curvature of the mirror. It is represented by, the symbol R., ,  INTRODUCTION : There are two types of, , spherical mirrors:, (i) Concave mirror :, A, Principal, axis, P, , F, , , , B, , Focal length : The distance between the pole, and principal focus of the mirror, is called the, focal length of the mirror. In diagram, PF is, the focal length of the mirror. It is represented, by the symbol f., , (ii) Convex mirror :, A, Principal, axis, F C, , P, , TERMS ASSOCIATED, SPHERICAL MIRRORS., , WITH, , , , Aperture. The diameter of the circular rim of, the mirror. In diagram AB is the aperture of, the mirror., , , , Pole : The centre of the spherical surface of, the mirror is called the pole of the mirror. It, lies on the surface. In diagram, P is the pole, of the mirror., , , , , , , , R, for convex, 2, , f= , , R, for concave, 2, , Principal section : A section of the spherical, , B, ,   SOME, , f , , Centre of curvature : The centre of the, spherical shell, of which the mirror is a, section, is called centre of curvature of the, mirror. It lies outside the surface. Every point, on mirror surface lies at same distance from, it. In diagram, C is the centre of curvature of, the mirror., Principal axis : The straight line passing, through the pole and the centre of curvature, of the mirror, is called principal axis of the, mirror., Principal focus : It is a point on the principal, axis of the mirror, such that the rays incident, on the mirror parallel to the principal axis, after reflection, actually meet at this point (in, case of a concave mirror) or appear to come, from it (in case of a convex mirror). In, diagram, F is the principal focus of the, mirror., , mirror cut by a plane passing through its, centre of curvature and the pole of the mirror,, is called a principal section of the mirror. It, contains the principal axis. In diagram, APB, is the principal section of the mirror cut by, the plane of the book page., , RULES FOR IMAGE FORMATION BY, RAY DIAGRAM METHOD, ,  RULES FOR IMAGE FORMATION FROM, CONCAVE MIRROR, , (a)When the light ray incident parallel to the, principal axis., A ray light parallel to, the principal axis, Principal axis, C, F, , P, , OR, When the light ray incident towards focus., Reflected ray goes parallel, to the principal axis, P, C, , F

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(b)When the light ray incident towards centre of, curvature., A ray of light passing, through the centre of, curvature, C, , F, , P, , (c)When the light ray incident on the pole of the, mirror., Incident ray, C, , Fi, r, , P, , Reflected ray, ,  RULES FOR IMAGE FORMATION FROM, ,  SIGN CONVENTION, , (a) Description : It is a convention which fixes, the signs of different distances measured. The, sign convention to be followed is the New, Cartesian sign convention. It gives the, following rules :, 1. All distances are measured from the pole of, the mirror., 2. The distances measured in the same direction, as the direction of incident light from pole are, taken as positive., 3. The distances measured in the direction, opposite to the direction on incident light, from pole are taken as negative., 4. Distances, measured, upward, and, perpendicular to the principal axis, are taken, as positive., 5. Distances, measured, downward, and, perpendicular to the principal axis, are taken, as negative., , CONVEX MIRROR, , (a)When the light ray incident parallel to the, principal axis., Incident ray, F, , OR, ,   When the light ray incident parallel to the, principal axis., Incident ray, Reflected ray, , F, , P, Principal axis, , (b)When the light ray incident on the mirror, directing towards centre of curvature., Rays traveling towards, C behind the mirror, 90°, P, , MIRROR IN DIFFERENT CASES, Introduction : From mirror formula, we find that, for a mirror of a fixed focal length f, as object, distance u changes, image distance  also, changes., , Reflected ray, P, Principal axis, , IMAGE FORMATION BY SPHERICAL, , F C, , (A) BY CONCAVE MIRROR :, (1) Object at Infinity, A point object lying on the principal axis., Rays come parallel to the principal axis and, after reflection from the mirror actually meet, at the focus F., The image is formed at F. It is real and point, sized (fig.), , C, , F, , P, , Fig. Concave mirror : point object at infinity,, image at focus., (2) Object Beyond Centre of Curvature, Real object AB has its image AB formed, between focus and centre of curvature. The, image is real-inverted and diminished.

