Page 1 : 4, , LIGHT : REFRACTION, , CHAPTER, , CONTENTS, , ,  Refraction of Light,  Law of refraction of light,  Refractive index,  Refraction through glass slab,  Spherical lens,  Rules for image formation by ray, diagram method, Image formation by lens, , ,  Numerical method in lens,  Total internal reflection,  Real & apparent depth & height,  REFRACTION OF LIGHT, , Incident ray, P, , N1, i, , X, Plane transparent, surface, , Q, , r, N2, , Normal, ,  DEFINITION : When light rays travelling in, a medium are incident on a transparent, surface of another medium they are bent as, they travel in second medium., , SOME ASSOCIATED TERMS,  Transparent surface : The plane surface, which refracts light, is called transparent, surface. In diagram, XY is the section of a, plane transparent surface.,  Point of incidence : The point on transparent, surface, where the ray of light meets it, is, called point of incidence. In diagram, Q is the, point of incidence.,  Normal : Perpendicular drawn on the, transparent surface at the point of incidence,, is called normal. In diagram, N1QN2 is the, normal on surface XY.,  Incident ray : The ray of light which strikes, the transparent surface at the point of, incidence, is called incident ray in diagram, PQ is the incident ray.,  Refracted ray : The ray of light which, travels from the point of incidence into the, other medium, is called refracted ray. In, diagram, QR is the refracted ray.,  Angle of incidence : The angle between the, incident ray and the normal on the transparent, surface at the point of incidence, is called the, angle of incidence. It is represented by the, symbol i. In diagram, angle PQN1 is the, , , Rarer, Y, Denser, R, Refracted ray, , Fig. Refraction of light from a plane, transparent denser surface., , angle of incidence., Angle of refraction : The angle between the, refracted ray and the normal on the, transparent surface at the point of incidence,, is called angle of refraction. It is represented, by symbol r. In diagram angle RQN2 is the, , angle of refraction.,  Plane of incidence : The plane containing the, normal and the incident ray, is called plane of, incidence. For the diagram, plane of book, page is the plane of incidence.

Page 2 : , , Plane of refraction : The plane containing, the normal and the refracted ray, is called, plane of refraction. For the diagram, plane of, book page is the plane of refraction., , LAW OF REFRACTION OF LIGHT, First Law : The incident ray, the normal to, the transparent surface at the point of, incidence and the refracted ray, all lie in one, and the same plane.,  Second Law : The ratio of sine of angle of, incidence to the sine of the angle of refraction, is constant and is called refractive index of, the second medium with respect to the first, medium., , , sin i, =µ, sin r, ,  REFRACTIVE INDEX, (a) Refractive Index in terms of Speed of, Light:, The refractive index of a medium may be, defined in terms of the speed of light as, follows :, Speed of light in vacuum, Refractive index =, Speed of light in medium, c, or µ =, v, (b) Refractive Index in terms of Wavelength :, Since the frequency () remains unchanged, when light passes from one medium to, another, therefore,,  , , c, µ=, = vac, = vac,  med  ,  med, v, (c) Relative Refractive Index :, The relative refractive index of medium, 2 with respect to medium 1 is defined as the, ratio of speed of light (v1) in the medium 1 to, the speed of light (v2) in medium 2 and is, denoted by 1µ2., v, , , Thus, 1µ2 = 1 = 1 = 2, v 2  2 1, As refractive index is the ratio of two similar, physical quantities, so it has no unit and, dimension.,  FACTORS ON WHICH THE REFRACTIVE, INDEX OF A MEDIUM DEPENDS ARE :, (i) Nature of the medium., (ii) Wavelength of the light used., (iii) Temperature, (iv) Nature of the surrounding medium., , It may be noted that refractive index is a, characteristic of the pair of the media and also, depends on the wavelength of light, but is, independent of the angle of incidence., , REFRACTION THROUGH GLASS SLAB, (a) Refraction through a rectangular glass, slab and principle of reversibility of light :, Consider a rectangular glass slab, as shown in, figure. A ray AE is incident on the face PQ at, an angle of incidence i. On entering the glass, slab, it bends towards normal and travels, along EF at an angle of refraction r. The, refracted ray EF is incident on face SR at an, angle of incidence r'. The emerged ray FD, bends away from the normal at an angle of, refraction e., Thus the emergent ray FD is parallel to the, incident rays AE, but it has been laterally, displaced with respect to the incident ray., There is shift in the path of light on emerging, from a refracting medium with parallel faces., Lateral shift :, Lateral shift is the perpendicular distance, between the incident and emergent rays when, light is incident obliquely on a refracting slab, with parallel faces., Factors on which lateral shift depends are :, (i) Lateral shift is directly proportional to the, thickness of glass slab., (ii) Lateral shift is directly proportional to the, incident angle., (iii) Lateral shift is directly proportional to the, refractive index of glass slab., (iv) Lateral shift is inversely proportional to the, wavelength of incident light., A, N1, , B, Incident, ray, , P, , air, , i, , µa, , E, , Glass, µg, , r, , N2, S, , Q, N1, , tThickness, , r', F, c, , air, a, , R, d, , Refracted, e, N2, ray, Lateral displacement, Emergent, ray, D, , Proof for i = e

