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. 9.11. 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 229incidence (i). As the angle of incidence increases, so does the angle of Reprint 2025-26 Physics refraction, till for the ray AO3, the angle of refraction is p/2. The 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. 9.11. 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. This is called total internal reflection. When light gets reflected by a surface, normally some FIGURE 9.11 Refraction and internal reflection fraction of it gets transmitted. The of rays from a point A in the denser medium reflected ray, therefore, is always less (water) incident at different angles at the interface intense than the incident ray, howsoever with a rarer medium (air). 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 90°, say ÐAO3N, is called the critical angle (ic ) for the given pair of media. We see from Snell’s law [Eq. (9.10)] that if the relative refractive index of the refracting medium is less than one then, since the maximum value 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 1 with respect to rarer medium 2 will be n12 = 1/sinic. Some typical critical angles are listed in Table 9.1. TABLE 9.1 CRITICAL ANGLE OF SOME TRANSPARENT MEDIA WITH RESPECT TO AIR 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 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. Add a few drops of milk or any other suspension to water and stir so that water 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. 230 Reprint 2025-26 Ray Optics and Optical Instruments 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. 9.12(a)]? Now direct the laser beam from one side of the beaker such that it strikes the upper surface of water more obliquely [Fig. 9.12(b)]. Adjust the direction of laser beam until you find the angle for which the refraction 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. 9.12(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. 9.4.19.4.19.4.19.4.19.4.1 TotalTotalTotalTotalTotal internalinternalinternalinternalinternal reflectionreflectionreflectionreflectionreflection ininininin naturenaturenaturenaturenature andandandandand itsitsitsitsits technologicaltechnologicaltechnologicaltechnologicaltechnological applicationsapplicationsapplicationsapplicationsapplications (i) Prism: Prisms designed to bend light by 90° or by 180° make use of total internal reflection [Fig. 9.13(a) and (b)]. Such a prism is also used to invert images without changing their size [Fig. 9.13(c)]. In the first two cases, the critical angle ic for the material of the prism FIGUREFIGUREFIGUREFIGUREFIGURE 9.129.129.129.129.12 must be less than 45°. We see from Table 9.1 that this is true for both Observing total internal crown glass and dense flint glass. reflection in water with (ii) Optical fibres: Nowadays optical fibres are extensively used for a laser beam (refraction transmitting audio and video signals through long distances. Optical due to glass of beaker fibres too make use of the phenomenon of total internal reflection. neglected being very Optical fibres are fabricated with high quality composite glass/quartz thin). 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. 9.14). Since light undergoes total internal reflection at each stage, there is no appreciable loss in the intensity of the light signal. Optical fibres are fabricated such that light reflected at one side of inner surface strikes the other at an 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. FIGUREFIGUREFIGUREFIGUREFIGURE 9.139.139.139.139.13 Prisms designed to bend rays by A bundle of optical fibres can be put to 90° and 180° or to invert image without changing several uses. Optical fibres are extensively its size make use of total internal reflection. used for transmitting and receiving 231 Reprint 2025-26 Physics 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 like esophagus, stomach and intestines. You might have seen a commonly available decorative lamp with fine FIGURE 9.14 Light undergoes successive total plastic fibres with their free ends forming a internal reflections as it moves through an fountain like structure. The other end of the optical fibre. 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 9.15 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, MN 232 tan ÐNOM = OM Reprint 2025-26 Ray Optics and Optical Instruments MN tan ÐNCM = MC MN tan ÐNIM = MI Now, for DNOC, i is the exterior angle. Therefore, i = ÐNOM + ÐNCM MN MN i = + (9.13) OM MC Similarly, FIGURE 9.15 Refraction at a spherical r = ÐNCM – ÐNIM surface separating two media. MN MN i.e., r = − (9.14) MC MI Now, by Snell’s law n1 sin i = n 2 sin r or for small angles n1i = n 2r Substituting i and r from Eqs. (9.13) and (9.14), we get n 1 n 2 n 2 − n 1 + = (9.15) OM MI MC 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 n 1 n 2 − n 1 − = (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 9.5 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 n2 = 1.5. We then have 1.5 1 0.5 + = v 100 20 or v = +100 cm The image is formed at a distance of 100 cm from the glass surface, EXAMPLE in the direction of incident light. 9.5 233 Reprint 2025-26 Physics 9.5.2 Refraction by a lens Figure 9.16(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. 