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B, A', A C, , B, A, C, F, , P, F, , Parallel rays, to infinity, , B', , Concave mirror : object beyond centre of, curvature, image between focus and centre of, curvature., (3) Object at Centre of Curvature, Real object AB, has its image AB formed at, centre of curvature., The image is real-inverted and has same size, as the object. (fig.)., B, , Concave mirror : object at focus image at, infinity., (6) Object between Focus and Pole, Real object AB has its image AB formed, behind the mirror. The image is virtual-erect, and enlarged., B', B, C, , A, C A', , P, , F, , P, , FA, , P, , A', , Fig. Concave mirror : Object between pole, and focus, image behind the mirror., , B', , Concave mirror : object at centre of, curvature, image at centre of curvature, (4) Object between Centre of Curvature and Focus, Real object AB has its image AB formed, beyond centre of curvature., The image is real-inverted and enlarged, (bigger in size than the object). (Fig.), , (B) BY CONVEX MIRROR :, , (1) Object at infinity, A point object lying on the principal axis., Rays come parallel to the principal axis and, after reflection from the mirror, appear to, diverge from focus F behind the mirror., The image is formed at F., The image is virtual and point sized. [fig.], , B, A', CA, , F, , P, , P, , F, , C, , B', , Concave mirror : object between centre of, curvature and focus, image beyond centre of, curvature., (5) Object at Focus, Real object AB has its image formed at, infinity., The image is imaginary inverted (reflected, rays go downward) and must have very large, size., , Fig. Convex mirror : point object at infinity,, virtual image at focus., (2) Object at anywhere on principle axis, , O, , IF C, , Image is virtual & point sized

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NUMERICAL METHOD IN SPHERICAL, (A) Mirror formula,  Definition : The equation relating the object, distance (u) the image distance (v) and the, mirror focal length (f) is called the mirror, formula., 1 1 1,  , v u f, , convex mirror, , MIRROR, , Between, P and F, , Behind, the, mirror, , Enlarged, , Virtual, and, erect, , at infinite, , at focus, , highly, virtual, diminished point, size, , anywhere between, on, pole &, principal focus, axis, , diminished virtual, erect, , , , Assumptions made :, (i) The mirror has a small aperture., (ii) The object lies close to principal axis of the, mirror., (iii) The incident rays make small angles with the, mirror surface or the principal axis., ,  SOLVED EXAMPLES , , , , Ex.1, , An object is placed in front of a plane mirror., If the mirror is moved away from the object, through a distance x, by how much distance, will the image move?, , Sol., , Suppose the object O was initially at a, distance d from the plane mirror M as shown, in fig. The image formed at O’ is at a distance, d behind the mirror. Now, the mirror is, shifted by a distance x to M’ such that the, distance of the object from M’ becomes d + x., The image now formed at O” which is also at, a distance d + x from M’., , (B) linear magnification For spherical mirrors,  Definition : The ratio of the size of the, image, as formed by reflection from the, mirror to the size of the object, is called linear, magnification produced by the mirror. It is, represented by the symbol m., v height of image, m , u height of object, (C) Power of mirror, Power of a mirror [in Diopters] =, , 1, f (in metre), , SUMMARY OF IMAGES BY SPHERICAL, , Concave mirror, , Position, of Image, , Size of, Image, , Nature, of, Image, , At infinity At focus, F, , Highly, Real and, diminished inverted, , Beyond C, , Between, F, and C, , Diminished Real and, inverted, , At C, , At C, , Same size, , Real and, inverted, , Between, F and C, , Beyond C Enlarged, , Real and, inverted, , At F, , At infinity Highly, enlarged, , Real and, inverted, , , , Ex.2, , M', O', , d, , MIRROR, Position, of object, , M, , O, , O'', , d, , So, OM = MO' = d, OM' = M'O" = d + x, Thus, OO" = OM' + M'O" = 2(d + x) ...(1), when OO' = OM + MO' = 2d, ...(2), , O'O" = OO" – OO', = 2(d + x) – 2d, = 2x, Thus, the image is shifted from O' to O" by a, distance 2x., An insect is at a distance of 1.5m from a, plane mirror. Calculate the following?, (i) Distance at which the image of the insect, is formed., (ii) distance between the insect and its image.