Page 3 : Case-I :, For light going from air to glass at point E., µa sin i = µg sin r, ..... (1), Case-II :, For light going from glass to air at point F., µg sin r = µa sin e, .....(2), From (1) & (2) we can say that i = e,  incident & emergent rays are parallel to, each other., ,  SPHERICAL LENS, Definition : A piece of a transparent medium, bounded by at least one spherical surface, is, called a spherical lens.,  Types : There are two types of spherical, lenses:, (i) Convex or Converging Lenses : These are, thick in the middle and thin at the edges., , , SOME ASSOCIATED TERMS :, (i) Centre of curvature (C) :, The centre of curvature of the surface of a, lens is the centre of the sphere of which it, forms a part, because a lens has two surfaces,, so it has two centres of curvature. In figure, (a) and (b) points, C1 and C2 are the centres of, curvature., (ii) Radius of curvature (R) :, The radius of curvature of the surface of a, lens is the radius of the sphere of which the, surface forms a part. R1 and R2 in the figure, (a) and (b) represents radius of curvature., (iii) Principal axis (C1C2) :, It is the line passing through the two centres, of curvature (C1 and C2) of the lens., Optical, Centre, Radius of, Curvature, , Centre of, Curvature, , (a), , (b), , C2 P2, O, C1, R2, Principal, axis, , (c), , Fig. Three types of convex lenses, (a) Double Convex Lens : It has both the, surfaces convex., (b) Plano–Convex Lens : It has one surface, plane and the other surface convex., (c) Concavo–Convex Lens : It has one surface, concave and the other surface convex., (ii) Concave or Diverging Lenses : These are, thin in the middle and thick at the edges., There are three types of concave lenses :, , (a), , (b), , R1, P1, , (c), , Fig. Three types of concave lenses, (a) Double Concave Lens : It has both the, surfaces concave. (Fig.), (b) Plano–Concave Lens : It has one surface, plane and the other surface concave. (fig.), (c) Convexo–Concave Lens : It has one surface, convex and the other surface concave. (fig.), , (a), Optical, Centre, Radius of, Curvature, , Centre of, Curvature, C1, , R2, R1, , P1 O P2, , (b), , C2, Principal, axis, , Figure : Characteristics of convex and, concave lenses, (iv) Optical centre :, If a ray of light is incident on a lens such that, after refraction through the lens the emergent, ray is parallel to the incident ray, then the, point at which the refracted ray intersects, the, principal axis is called the optical centre of, the lens., OP1 P1C1 R 1, =, =, OP2 P2 C 2 R 2, If the radii of curvature of the two surfaces, are equal, then the optical centre coincides, with the geometric centre of the lens.

Page 4 : A, C2, , P1, , R1, O, , F1, , O, , P2, C1, , R2, B, , (a), , O, , (b), , f, , Figure : Ray diagram showing first, principal focus, (B) Second principal focus and second focal, length :, It is a fixed point on the principal axis such, that the light rays incident parallel to the, principal axis, after refraction through the, lens, either converge to this point (in convex, lens) or appear to diverge from this point (in, concave lens). It is denoted by F2., , O, , F2, , O, f, , (c), (v) Principal foci and focal length :, (A) First principal focus and first focal length :, It is a fixed point on the principal axis such, that rays starting from this point (in convex, lens) or appearing to go towards this point, (concave lens), after refraction through the, lens, become parallel to the principal axis. It, is represented by F1., , F1, , O, , O, , F2, , f, , Figure : Ray diagram showing second, principal focus, If the medium on both sides of a lens is same,, then the numerical values of the first and, second focal lengths are equal. Thus, f = f, , f, , (vi) Aperture :, It is the diameter of the circular boundary of, the lens.

Page 5 : RULE FOR IMAGE FORMATION BY, RAY DIAGRAM METHOD, , ,  THREE SPECIAL RAYS FOR CONVEX LENS, When light ray incident parallel to principal, axis., , F1, , C, ,  CONVEX LENS IN DIFFERENT CASES, Case 1 : Object at Infinity,  A point object lying on the principal axis, Rays come parallel to the principal axis and, after refraction from the lens, actually meet at, the second principal focus F2., , F2, F2, , C, , F1, When light ray incident from focus., , F1 C, , F2, , When light ray incident on the pole., , F1, , C, , F2, , ,  THREE SPECIAL RAYS FOR CONCAVE LENS,  When light ray incident parallel to principal, axis., , Fig. Convex lens point object at infinity,, image at focus., The image is formed at focus F2. It is real and, point sized.,  A big size object with its foot on the, principal axis, Parallel rays come inclined to the principal, axis. Image of foot is formed at the focus., Image is formed at the second principal focus, F2. It is real inverted and diminished (smaller, in size than the object). (Fig.), Parallel rays, from infinity, F1, , F2, , C, , F1, , F2, , C, , F1, , F2, , C, , F1, , F2A', , C, , B', , Fig. Convex lens : big size object at infinity,, image at focus, Case 2 : Object at distance more than twice the, Focal Length, Real object AB has its image AB formed, between distance f and 2f., The image is real inverted and diminished, (smaller in size than the object), B, A2F F1, ,  IMAGE FORMATION BY LENS, , , Introduction : From lens formula, we find, that for a lens of a fixed focal length as object, distance u changes, image distance  also, changes. Moreover, as u decreases or, increases, this changes the position, the nature, and the size of the image., Different cases, are as given below with their, ray diagrams., , F2, , A' 2F, , C, B', , Fig. Convex lens : object beyond 2f, image, between f and 2f., Case 3. Object at distance twice the Focal, Lengths, Real object AB has its image AB formed at, distance 2f.