9.16(b)]. The image I1 acts as a virtual object for the second surface that forms the image at I [Fig. 9.16(c)]. Applying Eq. (9.15) to the first interface ABC, we get n 1  n 2  n 2  n 1 (9.17) OB BI1 BC1 A similar procedure applied to the second interface* ADC gives,  n 2  n 1  n 2  n 1 (9.18) DI1 DI DC 2 For a thin lens, BI1 = DI1. Adding Eqs. (9.17) and (9.18), we get n1 n1  1 1  + = (n 2 − n1 ) + (9.19) OB DI  BC1 DC 2  Suppose the object is at infinity, i.e., OB ® ¥ and DI = f, Eq. (9.19) gives n1  1 1  = (n 2 − n1 ) + (9.20) f  BC1 DC 2  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 F¢, on either side of it (Fig. 9.17). By the sign convention, BC1 = + R1, DC2 = –R2 So Eq. (9.20) can be written as (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 FIGURE 9.16 (a) The position of object, and the of suitable radii of curvature. Note that the image formed by a double convex lens, formula is true for a concave lens also. In (b) Refraction at the first spherical surface and that case R1is negative, R2 positive and (c) Refraction at the second spherical surface. therefore, f is negative. * Note that now the refractive index of the medium on the right side of ADC is n1 while on its left it is n2. Further DI1 is negative as the distance is measured234 against the direction of incident light. Reprint 2025-26 Ray Optics and Optical Instruments From Eqs. (9.19) and (9.20), we get n1 n1 n1 + = (9.22) OB DI f 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 1 − = (9.23) v u f 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 F¢, 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 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 F¢ (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) A ray of light passing through the first principal focus of a convex lens [Fig. 9.17(a)] emerges parallel to the principal axis after refraction. (b) A ray of light incident on a concave lens appearing to meet the principal axis at second focus point emerges parallel to the principal axis after refraction [Fig. 9.17(b)]. Figures 9.17(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 formula, Eq. (9.23), holds good for all cases. 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. FIGURE 9.17 Tracing rays through (a) Magnification (m) produced by a lens is convex lens (b) concave lens. defined, like that for a mirror, as the ratio of the 235 size of the image to that of the object. Proceeding Reprint 2025-26 Physics in the same way as for spherical mirrors, it is easily seen that for a lens h ′ v m = = (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. Example 9.6 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? Solution 9.6 The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n1 = n2. This gives 1/f =0 or f ® ¥. The lens in the liquid will act like a plane sheet of glass. No, EXAMPLE 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 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 parallel to the principal axis falling at unit distance from the optical centre (Fig. 9.18). FIGURE 9.18 Power of a lens. h 1 1 tan δ = ; if h = 1, tan δ = or δ = for small f f f value of d. Thus, 1 P = (9.25) f 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 9.7 (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 9.7 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 EXAMPLE index for air-glass = 1.5.) Reprint 2025-26 Ray Optics and Optical Instruments 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 f, R1 and R2. Substituting the values, we have 1  1 1  12 = (n − 1)  10 −−15  This gives n = 1.5. (iii) For a glass lens in air, n2 = 1.5, n1 = 1, f = +20 cm. Hence, the lens formula gives 1  1 1  = 0.5 − 20  R1 R 2  For the same glass lens in water, n2 = 1.5, n1 = 1.33. Therefore, 1 .33  1 1  = (1 .5 − 1 .33 ) − (9.26) EXAMPLE f  R1 R 2  Combining these two equations, we find f = + 78.2 cm. 9.7 9.5.4 Combination of thin lenses in contact 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 the first lens A (Fig. 9.19). 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 FIGURE 9.19 Image formation by a final image. In fact, the direction of rays emerging combination of two thin lenses in contact. from the first lens gets modified in accordance with 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 − = (9.27) v1 u f 1 For the image formed by the second lens B, we get 1 1 1 − = (9.28) v v1 f 2 Adding Eqs. (9.27) and (9.28), we get 1 1 1 1 − = + (9.29) v u f 1 f 2 If the two lens-system is regarded as equivalent to a single lens of focal length f, we have 237 Reprint 2025-26 Physics 1 1 1 − = v u f so that we get 1 1 1 = + (9.30) f f 1 f 2 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 1 1 1 = + + + … (9.31) f f 1 f 2 f 3 In terms of power, Eq. (9.31) can be written as P = P1 + P2 + P3 + … (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 m2 m3 ... (9.33) Such a system of combination of lenses is commonly used in designing lenses for cameras, microscopes, telescopes and other optical instruments. Example 9.8 Find the position of the image formed by the lens combination given in the Fig. 9.20. FIGURE 9.20 Solution Image formed by the first lens 1 1 1 − = 1 u 1 f 1 9.8 v 1 1 1 − = v1 −30 10 or v1 = 15 cm EXAMPLE Reprint 2025-26 Ray Optics and Optical Instruments 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 1 − = v 2 10 − 10 or v2 = ¥ 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 1 − = v 3 u 3 f 3 1 1 1 or = + v 3 ∞ 30 EXAMPLE or v3 = 30 cm The final image is formed 30 cm to the right of the third lens. 9.8

9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 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¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.

9.21 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.