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Sol., , (i) The distance of insect from the mirror, = 1.5 m, , 15 cm, , The distance of insect from the mirror, is also equal to 1.5 m. The image is, formed at 1.5 m behind the mirror., (ii) The distance between the insect and, image, = 1.5 + 1.5 = 3m, , B', , Sol., , Ex.4, , A, B, , A', , F, 10 cm, , 30 cm, , Sol., , or, , h' = – 2 × h = – 2 × 3, = – 6 cm, So the height of the image is 6 cm. The minus, sign shows that it is on the lower side of the, principal axis, i.e. the image is inverted., Ex.6, , A 1.4 cm long object is placed perpendicular, to the principal axis of a convex mirror of, focal length 15 cm at a distance of 10 cm, from it. Calculate the following :, (i) location of the image, (ii) height of the image, (iii) nature of the image, , We have u = –15 cm and f = –10 cm, , A, , 1, 1, 1, Using the relation,,, +, =, we get, f, v, u, , or, , 1, 1, 1, +, =, v,  15,  10, 1, 1, 1, 1, =, –, =–, v, 15, 10, 30, , or, v = –30 cm, So the image will be formed 30 cm from the, mirror. Since  has a negative sign, the image, is formed to the left of the mirror, i.e. in front, of the mirror as shown in fig., Ex.5, , A 3 cm long object is placed perpendicular to, the principal axis of a concave mirror. The, distance of the object from the mirror is 15, cm, and its image is formed 30 cm from the, mirror on the same side of the mirror as the, object . Calculate the height of the image, formed., , h', v, , h, u, h', (30), =–, =2, h, (15), , or, , An object is placed at a distance of 15 cm, from a concave mirror of focal length 10 cm., Find the position of the image., , B', , Here u = –15 cm and  = –30 cm, Size of the object, h = 2 cm, Magnification, m = m =, , 30cm, = 15cm, 2, , 15 cm, , F, , 30 cm, , A concave mirror is made up by cutting a, portion of a hollow glass sphere of radius 30, cm. Calculate the focal length of the mirror., The radius of curvature of the mirror = 30 cm, Thus, the focal length of the mirror, =, , B, , A', , Sol., Ex.3, , A, , A', B10 cm, , 6 cm B', , F, , C, , 15cm, , Sol. (i) For a convex mirror, focal length is positive., Therefore, f = +15 cm and u = –10 cm, Using the relation,, , 1 1 1, + = , we get, v u f, 1, 1, 1, +, =, v  10 15, , or, , 1, 1, 1, 5 1, =, + =, =, v, 15 10 30 6, , or, ,  = 6 cm, , Since  is positive, the image is formed to the, right of the mirror at a distance 6 cm from it.

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(ii) Magnification,, or, , Sol., m=, , Now, using the mirror formula,, , h' = + 0.6 × h0 = 0.6 × 1.4, = 0.84 cm., Thus, the height of the image is 0.84 cm., (iii) Since h' is positive, the image will be on the, same side of the principal axis as the object., Hence, the image is virtual, erect and, diminished., , 1, 1, 1, +, =, f,  60,  30, , or, , 1, 3, 1, =–, =–, f, 60, 20, , or f = – 20 cm, Focal length of the mirror = 20 cm, , An object is placed at a distance of 40 cm, from a convex mirror of focal length 30 cm., Find the position of image and its nature., A', F, , m=, , or, , h', (60), =–, (30), 3, , Sol., , or, h' = 3 × (–2) = –6 cm, The height of the image is 6 cm. The negative, sign shows that the image is inverted., , C, , 30cm, , Ex.9, , Here, object distance, u = –40 cm, Focal length of convex mirror, f = +30 cm, Now, using mirror formula,, , 1, 1, 1, +, =, we, f, v, u, , get, , Sol., , A 1 cm high object is placed at 20 cm in front, of a concave mirror of focal length 15 cm., Find the position and nature of the image., u = –20 cm, f = –15 cm, h0 = 1 cm, Using mirror formula,, , 1, 1, 1, +, =, v,  40 30, , or, , 1, 1, 1, 7, =, +, =, v, 40, 30, 120, , or, , =, , 120, 7, , The positive sign shows that the image is, formed on the right, i.e. behind the mirror., Now, magnification,, m = –, , Ex.8, , v, 120, 3, =–, =+, 7, , (, , 40, ), u, 7, , Since, the magnification is positive, the image, is erect. Thus, the image is formed 17.1 cm, behind the mirror. The image is virtual, erect, and diminished., A 3 cm high object is placed at a distance of, 30 cm from a concave mirror. A real image is, formed 60 cm from the mirror. Calculate the, focal length of the mirror and the size of the, image., , h', v, =–, h, u, , Magnification, , A, , B', , 1 1 1, + =, v u f, , we get, , or, , B 40 cm, , u = –30, , Image distance,, = –60, (real image is formed on the same side), , v, h', = –, h, u, , h', 6, =, = + 0.6, h,  10, , Ex.7, , Object distance,, , 1 1 1, + = we get, v u f, , 1, 1, 1, +, =, v,  20,  15, 1, 1, 1, 1, =–, +, =–, v, 15 2v, 60, , or, , ,  = – 60 cm, The image is formed 60 cm from the mirror., Since, the signs of u and  are the same, the, object and image are formed on the same side, of the mirror. Therefore, the image is real., Now magnification,, m=, , h' v, 60 cm, =, =, = –3, h, u  20 cm, ,  h’ = –3h = – 3 × 1 cm = – 3cm, The negative sign shows that the image is, inverted. Thus, the image is real, inverted and, of size 3 cm and formed 60 cm in front of the, mirror.

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Ex.10 An object 4 cm high is placed 25 cm in front, of a concave mirror of focal length 15 cm. At, what distance from the mirror should a screen, be placed in order to obtain a sharp image?, Find the nature and size of image., Sol., Here, u = –25 cm, f = –15 cm, h = + 4 cm, , or, , h', 60 / 7, 60, 3, =–, =+, =, (20), 5, 7  20, 7, , or, , h' = 5 ×, , 3 15, = cm, 7 7, , 1 1 1, + = , we get, v u f, , The height of the image is 2.1 cm. Positive, sign shows that the image is erect., , or, , 1, 1, 1, +, =, v  25,  15, , or, , 1, 1, 1, 1, 1, 2, =, –, =– +, =–, v  15,  25, 15 25, 75, , Ex.12 A convex mirror used on a automobile has 3, m radius of curvature. If a bus is located 5 m, from this mirror, find the position, nature and, size of image., , or, , =, , Using the mirror formula,, , 75, = –37.5 cm, 2, , Thus, the screen must be placed 37.5 cm from, the mirror on the same side as the object., v, h', Now, magnification, m = = , h, u, , Sol., , Here, u = –5 m, r = +3m, , , or h' = – 1.5 × 4 = –6 cm, Negative sign shows that the image is, inverted. Hence, the image is real, inverted, and of size 6 cm., ,  Focal length, f =, , r, 30, =+, = + 15 cm, 2, 2, , 1 1 1, Using the mirror formula, + = , we get, v u f, 1, 1, 1, +, =, v,  20  15, , or, or, , 1 1, 1, 7, = +, =, v 15 20 60, ,  =, , 60, cm, 7, , The image is formed 8.5 cm from the mirror., The positive sign shows that the image is, formed on the other side or behind the mirror., So the image is virtual., h', v, Magnification, m = = –, h, u, , r, 3, =+, 1.5 m, 2, 7, , Using the relation,, , (37.5), h', , = – 1.5, 4.0 cm, (25), , Ex.11 An object 5 cm high is placed at a distance of, 20 cm from a convex mirror of radius of, curvature 30 cm. Find the position, nature and, size of image., Sol., Here, u = –20 cm, h = 5 cm, Radius of curvature, r = +30 cm, , f=, , 1, 1 1, + = , we get, v, u f, , or, , 1, 1, 1, +, =, v,  5 1.5, , or, , 1, 1, 1, =, + = + 1.15 m, v 1.5 5, , The image is 1.15 m behind the mirror., Magnification, m =, , h', v, 1.15, =– =–, = + 0.23, h, (5), u, , Thus, the image is virtual, erect and smaller, in size than the object., IMPORTANT POINTS TO BE REMEMBER, , , Laws of reflection :, , (a) Angle of incidence is equal to the angle of, reflection ., (b) The incident ray, the reflected ray and the, normal all lie in the same plane., , , Mirror : A smooth, highly polished reflecting, surface is called a mirror. There are two types, of mirrors : (a) plane mirror (b) curved, mirrors, Curved mirrors are classified as spherical, mirrors or parabolic mirrors depending upon, the curvature of the mirror., ,   Concave mirror : A spherical mirror whose, inner hollow surface is the reflecting surface

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is called concave mirror. A concave mirror, forms a real, inverted image of an object if, the object is placed at any place outside F., However, when the object lies between F and, P, the image formed is erect and virtual., , , , , Convex mirror : A spherical mirror whose, outer bulging at surface is the reflecting, surface is called convex mirror. The image, formed by a convex mirror is always erect,, virtual and smaller than the object whatever, may be the position of the object. A convex, mirror is used as a side-mirror (driver’s, mirror) on automobiles., , , , Aperture of a mirror : The effective width, of a spherical mirror from which reflection, can take place is called its aperture., , , , Pole : The centre of a curved mirror is called, its pole., , , , screen is called a virtual image. A virtual, image is formed only when the reflected or, the refracted rays appear to come from a, point., , Centre of curvature : The centre of the, hollow sphere of which the spherical mirror is, a part is called its centre of curvature. The, centre of curvature of a concave mirror is in, front of it, while that of a convex mirror is, behind it., , Sign convention for spherical mirrors :, According to the sign convention for mirror,, the focal length of a concave mirror is, negative and that of a convex mirror is, positive., , , , Mirrors, , formula, , :, , The, , relationship,, , 1 1 1,   is called the mirror formula., f v u, , , Magnification : The ratio of the size of the, image to that of the object is called, magnification. For a mirror, magnification, (M) is given by,, M=–, , h, v, = i, u, ho, , power (in diopters) , , 1, f ( metre), , , CONCLUSIONS FROM THE SIGN CONVENTION, For spherical mirror :, , Radius of curvature : Radius of the hollow, , sphere of which the mirror is a part is called, its radius of curvature., , , Principal axis : A straight line passing, through the centre of curvature and pole of a, spherical mirror is called its principal axis., , Focus : A point on the principal axis of a, , mirror at which the rays coming from infinity, meet or appear to meet is called its focus., Focus is denoted by F., Focal length : The distance between the pole, , of a spherical mirror and the focus is called, the focal of a spherical mirror., , , Real image : The image which can be, obtained on a screen is called a real image. A, real image is formed only when the reflected, or refracted rays actually meet at a point., , Virtual image : The image which can be, , seen into a mirror but cannot be obtained on a, , Distance of the object,, Distance of the real image,, Distance of the virtual image, Focal length,, , Radius of curvature,, , Height of the object,, Height of the inverted (real) image,, Height of the erect (virtual) image,, , u is negative, v is negative, v is positive, f is negative for, concave & f is +, ve for convex, R is negative, for concave &, R is + ve for, convex, O is positive, I is negative, I is positive

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EXERCISE # 1, A. Very Short Answer Type Questions, Q.1, , A ray of light is incident on a plane mirror, i, being the angle of incidence. What is the, deviation suffered by the ray of light?, , Q.2, , A plane mirror reflects a pencil of light to, form a real image. What is the nature of the, pencil of light incident on the mirror?, , Q.3, , Define principal axis of a spherical mirror., , Q.4, , What is the focal length of a plane mirror?, , Q.5, , Two perpendicular plane mirror forms, .............. number of images of a point source, of light., , Q.6, , What is the magnification produced by a, plane mirror?, , Q.7, , Which mirror would you use for shaving?, , Q.8, , Suppose x and y are distances of object and, image respectively from a mirror. What shall, 1, be the shape of the graph between, and, x, 1, for a concave mirror ?, y, , Q.13, , We known that plane and convex mirrors, produce virtual images of objects. Can they, produce, real, images, under, some, circumstances ? Explain, , Q.14, , The wall of a room is covered with perfect, plane mirror. Two movie films are made, one, recording the movement of a man and the, other of his mirror image. From viewing the, films later, can an outsider tell which is, which?, , Q.15, , A concave mirror is held in water. What, would be the change in the focal length of the, mirror?, , Q.16, , What is the difference between the virtual, images produced by (i) plane mirror,, (ii) concave mirror, (iii) convex mirror?, , Q.17, , Show that if a ray of light is reflected, successively from two mirrors inclined at an, angle , the deviation of the ray does not, depend upon the angle of incidence., , Q.18, , Use the mirror equation to deduce that an, object placed between f and 2f of a concave, mirror produces a real image beyond 2f., , Q.19, , Show that a convex mirror always produces a, virtual image independent of the location of, the object., , B.  Short Answer Type Questions, Q.9, , An object is placed between two plane, parallel mirrors. Why do the distant images, get fainter and fainter?, , Q.20, , Prove that the virtual image produced by a, convex mirror is always diminished in size, and is located between the focus and the pole., , Q.10, , Why mirrors used in search light are, parabolic and not concave spherical?, , Q.21, , Q.11, , You read a newspaper because of the light, that it reflects. Then why do you not see even, a faint image of yourself in the newspaper?, , Show analytically that an object placed, between the pole and focus of a concave, mirror produces a virtual and enlarged image., , Q.22, , We know that a virtual image cannot be, obtained on a screen. But when we see a, virtual image, we are obviously bringing it on, the retina (may be regarded as a screen) of the, eye. Point out the contradiction, if any., , Q.12, , If you were driving a car, what type of mirror, would you prefer to use for observing traffic, at your back and why?

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Q.23, , Why a concave mirror of small aperture, forms a sharper image?, , Q.24, , What do you understand by the term, ‘parallax’?, , Q.25, , How can you distinguish between three, different mirrors just by looking at them?, , Q.26, , What is the effect of size of mirror on the, nature of image ?, , Q.27, , Is irregular reflection follows the laws of, reflections or not ?, , C.  Long Answer Type Questions, Q.28, , Prove that the radius of curvature of a, spherical mirror is equal to twice the focal, length of the mirror., , Q.29, , Derive mirror formula for a concave mirror, when image formed is (i) real (ii) virtual Also, give the sign convention used., , Q.30, , Find formulae for magnification produced in, the following cases : (i) concave mirror, when, image formed is real (ii) concave mirror,, when image formed is virtual (ii) convex, mirror., , Q.31, , Draw a ray diagram to show the formation of, image of an object placed between the pole, and centre of curvature of a concave mirror., Derive the formula connecting object distance, (u), image distance () and focal length (f) for, this particular case for the given concave, mirror. State clearly the assumptions and sign, conventions used., , Q.32, , Express magnification produced by a, spherical mirror in terms of (i) u and f(ii) , and f.