Page 6 : B, , 2F, A', , F2, A F1, 2F, , C, B', , Fig. Convex lens : object at distance 2f,, image at distance 2f., The image is real, inverted and has same size, as the object., Case 4 : Object at distance more than Focal, Length and less than twice is Focal Length, Real object AB has its image AB formed, beyond distance 2f., B, F2 2F A', 2F A, , C, , F1, , B', , Fig. Convex lens : object at distance between, f and 2f, image beyond 2f., The image is real inverted and enlarged, (bigger in size than the object)., Case 5 : Object at Focus, Real object AB has its image formed at, infinity., B, A, F1, ,  CONCAVE LENS IN DIFFERENT CASES, Case 1 : Object at infinity,  A point object lying on the principal axis., Rays come parallel to the principal axis and, after refraction from the lens, appears to come, from the second principal focus F2., , F2, , F1, , C, , Fig. Concave lens point object at infinity,, image at focus., The image is formed at focus F2. It is virtual, and point sized (fig.),  A big size object with its foot on the principal, axis., Parallel rays come inclined to the principal, axis. Image of foot is formed at focus., The image is formed at the second principal, focus F2., It is virtual–erect and diminished (fig.), Parallel rays, from infinity, , B', A' F2, , C, , F1, , F2, , C, , Parallel rays, to infinity, , Fig. Convex lens : object at focus, image at, infinity., The image is imaginary inverted (refracted, rays to downward) and must have very large, size., Case 6 : Object between Focus and Optical, Centre, Real object AB has its image AB formed in, front of the lens., , Fig. Concave lens : big size object at infinity, image at focus., Case 2 : Object at a Finite Distance, Real object AB has its image AB formed, between second principal focus F2 optical, centre C., The image is virtual–erect and diminished., ,  NUMERICAL METHOD IN LENS, , B', B, A' F1 A C, , F2, , (A) LENS FORMULA,  Definition : The equation relating the object, distance (u), the image distance (v) and the, focal length (f) of the lens is called the lens, formula.

Page 7 : Assumptions made :, The lens is thin., The lens has a small aperture., The object lies close to principal axis., The incident rays make small angles with the, lens surface or the principal axis., 1 1 1, Lens Formula:  , v u f, , , 1., 2., 3., 4., , (B) LINEAR MAGNIFICATION FOR LENS,  LINEAR MAGNIFICATION,  Definition : The ratio of the size of the, image formed by refraction from the lens to, the size of the object, is called linear, magnification produced by the lens. It is, represented by the symbol m., If I be the size of the image and O be the, size of the object, then, v, I, m= =, O u, (C) POWER OF LENS,  Definition : It is the capacity or the ability of, a lens to deviate (converge or diverge) the, path of rays passing through it. A lens, producing more converging or more, diverging, is said to have more power, 1, Power of lens (in diopter) , , f (in metre), ,  TOTAL INTERNAL REFLECTION, , , Definition : When light travels from a denser, medium to a rarer medium and is incident at, an angle more than the critical angle for that, medium, it is completely returned inwardly in, the denser medium. This complete inward, return of light is called total (complete), internal (inward) reflection (return)., , Air, Water i, O, , r1, 1, , r2, i2, ,  CRITICAL ANGLE, The angle of incidence in denser medium for, which angle of refraction is 90º, is called the, critical angle. It is represented by the symbol C.,  w sin i3 = a sin 90°, 1,   sin i 3 , , w, [Note. More is the value of µ, lesser will be, angle C].,  Condition, (i) Light must travel from denser to rarer, medium., (ii) Light must be incident at an angle more than, the critical angle for the denser medium., Merit : In total internal reflection 100% light, is reflected, hence images formed are more, bright., In ordinary reflection from mirrors, only 85%, light is reflected, rest 15% is either absorbed, by mirror glass or transmitted due to poor, polish. Images formed by ordinary reflection, are less bright., , REAL & APPARENT DEPTH & HEIGHT, (A) Seeing from air to liquid :, , apparent, , dactual, actual, , apparent depth from surface =, , d actual, , , (B) Seeing from liquid to air, apparent, , r3 D, i3 i4 r, 4, , actual, dactual, , Total internally, reflected ray, , Fig. Total internal reflection., apparent height from surface = Hactual × , , 

