10.2 What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.

10.2 HUYGENS PRINCIPLE We would first define a wavefront: when we drop a small stone on a calm pool of water, waves spread out from the point of impact. Every point on the surface starts oscillating with time. At any instant, a photograph of the surface would show circular rings on which the disturbance is maximum. Clearly, all points on such a circle are oscillating in phase because they are at the same distance from the source. Such a locus of points, which oscillate in phase is called a wavefront; thus a wavefront is defined as a surface of constant FIGURE 10.1 (a) Aphase. The speed with which the wavefront moves outwards from the diverging spherical source is called the speed of the wave. The energy of the wave travels wave emanating from in a direction perpendicular to the wavefront. a point source. The If we have a point source emitting waves uniformly in all directions, wavefronts are then the locus of points which have the same amplitude and vibrate in spherical. the same phase are spheres and we have what is known as a spherical wave as shown in Fig. 10.1(a). At a large distance from the source, a small portion of the sphere can be considered as a plane and we have what is known as a plane wave [Fig. 10.1(b)]. Now, if we know the shape of the wavefront at t = 0, then Huygens principle allows us to determine the shape of the wavefront at a later time t. Thus, Huygens principle is essentially a geometrical construction, which given the shape of the wafefront at any time allows us to determine the shape of the wavefront at a later time. Let us consider a diverging FIGURE 10.1 (b) At a wave and let F1F2 represent a portion of the spherical wavefront at t = 0 large distance from (Fig. 10.2). Now, according to Huygens principle, each point of the the source, a small wavefront is the source of a secondary disturbance and the wavelets portion of the emanating from these points spread out in all directions with the speed spherical wave can of the wave. These wavelets emanating from the wavefront are usually be approximated by a plane wave.referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. FIGURE 10.2 F1F2 represents the spherical wavefront (with O as centre) at t = 0. The envelope of the secondary wavelets emanating from F1F2 produces the forward moving wavefront G1G2. The backwave D1D2 does not exist. 257 Reprint 2025-26 Physics Thus, if we wish to determine the shape of the wavefront at t = t, we draw spheres of radius vt from each point on the spherical wavefront where v represents the speed of the waves in the medium. If we now draw a common tangent to all these spheres, we obtain the new position of the wavefront at t = t. The new wavefront shown as G1G2 in Fig. 10.2 is again spherical with point O as the centre. The above model has one shortcoming: we also have a backwave which is shown as D1D2 in Fig. 10.2. Huygens argued that the amplitude of the secondary wavelets is maximum in the forward direction and zero in the backward direction; by making this adhoc assumption, Huygens could explain the absence of the backwave. However, this adhoc assumption is not satisfactory and the absence of the backwave is really justified from more rigorous wave theory. In a similar manner, we can use Huygens principle to determine the shape of the wavefront for a plane wave propagating through a medium (Fig. 10.3). FIGURE 10.3 Huygens geometrical construction for a 10.3 REFRACTION AND REFLECTION OF plane wave PLANE WAVES USING HUYGENS PRINCIPLE propagating to the right. F1 F2 is the 10.3.1 Refraction of a plane wave plane wavefront at t = 0 and G1G2 is the We will now use Huygens principle to derive the laws of refraction. Let PP¢ wavefront at a later represent the surface separating medium 1 and medium 2, as shown in time t. The lines A1A2, Fig. 10.4. Let v1 and v2 represent the speed of light in medium 1 and B1B2 … etc., are medium 2, respectively. We assume a plane wavefront AB propagating in normal to both F1F2 the direction A¢A incident on the interface at an angle i as shown in the and G1G2 and figure. Let t be the time taken by the wavefront to travel the distance BC. represent rays. Thus, BC = v1 t FIGURE 10.4 A plane wave AB is incident at an angle i on the surface PP¢ separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. The figure corresponds to v2 < v1 so that the refracted waves bends towards the 258 normal. Reprint 2025-26 Wave Optics In order to determine the shape of the refracted wavefront, we draw a sphere of radius v2t from the point A in the second medium (the speed of the wave in the second medium is v2). Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2 t and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain BC v1τ sin i = = (10.1) AC AC and AE v 2τ sin r = = (10.2) AC AC ChristiaanChristiaanChristiaanChristiaanChristiaan HuygensHuygensHuygensHuygensHuygens where i and r are the angles of incidence and refraction, (1629(1629(1629(1629(1629 ––––– 1695)1695)1695)1695)1695) Dutch respectively. Thus we obtain physicist, astronomer, mathematician and the sin i v1 founder of the wave = sin r v 2 (10.3) theory of light. His book, Treatise on light, makes From the above equation, we get the important result CHRISTIAAN fascinating reading even that if r < i (i.e., if the ray bends toward the normal), the today. He brilliantly speed of the light wave in the second medium (v2) will be explained the double less then the speed of the light wave in the first medium refraction shown by the (v1). This prediction is opposite to the prediction from mineral calcite in this the corpuscular model of light and as later experiments work in addition to HUYGENSshowed, the prediction of the wave theory is correct. Now, reflection and refraction. if c represents the speed of light in