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EXERCISE # 2, Q.8, , Single Correct Answer type Questions, Q.1, , Q.2, , Q.4, , (A) is nearly a perfectly transparent, , A child walks towards a fixed plane mirror at, , (B) neither absorbs nor reflects light, , a speed of 5 km h–1. The velocity of the, image with respect to mirror is -, , (C) transmits whole of light, , (A) 5 km h–1, , (B) –5 km h–1, , (C) 10 km h–1, , (D) –10 km h–1, , (D) all of the above are correct, Q.9, , (B) M, , (C) O, , (B) Angle of incidence is less than the angle, of reflection, , (D) W, , (C) Angle of incidence is greater than the, angle of reflection, , In a plane mirror, an object is 0.5 m in front, of the mirror. The distance between object, and image is (A) 0.5 m, , (B) 1 m, , (C) 0.25 m, , (D) 0.75 m, , (D) None of these, Q.10, , An object 0.5 m tall is in front of a plane, mirror at a distance of 0.2 m. The size of the, image formed is(A) 0.2 m (B) 0.5 m (C) 0.1 m (D) 1 m, , Q.5, , According to laws of reflection of light (A) Angle of incidence is equal to the angle, of reflection, , The letter that does not show lateral, inversion(A) Z, , Q.3, , Air is not visible because it-, , Which of the following correctly represents, graphical relation between angle of incidence, (i) and angle of reflection (r) ?, , i, , i, , O, , A plane mirror is approaching you at 10 cm, , (A) + 10 cm, , s–1, , (C) + 20 cm s–1, , Q.7, , y, , (A), , s–1. Your image shall approach you with a, speed of-, , Q.6, , y, , (B) – 10 cm, , s–1, , (A) beam of light, , (B) ray of light, , (C) pencil of light, , (D) none of these, , A thin layer of water is transparent but a very, thick layer of water is(A) translucent, , (B) opaque, , (C) most transparent (D) none of these, , i, , i, r, , x, , r, , (D), , O, , r, , x, , x, , A light ray falls on a mirror and deviates by, 60° then the angle of reflection will be, (A) 30°, (C) 60°, , Q.12, , O, y, , O, Q.11, , (B), , y, , (C), , (D) – 20 cm s–1, , The path along which light travels in a, homogeneous medium is called a-, , r, , x, , (B) 90°, (D) 180°, , A ray of light is incident on a plane mirror at an, angle . If the angle between the incident and, reflected rays is 80°, what is the value of ., (A) 40°, , (B) 50°, , (C) 45°, , (D) 55°

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Q.13, , Light shows -, , Q.18, , (A) Random propagation, , The magnification produced by a concave, mirror -, , (B) Curvilinear propagation, , (A) is always more than one, , (C) Rectilinear propagation, , (B) is always less than one, , (D) None of these, , (C) is always equal to one, (D) may be less than or greater than one, , Q.14, , Q.15, , Q.16, , Q.17, , The image of the moon is formed by a, concave mirror whose radius of curvature is, 4.8 m at a time when distance from the moon, is 2.4 × 108 m . if the diameter is of the image, is 2.2 cm, the diameter of the moon is(A) 1.1 × 106 m, , (B) 2.2 × 106 m, , (C) 2.2 × 108 m, , (D) 2.2 × 1010 m, , The focal length of a concave mirror is f and, the distance of the object from the principal, focus is a. The magnitude of magnification, obtained will be-, , Q.19, , (A) R , , 2uv, uv, , (B) R , , 2, uv, , (C) R , , 2( u  v ), (uv), , (D) none of these, , (A) (f + a)/f, , (B) f /a, , The image formed by a concave mirror is, real, inverted and of the same size as that of, the object. The position of the object should, be-, , (C), , (D) f2/a2, , (A) Beyond C, , (B) Between C and F, , (C) At C, , (D) At F, , f/ a, , The magnification of an object placed 10 cm, from a convex mirror of radius of curvature, 20 cm will be-, , Q.20, , Choose the correct relation between u, v and, R-, , (A) 0.2, , (B) 0.5, , A boy is standing in front of a plane mirror at, a distance of 3m form it. What is the distance, between the boy and his image ?, , (C) 1, , (D) infinity, , (A) 3 m, , (B) 4.5 m, , (C) 6 m, , (D) none of these, , The image formed by a concave mirror is, observed to be virtual, erect and larger than, the object. then the position of the object, should be(A) between the focus and the centre of, curvature, (B) at the centre of curvature, (C) beyond the centre of curvature, (D) between the pole of the mirror and the, focus, , Q.21

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ANSWER KEY, EXERCISE-1, 1.  – 2i, , 2. virtual, , 6. 1, , 7. plane mirror, , 1, r2, 26. Image becomes brighter, , 9. as I , , 5. 3, 1 1 1, 8.,    linear, v u f, , 10. They reflect light parallel, , 15. No change, , 27. Yes, , EXERCISE-2, Ques, Ans, Ques, Ans, , 1, B, 16, B, , 2, A, 17, D, , 3, B, 18, D, , 4, B, 19, A, , 5, C, 20, C, , 6, B, 21, C, , 7, A, , 8, D, , 9, A, , 10, D, , 11, C, , 12, B, , 13, C, , 14, B, , 15, B