Page 8 :  SOLVED EXAMPLES , Ex.1, Sol., , Speed of light in water is 2.25 × 108 m/s., Calculate the refractive index of water., Refractive index is given by, n=, =, , Ex.4, , Sol., , speed of light in vaccum (c), speed of light in water ( ), 3 108 m / s, 2.25 108 m / s, , If the refractive index of water is 4/3 and that, of glass is 3/2. Calculate the refractive index, of glass with respect to water., We known that, w, , g =, , where, , = 1.33, , g, w, w, , g = refractive index of glass with, respect to water, , g = refractive index of glass = 3/2, Ex.2, Sol., , Refractive index of diamond is 2.42., Calculate the speed of light in diamond., We know that refractive index,, c, speed of light in vaccum, =, speed of light in diamond, v, , n=, , Ex.3, , or, , 2.42 =, , or, , =, , 3  10, v, , w = refractive index of water = 4/3,  , , Ex.5, , 8, , 3  108, = 1.24 × 108 m/s., 2.42, , Sol., , A ray of light travelling in air falls on the, surface of water. The angle of incidence is, 60° with the normal to the surface. The, refractive index of water = 4/3. Calculate the, angle of refraction., , w, , g =, , 3/ 2 9, = =1.1, 4/3 8, , A ray of light is incident on the plane surface, of a transparent medium at an angle 60° with, the normal. The angle of refraction is 30°., Calculate the refractive index of the, transparent material., Here,, Angle of incidence,, i = 60°, Angle of refraction,, r = 30°, Refractive index,, n=, , sin i, sin 60, 3/2, =, =, =, sin r, sin 30, 1/ 2, , 3, , Sol., , Air, , Ex.6, , 60°, Water, , We know that, , sin i, =n, sin r, , Here, i = 60°, n = 4/3, , , sin 60 4, =, sin r, 3, , or, , 3/2, 4, =, sin r, 3, , or sin r =, , 3 3, = 0.65, 8, ,  r = sin 0.65, –1, , Sol., , A ray of light travelling in air falls on the, surface of a glass slab at an angle 45° with the, normal. The refractive index of glass is 1.5., Calculate the angle of refraction., Angle of incidence = 45°, Refractive index of glass,, n = 1.5, Since, , n=, sin 45, sin r, , or, , 1.5 =, , or, , sin r =, =, , sin i, sin r, , sin 45, 1/ 2, =, 1 .5, 1.5, , 1, , 2  1 .5, 1, 1, =, =, = 0.47, 1.41 1.5 2.115, , , , r = sin–10.47

Page 9 : Ex.7, , Sol., , The refractive index of diamond is 2.42 and, that of carbon disulphide is 1.63. Calculate, the refractive index of diamond with respect, to carbon disulphide., Refractive index of carbon disulphide,, n1= 1.63, Refractive index of diamond, n2 = 2.42, Refractive index of diamond with respect, for carbon disulphide,, 1, , n2 =, , n2, 2.42, =, = 1.48, n1, 1.63, , Now,, , sin ic =, =, , 1.5, = 0.6198, 2.42, , , , ic = sin–1 0.62, Ex.11 Refractive index of glass is 3/2. A ray of light, travelling in glass is incident on glass-water, surface at an angle 30° with normal. Will it be, able to come out into the water Refractive, index of water = 4/3., Sol., Refractive index of glass, n1 = 3/2, Refractive index of water,, , Ex.8, , Sol., , A coin is placed in a tumbler, water is then, filled in the tumbler to a height of 20 cm. If, the refractive index of water is 4/3, calculate, the apparent depth of the coin., Here,, Real depth,, h = 20 cm, Refractive index, n = 4/3, real depth, apparent depth, , Now,, , n=, , or, , 4, 20, =, apparent depth, 3, , or Apparent depth =, Ex.9, , Sol., , 20 3, = 15 cm, 4, , There is a black spot on a table. A glass slab, of thickness 6 cm is placed on the table over, the spot. Refractive index of glass is 3/2. At, what depth from the upper surface will the, spot appear when viewed from above?, Real depth of the spot = 6 cm, Refractive index of glass, n =, , 3, 2, , Now,, , n=, , real depth, apparent depth, , or, , 3, 6, =, 2 apparent depth, , Apparent depth =, , 6 2, = 4 cm, 3, , Ex.10 Refractive index of diamond is 2.42 and that, of glass is 1.5. Calculate the critical angle for, diamond-glass surface., Sol., Refractive index of diamond, n1 = 2.42, Refractive index of glass,, n2 = 1.5, , n2, n1, , Now,, , sin ic =, , n2 = 4/3, , n2, 4/3 8, =, = = 0.88, n1, 3/ 2 9, , , , ic = 62°, Since, the angle of incidence (30°) is less than, the critical angle, the ray will be refracted, into the water., Ex.12 The refractive index of dense flint glass is, 1.65 and that of alcohol is 1.36, both with, respect to air. What is the refractive index of, flint glass with respect to alcohol ?, Sol., Refractive index of flint glass, n2 = 1.65, Refractive index of alcohol, n1 = 1.36, Refractive index of flint glass with respect, to alcohol is given by, 1, , n2 =, , n2, 1.65, =, = 1.21, n1, 1.36, , Ex.13 An object is placed 36 cm from a convex, lens. A real image is formed 24 cm from the, lens. Calculate the focal length of the lens., Sol., According to the sign convention the object is, placed on the left-hand side of the lens. So, object distance (u) is negative. Real image is, formed on the other side of the lens. So the, image distance () is positive. Thus, u = –36, cm,  = +24 cm, f = ?, 1 1 1, – = , we get, v u f, 1, 1, 1, –, =,  24  36, f, 1, 1, 1, 5, =, +, =, f, 24, 36, 72, 72, f=, = 14.4 cm, 5, , Using lens formula,, , or,  , , , , 

Page 10 : Ex.14 A 2 cm long pin is placed perpendicular to, the principal axis of a lens of focal length 15, cm at distance of 25 cm from the lens. Find, the position of image and its size., Sol., Here,, u = –25 cm, f = +15, Using the lens formula,, or, or, or, , 1 1, 1, –, = we get, f, v u, , 1, 1, 1, –, =, v  25  15, 1, 1, 1, 2, =, –, =, v, 15, 25 75, 75, =, = 37.5 cm, 2, , The positive sign shows that the image is, formed on the right-hand side of the lens., Magnification is given by, h', v, =, h, u, h ' 37.5, = =, = – 1.5, h,  25, , m=, or,  , , h = – 1.5 × h = –1.5 × 2 cm, = –3 cm, The image of the pin is 3 cm long. The, negative sign shows that it is formed below, the principal axis, i.e. the image is inverted., Ex.15 A point object is placed at a distance of 18 cm, from a convex lens on its principal axis. Its image, is formed on the other side of the lens at 27 cm., Calculate the focal length of the lens., Sol., According to the sign convention, the object, is placed on the left-hand side of the lens,, therefore object-distance is negative,, i.e. u = –18 cm. Since the image is formed on, the other side, the image-distance is positive,, i.e., v = +27 cm. Using lens formula,, , or, or, , 1 1 1, – = , we have, v u f, 1, 1, 1, –, =, f,  27  18, 1, 1, 5, 1, +, =, =, f, 27 18, 54, 54, f =, = 10.8 cm, 5, , Ex.16 A convex lens forms an image of the same, size as the object at a distance of 30 cm from, the lens. Find the focal length of the lens., Also find power of the lens. What is the, distance of the object from the lens ?, Sol., A convex lens forms the image of the same, size as the object only when the object is, , placed at a distance 2f from the lens. In this, case the image is also equal to 2f from the, lens., Hence,, 2f = 30 cm, or, f = 15 cm = 0.15 m, Power of the lens,, , P=, , 1, 1, =, D = 6.6D, f, 0.15, , The distance of the object from the lens is, also 2f = 30 cm., Ex.17 A 3 cm high object is placed at a distance of, 80 cm from a concave lens of focal length 20, cm. Find the position and size of the image., Sol., Here, u = –80 cm, f = – 20 cm, 1, 1, 1, –, =, , we get, f, v, u, 1, 1, –, =,  80  20, 1, 1 5, 1, =–, –, =, =–, 20, 80 80, 16, , Using the lens formula,, 1, v, 1, , , or, or, ,  = –16 cm, , or, , h' v, 16 1, = =, =, h, u  80 5, h 3 .0, h’ = =, = 0.6 cm, 5, 5, , Magnification, m =, or, , Length of image is 0.6 cm. Positive sign, shows that the image is erect., Ex.18 An object is placed on the principal axis of a, concave lens at a distance of 40 cm from it. If, the focal length of the lens is also 40 cm, find, the location of the image and the, magnification., Sol., For a concave lens focal length f is negative,, i.e. f = –40 cm. Since by convention, object is, placed on the left of the lens, so u = – 40 cm., 1 1, 1, –, = , we get, f, v u, 1, 1, –, =,  40  40, 1, 1, 1, =–, –, =–, 40, 40, 20, , Using the lens formula,, or, or, , 1, v, 1, v, , or, = – 20 cm, The image is formed 20 cm from the lens., Minus sign shows that the image is formed on, the same side of the lens as the object., Now, magnification, m =, , h', v, 20 1, = =, =, h, u  40 2, , Positive sign shows that the image is erect.

Page 11 : Ex.19 A beam of light travelling parallel to the, principal axis of a concave lens appears to, diverge from a point 25 cm behind the lens, after refraction. Calculate the power of the, lens., Sol., When a parallel beam after refraction through, the lens is incident on a concave lens, it, appears to diverge from the focus of the lens., Hence, the focal length of the lens is 25 cm., According to sign convention, focal length of, a concave lens is negative., , f = –25 cm = –0.25 m,  Power,, , P=, , 1, 1, =, = – 4D, f,  0.25, , Ex.20 A convex lens of power 5D is placed at a, distance of 30 cm from a screen. At what, distance from the lens should the screen be, placed so that its image is formed on the, screen?, Sol., Power of the lens, P = +5D, 1, 1, Focal length, f =, = = 0.20 m = 20 cm, 5D 5, , Here, the screen is placed 30 cm from the lens., v = +30 cm, f = +20 cm, u = ?, 1 1, 1, –, = , we get, f, v u, 1, 1 1, – =, 30 u 20, 1, 1, 1, 1, =, –, =–, u 30 20, 60, , Using the lens formula,, , or, , or, u = – 60 cm, Therefore, the screen should be placed at 60, cm from the lens., Ex.21 A pin 3 cm long is placed at a distance of, 24 cm from a convex lens of focal length 18, cm. The pin is placed perpendicular to the, principal axis. Find the position, size and, nature of the image., Sol., Here, u = –24 cm, f = +18 cm, v = ?, 1, 1 1, – = , we get, v, u f, 1, 1, –, =,  24,  18, 1, 1, 1, =, –, =, 18 24, 72, , Using the lens formula,, , or, or, , 1, v, 1, v, , v = 72 cm, , The image is formed 72 cm from the lens on, the other side. So the image is real., Magnification,, , h' v, 72, = =, = –3, h, u  24, , m=, , or, h' = –3 × h = –3 × 3.0 = – 9 cm, The image is 9 cm in size. Negative sign, shows that the image is inverted., Ex.22 A convex lens of focal length 40 cm and a, concave lens of focal length 25 cm are placed, in contact in such a way that they have the, common principal axis. Find the power of the, combination., Sol., Focal length of the convex lens,, f1 = 40 cm = +0.4 m, Power of the convex lens,, 1, , P1=, = +2.5D,  0.40, Focal length of the concave lens,, f2 = –25 cm = –0.25 m,  Power of the concave lens,, P2 =, , 1, = – 4D,  0.25, , Power of the combination,, P = P1 + P2 = 2.5 – 4D = – 1.5D, Ex.23 A concave lens has a focal length of 15 cm., At what distance should the object be from, the lens placed so that it forms an image 10, cm from the lens ? Also find the, magnification., Sol., A concave lens always forms a virtual, erect, image on the same side as the object., Image distance,, v= –10 cm, Focal length, f = –15 cm, Object distance,, u=?, Using, the lens formula,, or, or, , 1 1 1, – = , we get, v u f, , 1, 1, 1, –, =,  10, u,  15, 1, 23, 1, –, =, =–, 10, 30, 30, , or, u = –30 cm, Thus, the object should be placed 30 cm on, the lens., Magnification, m =, , v, 10 1, =, = = 0.33, u  30 3, , The positive sign shows that the image is, erect and virtual. The size of the image is, one-third of that of the object.

Page 12 : Ex.24 A 2 cm tall object is placed perpendicular to, the principal axis of a convex lens of focal, length 10 cm. The distance of object from the, lens is 15 cm. Find the position, nature and, size of the image. Calculate the magnification, of the lens., Sol., Object distance,, u = –15 cm, Focal length,, f = +10 cm, Object height,, h = +2 cm, Image distance,, v=?, Image height,, h' = ?, Using the lens formula,, , 1, 1, 1, –, = , we get, f, v, u, , 1, 1, 1, =, =, v,  15,  10, 1, 1, 1, 1, =, –, =, v, 10, 15, 30, , or, , or,  = +30 cm, Positive sign of v shows that the image is, formed at a distance of 30 cm on the right, side of the lens. Therefore the image is real, and inverted, Magnification,, , m=, , h', v, =, h, u, , h', 30, =, = –2, 2.0  15, , or, , h' = –2 × 2 = –4 cm, , Magnification, m =, , v, 30, =, = –2, u,  15, , Negative sign with the magnification and, height of the image shows that the image is, inverted and real. Thus, a real image of height, 4 cm is formed at a distance of 30 cm on the, right side of the lens. Image is inverted and, twice the size of the object.,  IMPORTANT POINTS TO BE REMEMBER,  Refraction : The bending of a ray of light as, it passes from one medium to another is, called refraction., (a) A ray of light travelling from a rarer medium, to denser medium (say, water or glass) bends, towards the normal., , (b) A ray of light travelling from a denser to a, rarer medium bends (or refractive) away from, the normal., Laws of refraction :, (a) The ratio of sine of the angle of incidence to, the sine of the angle of refraction for a, particular pair of media is constant, i.e., sin i / sin r = constant is equal to the, refractive index of the medium into which, the light is entering., (b) The incident ray, the refracted ray and the, normal all lie in the same plane.,  Lateral displacement : The perpendicular, distance of separation between the emergent, ray and the original path of the incident ray is, called lateral displacement.,  Lens : A piece of any transparent material, bound by two curved surfaces is called a lens., A lens which is thicker in the middle and, thinner at the edges is called a convex lens. A, convex lens is also called converging lens., A lens which is thicker at the edges and, thinner at the centre is called a concave lens., A concave lens is also called a diverging lens.,  Optical centre of a lens : The centre point of, a lens is called its optical centre. A ray of, light passing through the optical centre does, not suffer any deviation.,  Image formed by the lenses : A convex lens, forms real and inverted images for all the, positions of an object at and outside the focus, (F). However, when the object is placed, between F and O, the image formed by a, convex lens is virtual and erect., A concave lens always forms a virtual, erect, and a diminished image whatever may be the, distance of the object from the lens.,   Lens formula :, , , The lens formula is, , 1, 1, 1, =, – ., f, v, u, , Power of a lens : Reciprocal of the focal, , length of a lens measured in metres is called, its power. Power of a lens is described in, dioptre (D) unit.

Page 13 : Refractive indices of various substances relative, to vacuum with light of wavelength 589 mm, Solids (at 20ºC), , Substance, , Refractive index, , Diamond, , 2.42, , Ruby, , 1.71, , Sapphire, , 1.77, , Quartz (fused), , 1.46, , Canada Balsam, , 1.53, , Rock salt, , 1.54, , Glass (crown), , 1.52, , Glass (flint), , 1.66, , Ice, , 1.31, , Liquids (at 20ºC), , Substance, , Refractive index, , Water, , 1.33, , Ethyl alcohol, , 1.36, , Kerosene, , 1.44, , Turpentine oil, , 1.47, , Glycerine, , 1.47, , Benzene, , 1.5, , Carbon disulphide, , 1.63, , Gases (at 0ºC, 1 atm), , Substance, , Refractive index, , Air, , 1.00029, , Carbon dioxide, , 1.00045

Page 14 : EXERCISE # 1, A. Very Short Answer Type Questions, Q.1, , Q.2, , What is the maximum angle of refraction, when a ray of light is refracted from glass, into air ?, What should be the position of an object, relative to biconvex lens so that this lens, behaves like a magnifying glass?, , Q.3, , Can the absolute refractive index of a, medium be less than unity?, , Q.4, , To a fish under water viewing obliquely a, fisherman standing on the bank of a lake,, does the man look taller or shorter than what, actually he is?, , Q.5, , Does the apparent depth of a tank of water, change if viewed obliquely? If so, does the, apparent depth increase or decrease?, , Q.6, , A substance has critical angle of 45° for, yellow light. What is its refractive index?, , Q.7, , What is critical angle for a material of, refractive index, , A ray of light is incident normally on a glass, slab. What is the angle of refraction?, , Q.9, , What is the power of the combination of a, convex lens and a concave lens of the same, focal length?, , Q.11, , Q.12, , Explain the shining of an air bubble in water., , Q.13, , For the same angle of incidence, the angles of, refraction in media P, Q and R are 35°,, 25°,15° respectively, In which medium will, the velocity of light be minimum ?, , Q.14, , Define focus and principal focus of a lens., , Q.15, , 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., retina of our eye. Is there a, contradiction?, , Q.16, , A convex lens is held in water. What would, be the change in the focal length ?, , Q.17, , Why goggles (Sun glasses) have zero power, even though their surfaces are curved?, , Q.18, , The lens shown in fig. is made of two, different materials. A point objects is placed, on the principal axis of this lens. How many, images will be obtained?, , Q.19, , Refer to fig. (a), (b) and (c). Give relationship, , 2?, , Q.8, , Q.10, , B.  Short Answer Type Questions, , How is power of a lens related to its focal, length?, Define critical angle for total internal, reflection., , between 1 and 2 in each case., 1, 2, , 1, 2, , (a), , 1, 2, , (b), , (c)

Page 15 : Q.20, , Images formed by totally reflected light are, brighter than the images formed by ordinary, reflected light. why ?, , Q.21, , Can light travelling from air to glass suffer, total internal reflection? Justify your answer., , Q.22, , What are the five general features of the, image formed by a plane mirror ?, , Q.33, , Q.34, Q.35, , C.  Long Answer Type Questions, Q.23, , Q.24, , (a) What is total internal reflection? How is, critical angle related to refractive index?, (b) A ray of light while travelling from a, denser to a rarer medium undergoes total, internal reflection. Derive the expression, for the critical angle in terms of the speed, of light in the respective media., Discuss in detail refraction at, (i) convex surface (ii) concave surface., , Q.25, , Derive lens formula for a thin lens., , Q.26, , What is critical angle? Give one application, of total internal reflection., , Q.27, , State Snell’s law of refraction., , Q.28, , What is meant by power of a lens? What is, one dioptre ?, , D.  Numerical Problem, Q.29, , The speed of light in air is 3 × 108 m/s., Calculate the speed of light in glass given, that the refractive index of glass is 1.5., , Q.30, , The refractive index of water with respect to, air is 4/3. Calculate the refractive index of air, with respect to water., , Q.31, , The refractive index of glass is 3/2. What is, the critical angle for the glass-water surface?, , Q.32, , A ray of light travelling in air falls on the, surface of a glass slab at an angle of, incidence 45°. Find the angle made by the, refracted ray with the normal within the slab, where refractive index for glass is 3/2., , A ray of light travelling in air is incident on, the surface of a transparent material of, refractive index 3 . If the angle of refraction, is 30°, calculate the angle of incidence., Focal length of a convex lens is 50 cm., Calculate its power., A point object is placed at a distance of 12 cm, from a convex lens on its principal axis. Its, image is formed 18 cm from the lens on the, other side. Calculate the focal length of the, lens., , Q.36, , An object is placed at a distance of 20 cm, from a concave lens on its principal axis. If, the focal length of the lens is 20 cm, find the, position of the image., , Q.37, , A beam of light incident parallel to the, principal axis of a concave lens appears to, diverge from a point 20 cm behind the lens, after refraction through the lens. Calculate the, power of the lens., , Q.38, , A pin 2 cm long is placed at a distance of 16, cm from a convex lens of focal length 12 cm, perpendicular to the principal axis. Find the, position, nature and size of the image., , Q.39, , A convex lens of focal length 20 cm and a, concave lens of focal length 12.5 cm are, placed in contact having the same principal, axis. Calculate the power of the combined, lens., , Q.40, , Two thin lenses of power +3.5D and –2.5D, are placed in contact. Find the power and, focal length of the lens combination., , Q.41, , An illuminated slit is kept at a distance of 40, cm in front of a convex lens of focal length, 15 cm. Find the position of the screen to, obtain the image., , Q.42, , A ray incident at a slab at angle 10° as shown, in figure. Find angle of emergent , µ = 1.2 µ = 2, , Liquid, µ = 4.1, , 10°, , µ=2, , 3, , µ=4, , , Liquid, µ = 4.1

Page 16 : EXERCISE # 2, Single Correct Answer type Questions, Q.1, , Q.3, , Q.4, , Q.5, , Refractive index of glass with respect to air is, 1.5 and refractive index of water with respect, 4, to air is . What will be the refractive index, 3, of glass with respect to water ?, (A) 1, (B) 1.5, (C) 1.125, (D) –10, , Q.7, , The refractive index of a medium depends, upon(A) Nature of material of the medium, (B) Optical density of the medium, (C) Wavelength of light, (D) All of these, , How will the image formed by a convex lens, be affected, if the central portion of the lens is, wrapped in black paper, as shown in the fig., , (A) No image will be formed, (B) Full image will be formed but it is less, bright, (C) Full image will be formed but without the, central portion, (D) Two images will be formed, one due to, each exposed half., Q.2, , Q.6, , Q.8, , The critical angle for light going from, medium X into medium Y is . The speed of, light in medium X is . The speed of light in, medium Y is(A)  (1 – cos ), , (B) /cos , , (C) cos , , (D) /sin , , One surface of a lens is convex and the other, is concave. If the radii of curvature are r1 and, r2 respectively, the lens will be convex, if(A) r1 > r2, , (B) r1 = r2, , (C) r1 < r2, , (D) r1 = 1/r2, , An object is immersed in a fluid. In order that, the object becomes invisible, it should, (A) behave as a perfect reflector, (B) absorb all light falling on it, (C) have refractive index one, (D) have refractive index exactly matching, with that of the surrounding fluid., R.I. of glass w.r.t. air is, air w.r.t. glass is3, 2, (A), (B), 4, 3, , (C), , 3, , then the R.I. of, 2, 1, 3, , (D) 3, , 4, ,, 3, then refractive index of air w.r.t. water will, be-, , If refractive index of water w.r.t. air is, , (A) 4 × 3, , (B), , 3, 4, , (C), , 4, 3, , (D), , 3, 4, , Q.9, , A ray of light is incident normally on a, rectangular piece of glass. The value of angle, of refraction will be(A) 180° (B) 90° (C) 45°, (D) 0°, , Q.10, , What is the angle of deviation ?, (A) Angle between the reflected, incident ray, (B) Angle between the reflected, refracted ray, (C) Angle between the incident, refracted ray, (D) angle between the incident, emergent ray, , Q.11, , ray and, ray and, ray and, ray and, , The speed of light in vacuum is 3.0 × 108 m/s., If the refractive index of a transparent liquid, is 4/3, then the speed of light in the liquid is(A) 2.25 × 108 m/s, (B) 3 × 108 m/s, (C) 4 × 108 m/s, (D) 4.33 × 108 m/s

Page 17 : Q.12, , A swimming pool appears to be 2m deep. Its, actual depth is ( for water = 1.33)(A) 2.66 m, (B) 2 m, (C) 2.34, (D) 2.54 m, , Q.13, , To get a real and inverted image of the same, size as that of the object the object should be, placed in front of the convex lens at(A) F, (B) 2F, (C) between F and 2F, (D) away from 2F, where F is focus, , Q.14, , A spherical mirror and a spherical lens each, have focal length of –10 cm. The mirror and, lens are(A) both convex, (B) both concave, (C) mirror is convex and lens is concave, (D) mirror is concave and lens is convex, , Q.15, , The power of a lens having focal length 50, cm is1, (A) D (B) 2D, (C) 3D, (D) 0.2 D, 2, , Q.16, , The focal length of a lens of power –2.0 D is(A) –2.0 m, (B) 0.2 m, (C) –0.5 m, (D) 0.5 m, , Q.17, , Two lenses of power + 5D and –5D are, placed in close contact. The focal length of, the combination is(A) Zero, (B) , (C) Zero or , (D) None of these, , Q.18, , A student needs a lens of power –2.0 diopter, to correct his distant vision. The focal length, of the given lens is(A) +50 cm, (B) –50 cm, (C) 100 cm, (D) –100 cm, , Q.19, , Focal length of coloured goggles (Without, number) is(A) zero, (B) infinity, (C) between zero and infinity, (D) None of these, , Q.20, , Where should an object be placed so that a, real and inverted image of very large size is, obtained, using a convex lens ?, (A) At the focus, (B) At 2F, (C) Between F and 2F (D) Beyond 2F, , Q.21, , A convex lens is –, (A) Thicker at the middle, thinner at the edges, (B) Diverging, (C) Thicker at the edges thinner in the middle, (D) Of uniform thickness everywhere, , Q.22, , A glass rod of refractive index 1.42 is, immersed in kerosene. The refractive index of, kerosene is 1.42. Then the rod will(A) appear bent, (B) appear raised above the liquid, (C) become invisible, (D) none of the above, , Q.23, , The power of a lens whose focal length is 25, cm is(A) 4 Diopter, (B) 25 Diopter, (C) 0.04 Diopter, (D) 2.5 Diopter, , Q.24, , A thin lens is made with a material having, refractive index  = 1.5. Both the sides are, convex. It is dipped in water ( = 1.33). It, will behave like(A) convergent lens (B) a divergent lens, (C) a rectangular slab (D) a prism, , Q.25, , Choose the correct option(A) If the final rays are converging, we have, a real image, (B) If the incident rays are converging, we, have a real image, (C) If the image is virtual, the corresponding, object is called a virtual object, (D) The image of a virtual object is called a, virtual image, , Q.26, , A convex lens forms a real image of a point, object placed on its principal axis. If the, upper half of the lens is painted black., (A) the image will be shifted backward, (B) the image will not be shifted, (C) the intensity of the image will decrease, (D) both (B) and (C), , Q.27, , The minimum distance between an object and, its real image formed by a convex lens of, focal length f is(A) f, (B) 2f, (C) 3f, (D) 4f

Page 18 : ANSWER KEY, EXERCISE-1, 29. 2 × 108 m/s, , 30. 3/4, , 35. 7.2 cm, , 36. 10 cm on the same side of lens, , 31. 42º, , 32. 28º, , 33. 60º, , 34. + 2D, , 37. –5D, , 38. At 48 cm from the lens on the other side. Image is real, inverted and of size 6 cm., 39. –8D, , 40. + 1D, 100 cm, , 42.  = 10°, , 41. 24 cm, , EXERCISE-2, Ques, Ans, Ques, Ans, , 1, B, 16, C, , 2, D, 17, B, , 3, C, 18, B, , 4, D, 19, B, , 5, B, 20, A, , 6, C, 21, A, , 7, D, 22, C, , 8, B, 23, A, , 9, D, 24, A, , 10, D, 25, A, , 11, A, 26, D, , 12, A, 27, D, , 13, B, , 14, B, , 15, B