vacuum, then, He was the first to analyse circular and c simple harmonic motion (1629 n1 = – v1 (10.4) and designed and built improved clocks and and telescopes. He discovered c the true geometry of 1695) n2 = v 2 (10.5) Saturn’s rings. are known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can be written as n1 sin i = n2 sin r (10.6) This is the Snell’s law of refraction. Further, if l1 and l 2 denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to l1 then the distance AE will be equal to l2 (because if the crest from B has reached C in time t, then the crest from A should have also reached E in time t ); thus, λ1 BC v1 = = λ2 AE v 2 or v1 v 2 = (10.7) 259 λ1 λ2 Reprint 2025-26 Physics The above equation implies that when a wave gets refracted into a denser medium (v1 > v2) the wavelength and the speed of propagation decrease but the frequency n (= v/l) remains the same. 10.3.2 Refraction at a rarer medium We now consider refraction of a plane wave at a rarer medium, i.e., v2 > v1. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. 10.5. The angle of refraction will now be greater than angle of incidence; however, we will still have effect n1 sin i = n2 sin r . We define an angle ic by the following equation n 2 sin i c = (10.8) Doppler n1 and Thus, if i = ic then sin r = 1 and r = 90°. Obviously, for i > ic, there can not be any refracted wave. The angle ic is known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection. The phenomenon of total internal reflection and its resonance applications was discussed in Section 9.4. refraction, diffraction, interference, of FIGURE 10.5 Refraction of a plane wave incident on a rarer medium for which v2 > v1. The plane wave bends away from the normal. Demonstration http://www.falstad.com/ripple/ 10.3.3 Reflection of a plane wave by a plane surface We next consider a plane wave AB incident at an angle i on a reflecting surface MN. If v represents the speed of the wave in the medium and if t represents the time taken by the wavefront to advance from the point B to C then the distance BC = vt In order to construct the reflected wavefront we draw a sphere of radius vt from the point A as shown in Fig. 10.6. Let CE represent the tangent plane drawn from the point C to this sphere. Obviously 260 AE = BC = vt Reprint 2025-26 Wave Optics FIGURE 10.6 Reflection of a plane wave AB by the reflecting surface MN. AB and CE represent incident and reflected wavefronts. If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles i and r (as shown in Fig. 10.6) would be equal. This is the law of reflection. Once we have the laws of reflection and refraction, the behaviour of prisms, lenses, and mirrors can be understood. These phenomena were discussed in detail in Chapter 9 on the basis of rectilinear propagation of light. Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave passing through a thin prism. Clearly, since the speed of light waves is less in glass, the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we consider a plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F. In a similar manner, we can understand refraction and reflection by concave lenses and convex mirrors. From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. For example, when a convex lens focusses light to form a real image, although the ray going through the centre traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens. FIGURE 10.7 Refraction of a plane wave by (a) a thin prism, (b) a convex lens. 261 (c) Reflection of a plane wave by a concave mirror. Reprint 2025-26 Physics Example 10.1 (a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why? (b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave? (c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light. Solution (a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light. (b) No. Energy carried by a wave depends on the amplitude of the 10.1 wave, not on the speed of wave propagation. (c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per EXAMPLE unit time. 10.4 COHERENT AND INCOHERENT ADDITION OF WAVES In this section we will discuss the interference pattern produced by the superposition of two waves. You may recall that we had discussed the superposition principle in Chapter 14 of your Class XI textbook. Indeed the entire field of interference is based on the superposition (a) (b) principle according to which at a particular point in the medium, the resultant FIGURE 10.8 (a) Two needles oscillating in displacement produced by a number ofphase in water represent two coherent sources. (b) The pattern of displacement of water waves is the vector sum of the displace- molecules at an instant on the surface of water ments produced by each of the waves. showing nodal N (no displacement) and Consider two needles S1 and S2 moving antinodal A (maximum displacement) lines. periodically up and down in an identical fashion in a trough of water [Fig. 10.8(a)]. They produce two water waves, and at a particular point, the phase difference between the displacements produced by each of the waves does not change with time; when this happens the two sources are said to be coherent. Figure 10.8(b) shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time. Consider a point P for which 262 S1 P = S2 P Reprint 2025-26 Wave Optics Since the distances S1 P and S2 P are equal, waves from S1 and S2 will take the same time to travel to the point P and waves that emanate from S1 and S2 in phase will also arrive, at the point P, in phase. Thus, if the displacement produced by the source S1 at the point P is given by y1 = a cos wt then, the displacement produced by the source S2 (at the point P) will also be given by y2 = a cos wt Thus, the resultant of displacement at P would be given by y = y1 + y2 = 2 a cos wt Since the intensity is proportional to the square of the amplitude, the resultant intensity will be given by I = 4 I0 where I0 represents the intensity produced by each one of the individual sources; I0 is proportional to a2. In fact at any point on the perpendicular bisector of S1S2, the intensity will be 4I0. The two sources are said to FIGURE 10.9interfere constructively and we have what is referred to as constructive (a) Constructive interference. We next consider a point Q [Fig. 10.9(a)] interference at a for which point Q for which the S2Q –S1Q = 2l path difference is 2l. (b) Destructive The waves emanating from S1 will arrive exactly two cycles earlier interference at a than the waves from S2 and will again be in phase [Fig. 10.9(a)]. Thus, if point R for which the the displacement produced by S1 is given by path difference is 2.5 l. y1 = a cos wt then the displacement produced by S2 will be given by y2 = a cos (wt – 4p) = a cos wt where we have used the fact that a path difference of 2l corresponds to a phase difference of 4p. The two displacements are once again in phase and the intensity will again be 4 I0 giving rise to constructive interference. In the above analysis we have assumed that the distances S1Q and S2Q are much greater than d (which represents the distance between S1 and S2) so that although S1Q and S2Q are not equal, the amplitudes of the displacement produced by each wave are very nearly the same. We next consider a point R [Fig. 10.9(b)] for which S2R – S1R = –2.5l The waves emanating from S1 will arrive exactly two and a half cycles later than the waves from S2 [Fig. 10.10(b)]. Thus if the displacement FIGURE 10.10 Locus produced by S1 is given by of points for which y1 = a cos wt S1P – S2P is equal to zero, ±l, ± 2l, ± 3l. then the displacement produced by S2 will be given by y2 = a cos (wt + 5p) = – a cos wt 263 Reprint 2025-26 Physics where we have used the fact that a path difference of 2.5l corresponds to a phase difference of 5p. The two displacements are now out of phase and the two displacements will cancel out to give zero intensity. This is referred to as destructive interference. To summarise: If we have two coherent sources S1 and S2 vibrating in phase, then for an arbitrary point P whenever the path difference, S1P ~ S2P = nl (n = 0, 1, 2, 3,...) (10.9) we will have constructive interference and the resultant intensity will be 4I0; the sign ~ between S1P and S2 P represents the difference between S1P and S2 P. On the other hand, if the point P is such that the path difference, 1 S1P ~ S2P = (n+ ) l (n = 0, 1, 2, 3, ...) (10.10) 2 we will have destructive interference and the resultant intensity will be zero. Now, for any other arbitrary point G (Fig. 10.10) let the phase difference between the two displacements be f. Thus, if the displacement produced by S1 is given by y1 = a cos wt then, the displacement produced by S2 would be interference y2 = a cos (wt + f) wave and the resultant displacement will be given by on y = y1 + y2 = a [cos wt + cos (wt +f)] = 2 a cos (f/2) cos (wt + f/2) experiments Tank The amplitude of the resultant displacement is 2a cos (f/2) and therefore the intensity at that point will be I = 4 I0 cos2 (f/2) (10.11) Ripple http://phet.colorado.edu/en/simulation/legacy/wave-interference If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by Eq. (10.9) we will have constructive interference leading to maximum intensity. On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to the condition given by Eq. (10.10)] we will have destructive interference leading to zero intensity. Now if the two sources are coherent (i.e., if the two needles are going up and down regularly) then the phase difference f at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time. However, if the two needles do not maintain a constant phase difference, then the interference pattern will also change with time and, if the phase difference changes very rapidly with time, the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. When this happens, we will observe an average intensity that will be given by I = 2 I0 (10.12) 264 at all points. Reprint 2025-26 Wave Optics When the phase difference between the two vibrating sources changes rapidly with time, we say that the two sources are incoherent and when this happens the intensities just add up. This is indeed what happens when two separate light sources illuminate a wall.

9.20 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. f1 = 30 cm and u1 = – , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives, v2 = – 220 cm. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm. This image becomes a real object for the second lens: f2 = + 30 cm, u2 = – (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2, and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); 112 × 20 f2 = – 20 cm which gives v2 = − cm 92 Magnitude of magnification due to the second (concave) 347 Reprint 2025-26 Physics lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.21 If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 1 1 1 9.22 (a) + = v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u)| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 9.23 (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. 9.24 Magnification = ( 6.25 / 1) = 2.5 v = +2.5u 1 1 1    2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. 9.25 (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The Reprint 2025-26 Answers effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations (both spherical and chromatic) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which increases if fe is smaller. Further, magnification of the objective v O 1 = is given by | u O | (| u O |/ f O ) − 1 which is large when |u O | is slightly greater than fO. The micro- scope is used for viewing very close object. So |u O | is small, and so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument.