NCERT Exemplar Solutions Class 12 Physics Wave Optics — Very Short, Short and Long Answer Questions Q10–23

NCERT Exemplar Problems, Class 12 Physics — Wave Optics, Very Short Answer Questions 10 to 15, Short Answer Questions 16 to 18 and Long Answer Questions 19 to 23. All 14 questions solved. The book numbers them 10.10 to 10.23.

Three ideas carry this page. Huygens’ principle and the shape of a wavefront explain why waves bend round obstacles and why a distant source gives a plane front. Interference depends on path difference and on coherence: add amplitudes with the right phases, then square. Polarisation changes the intensity by $\cos^2\theta$ (Malus’s law), and a polariser passes only one component of unpolarised light, which also decides whether two beams can interfere fully.

Key insight. Convert every situation into a path difference or a phase and add amplitudes before squaring. In Questions 18, 20 and 21 the answer comes from writing the path difference carefully, and in Questions 15 and 19 from tracking which component of the light each polariser lets through.

Very Short Answer Questions

Question 10

Is Huygens’ principle valid for longitudinal sound waves?

Solution. Yes. Huygens’ principle is a geometrical rule about how a wavefront advances: every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is their envelope. It uses only the fact that the disturbance spreads out with a definite speed, so it holds for any wave, longitudinal or transverse, and for sound as much as for light.

Question 11

Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?

Solution. A point source at the focus of the first lens gives a parallel beam, that is, plane wavefronts. The second convergent lens brings that parallel beam to a point, the final image. Beyond the image the light spreads out again from that point. A wavefront diverging from a point is a sphere centred on it, so the emerging wavefronts are spherical.

Question 12

What is the shape of the wavefront on earth for sunlight?

Solution. The Sun is so far away that the wavefront reaching us is a sphere of radius about $1.5\times10^{11}$ m. Over the size of the Earth, or of any experiment, such a small piece of so large a sphere is flat to a very good approximation. So the wavefront is, for all practical purposes, plane.

Question 13

Why is the diffraction of sound waves more evident in daily experience than that of light waves?

Solution. Diffraction is noticeable when the opening or obstacle is comparable to the wavelength. Audible sound has frequencies of 20 Hz to 20 kHz and speed about $330$ m/s, so its wavelength $\lambda = v/f$ runs from about $15$ mm to $15$ m. That matches doors, windows, corners and walls, so sound bends round them. Light has $\lambda \sim 10^{-7}$ m, and one needs apertures of that size to see it diffract. Everyday openings are millions of times larger, so light travels in straight lines.

Question 14

The human eye has an approximate angular resolution of $\phi = 5.8\times10^{-4}$ rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots.

Solution. The separation between neighbouring dots is $$l = \frac{2.54\ \text{cm}}{300} = 8.47\times10^{-3}\ \text{cm}.$$ Dots at distance $z$ subtend an angle $l/z$ at the eye. They merge when this is no larger than the eye’s resolution $\phi$: $$z = \frac{l}{\phi} = \frac{8.47\times10^{-3}}{5.8\times10^{-4}}\ \text{cm} \approx 14.6\ \text{cm}.$$ So the page should be held at about $14.5$ cm or more. (The key rounds $l$ to $0.84\times10^{-2}$ cm and quotes $14.5$ cm.)

Question 15

A polaroid (I) is placed in front of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.

Solution. Yes, in general. With I and II crossed, the light that leaves I is polarised along I’s axis and has no component along II’s axis, so nothing passes. Let III be placed with its axis at angle $\theta$ to I’s axis, so its axis is at $90^\circ-\theta$ to II’s. Let $I_1$ be the intensity after I. By Malus’s law, after III the intensity is $I_1\cos^2\theta$, and after II $$I = I_1\cos^2\theta\,\cos^2(90^\circ-\theta) = I_1\cos^2\theta\sin^2\theta = \frac{I_1}{4}\sin^22\theta.$$ This is non-zero unless $\theta = 0^\circ$ or $90^\circ$, that is, unless III’s axis is parallel to that of I or of II. III’s job is to rotate the plane of polarisation of the light so that it acquires a component along II’s axis. The emergent intensity is greatest, $I_1/4$, at $\theta = 45^\circ$.

Short Answer Questions

Question 16

Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?

Solution. Yes. Light is fully plane polarised on reflection at the Brewster angle $\theta_B$, given by $\tan\theta_B = n_2/n_1$, and this does not depend on which side is denser. Here the light starts in the denser medium, so $n_1 > n_2$ and $\tan\theta_B = n_2/n_1 < 1$, giving $\theta_B < 45^\circ$.

The only worry is total internal reflection, which sets in at the critical angle with $\sin\theta_c = n_2/n_1$. For every angle between $0^\circ$ and $90^\circ$, $\tan\theta > \sin\theta$. So $\tan\theta_B = \sin\theta_c$ implies $\theta_B < \theta_c$. The Brewster angle is therefore reached before total internal reflection begins, and the reflected light is plane polarised there.

Question 17

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Å and electrons accelerated through 100 V used as the illuminating substance.

Solution. For a given objective the least separation that can be resolved is $$d_{\min} = \frac{1.22\lambda}{2\sin\beta},$$ so with the same objective $d_{\min}\propto\lambda$, and the ratio of separations is the ratio of wavelengths.

The de Broglie wavelength of an electron accelerated through $V$ volts is $$\lambda_e = \frac{12.27}{\sqrt V}\ \text{Å} = \frac{12.27}{\sqrt{100}}\ \text{Å} = 1.227\ \text{Å}.$$ Therefore $$\frac{d_e}{d_{\text{light}}} = \frac{1.227}{5000} \approx 2.5\times10^{-4},$$ or $d_{\text{light}}/d_e \approx 4.1\times10^{3}$. An electron microscope resolves detail about four thousand times finer.

Printed key note. The key gets about $0.2\times10^{-3}$, the same order as ours, but it takes $5500$ Å for the light although the question says $5000$ Å and rounds $1.227$ Å to $1.3$ Å.

Question 18

Consider a two slit interference arrangement (Fig 10.4) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre O.

Solution. In the figure the slits $S_1$ and $S_2$ are on a line perpendicular to the axis, with $S_1C = CS_2 = D$, so the slit separation is $2D$. The screen is at distance $D$ from the slits, which is half the separation, as the question states. The point $P$ of the first minimum is at $OP = D$ from the centre, taken on the side of $S_1$.

Distances from the slits to $P$: $P$ is at height $D$ above the axis, $S_1$ is at height $D$, so $S_1$ is level with $P$ and $$S_1P = D.$$ $S_2$ is $D$ below the axis, so it is $2D$ below $P$ vertically, and $$S_2P = \sqrt{D^2 + (2D)^2} = \sqrt5\,D.$$

The first minimum needs a path difference of $\lambda/2$: $$S_2P – S_1P = (\sqrt5 – 1)D = \frac{\lambda}{2},$$ so $$D = \frac{\lambda}{2(\sqrt5-1)} \approx 0.40\,\lambda.$$

Printed key error. The key gives $D = \lambda/(2\sqrt5)$. It sets $\left[D^2 + (D-x)^2\right]^{1/2}$ to zero at $x = D$ and so drops the $-D$ from the other path. The path $S_1P$ is $D$, not zero, and the path difference is $(\sqrt5-1)D$, not $\sqrt5 D$.

Long Answer Questions

Question 19

Figure 10.5 shows a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If $I_0$ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

Solution. The source feeds both slits equally. A polariser P is placed in the path of beam 2 only (the path to $S_2$); beam 1 is unpolarised. Let each unpolarised beam have intensity $I$ at the screen. Without P the two waves are in phase at the principal maximum, so the amplitude doubles and the intensity is $4I = I_0$, that is $I = I_0/4$.

With P in beam 2, that beam becomes plane polarised along P’s axis, whatever that axis is, and its intensity is halved to $I/2$.

Beam 1 is unpolarised. Resolve it into two independent, incoherent, perpendicular components, one along P’s axis and one perpendicular to it, each of intensity $I/2$.

• The components along P’s axis, one from each slit, are coherent. Each has intensity $I/2$.
• The component of beam 1 perpendicular to P’s axis has nothing from beam 2 to interfere with. It adds a constant $I/2$ everywhere on the screen.

At the principal maximum the two parallel components are in phase, so their combined intensity is $\left(\sqrt{I/2}+\sqrt{I/2}\right)^2 = 2I$. Adding the perpendicular part, $$I_{\max} = 2I + \frac{I}{2} = \frac{5I}{2} = \frac{5}{2}\cdot\frac{I_0}{4} = \frac{5I_0}{8}.$$

At the first minimum the parallel components are in antiphase and equal in size, so they cancel completely. Only the perpendicular part is left: $$I_{\min} = \frac{I}{2} = \frac{I_0}{8}.$$

So the fringes are still there but less contrasted: the visibility is $\dfrac{5-1}{5+1} = \dfrac23$, not $1$.

Question 20

A small transparent slab containing material of $\mu = 1.5$ is placed along $AS_2$ (Fig 10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

Setup from the figure: $AC = CO = D$, $S_1C = S_2C = d \ll D$, the slab has thickness $L = d/4$ and is on the path $AS_2$. The slit separation is $S_1S_2 = 2d$, and $\theta$ is the angle of $P$ above the axis.

Solution. The slab lengthens the optical path on the $S_2$ side by $$(\mu-1)L = 0.5\times\frac d4 = \frac d8.$$ Without the slab, a point at angle $\theta$ above O has path difference $S_2P – S_1P = 2d\sin\theta$ (positive for $P$ on the $S_1$ side of O). With the slab, the total path difference between the two routes becomes $$\Delta = 2d\sin\theta + \frac d8.$$

Principal maximum: $\Delta = 0$, so $$\sin\theta_0 = -\frac1{16},\qquad OP = D\tan\theta_0 \approx -\frac{D}{16}.$$ The principal maximum moves a distance $D/16$ from O, towards the side of $S_2$, the side with the slab, because the slab delays that wave and the point where the two waves match has moved to the side from which it is nearer to $S_2$ relative to $S_1$.

First minima: $\Delta = \pm\lambda/2$, so $$2d\sin\theta + \frac d8 = \pm\frac\lambda2\quad\Rightarrow\quad \sin\theta_\pm = \pm\frac{\lambda}{4d} – \frac1{16}.$$ Their distances from O are $D\tan\theta_\pm$, which for small angles is $D\sin\theta_\pm$. The two minima are not symmetric about O any more, but they are symmetric about the new principal maximum, at $\pm\dfrac{\lambda D}{4d}$ from it.

Without the slab the principal maximum is at $O$ and the first minima are at $\sin\theta = \pm\lambda/4d$, that is $\pm\lambda D/(4d)$ from O.

The general result needs $\lambda/d$. The key’s numbers, $\sin\theta_+ = 3/16$ and $\sin\theta_- = -5/16$, follow if $d = \lambda$, which is not stated in the question. For $d = \lambda$: $$\sin\theta_+ = \frac14-\frac1{16} = \frac3{16},\quad \sin\theta_- = -\frac14-\frac1{16} = -\frac5{16},$$ giving distances $\dfrac{3D}{\sqrt{247}}$ on one side and $\dfrac{5D}{\sqrt{231}}$ on the other, which are the key’s figures.

Question 21

Four identical monochromatic sources A, B, C, D as shown in Fig 10.7 produce waves of the same wavelength $\lambda$ and are coherent. Two receivers $R_1$ and $R_2$ are at great but equal distances from B. (i) Which of the two receivers picks up the larger signal? (ii) Which of the two receivers picks up the larger signal when B is turned off? (iii) Which of the two receivers picks up the larger signal when D is turned off? (iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

Setup from the figure: $AB = BC = BD = \lambda/2$. A, B and C lie on a horizontal line with $R_1$ beyond A on the same line; D is directly below B and $R_2$ is directly above B. $R_1B = R_2B = d$, very large.

Solution. Because $d \gg \lambda$, each source sends an equal amplitude $a$ to a receiver and only the path differences matter. Measure each wave’s phase relative to the one from A at $R_1$.

At $R_1$ (on the line ABC, beyond A): A is the nearest, B is $\lambda/2$ farther, C is $\lambda$ farther. D is on the perpendicular through B, so its distance is about $d$, again $\lambda/2$ farther than A. The signs are therefore A $= +1$, B $= -1$, C $= +1$, D $= -1$ (in units of $a$), because $\lambda/2$ is a phase of $\pi$ and $\lambda$ a phase of $2\pi$. The total is $1-1+1-1 = 0$.

At $R_2$ (on the line through D and B, above B): the distances from A, B, C are all about $d$, so these three waves arrive in phase. D is $\lambda/2$ farther. The total is $1+1+1-1 = 2$, so the amplitude is $2a$.

(i) $R_2$ picks up the larger signal: amplitude $2a$ against zero at $R_1$.

(ii) With B off: $R_1$ gets $A + C + D = +1+1-1 = 1$ and $R_2$ gets $A + C + D = 1+1-1 = 1$. The two receivers pick up equal signals ($a$ each).

(iii) With D off: $R_1$ gets $A+B+C = 1-1+1 = 1$, and $R_2$ gets $1+1+1 = 3$. $R_2$ picks up the larger signal, with amplitude $3a$ against $a$ (intensities $9a^2$ against $a^2$).

(iv) $R_2$ can distinguish them: with B off it reads $a$, with D off it reads $3a$. $R_1$ reads $a$ in both cases, since removing B or D leaves a net $\pm1$ of magnitude $1$, so it cannot tell the cases apart.

Printed key errors. In (i) the key’s own working gives $R_1 = 0$ and $R_2 = 2a$, but it then says $R_1$ picks up the larger signal; that is a slip, the answer is $R_2$. In (iv) the key says that a signal at $R_1$ means B is off, but its own part (iii) gives $R_1$ the same signal when D is off, so $R_1$ cannot tell.

Question 22

The optical properties of a medium are governed by the relative permittivity $(\varepsilon_r)$ and relative permeability $(\mu_r)$. The refractive index is defined as $\sqrt{\mu_r\varepsilon_r} = n$. For ordinary material $\varepsilon_r > 0$ and $\mu_r > 0$ and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with $\varepsilon_r < 0$ and $\mu_r < 0$. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials $n = -\sqrt{\mu_r\varepsilon_r}$. As light enters a medium of such refractive index the phases travel away from the direction of propagation. (i) According to the description above show that if rays of light enter such a medium from air (refractive index $=1$) at an angle $\theta$ in 2nd quadrant, then the refracted beam is in the 3rd quadrant. (ii) Prove that Snell’s law holds for such a medium.

Solution. Take the interface along the $x$-axis, with the metamaterial in the region $y<0$. The incident ray lies in the 2nd quadrant, so it travels towards the origin from the upper left, moving to the right and downward. The key physical statement is that the phase of the wave must match along the interface. The wave vector $\vec k$ points along the direction in which the phase travels, and for light in air it is parallel to the ray.

(i) Direction of the refracted ray. Along the interface the phase of the incident wave and the phase of the refracted wave must agree at every point, otherwise the field would not be continuous across the boundary. So the $x$-component of $\vec k$ is the same on both sides: $$k_x = k_0\sin\theta > 0.$$ In a normal medium the energy (the ray) and the phase travel together, so the ray also has positive $x$-component and goes into the 4th quadrant. In the metamaterial, however, the phases travel away from the direction of propagation, so $\vec k$ points opposite to the direction of energy flow. The ray therefore has a component along $x$ opposite to $k_x$, that is negative, and it still enters the medium, going down ($y<0$). A ray moving left and down is in the 3rd quadrant. So the refracted beam is in the 3rd quadrant, on the same side of the normal as the incident ray.

(Equivalently, in terms of wavefronts: points on the wavefront reaching the interface later must still have the phase of the points reached earlier, and with the phase running backwards the wavefront tilts the opposite way, so the ray bends the ‘wrong’ way relative to the normal.)

(ii) Snell’s law. Continuity of the phase along the interface gives $k_x$ equal on both sides. In air $k_x = k_0\sin\theta_i$. In the medium $|\vec k| = |n|k_0$, and with the angle of the wave vector from the normal written as $\theta_r$, $$k_0\sin\theta_i = |n|k_0\,\sin|\theta_r|.$$ Now the wave vector and the ray point in opposite directions, so the angle of the ray is measured on the other side of the normal from the incident one. If we measure the refraction angle $\theta_r$ with a sign, negative on the opposite side of the normal, then $\sin\theta_r = -\sin|\theta_r|$, and with $n = -|n|$, $$n\sin\theta_r = (-|n|)(-\sin|\theta_r|) = |n|\sin|\theta_r| = \sin\theta_i.$$ So Snell’s law, $\sin\theta_i = n\sin\theta_r$, holds with the negative value of $n$ and a negative refraction angle. A negative angle is just what puts the ray in the 3rd quadrant, consistent with (i).

Question 23

To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediate between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF$_2$ ($n = 1.38$). What should be the thickness of the film so that at the centre of the visible spectrum (5500 Å) there is maximum transmission.

Solution. The film has $n = 1.38$, between air ($1$) and glass (about $1.5$). Light is reflected at the air–film surface and at the film–glass surface. Both reflections are from a rarer to a denser medium, so both reflected rays suffer the same phase change of $\pi$, and there is no relative phase change between them.

For maximum transmission, the reflected light must be as weak as possible, so the two reflected rays must interfere destructively. At normal incidence the extra path of the second ray is $2nd$ in the film, so $$2nd = \frac\lambda2\quad\Rightarrow\quad d = \frac{\lambda}{4n} = \frac{5500}{4\times1.38}\ \text{Å} = \frac{5500}{5.52}\ \text{Å} \approx 996\ \text{Å}.$$ So the film should be about $1000$ Å, that is about $1.0\times10^{-7}$ m, a quarter of the wavelength in the film.

Common mistakes

• Treating Huygens’ principle as only for light in Question 10. It is a statement about any wave front.
• Forgetting that the second lens makes the beam converge in Question 11. The wavefronts after the image are spherical, centred on the image.
• Ignoring the phase change on reflection in Question 23. Here both reflections are rarer to denser, so there is no relative phase change and the condition is $2nd = \lambda/2$, not $\lambda$.
• Taking the slit separation as $D$ in Question 18. $S_1C = CS_2 = D$ makes the separation $2D$, and $S_1P$ is $D$, not zero.
• Missing the extra path in Question 20. A slab adds optical path $(\mu-1)L$, not $\mu L$, because the same length of air would have been there anyway.
• Forgetting the unpolarised half in Question 19. The component of beam 1 perpendicular to P’s axis cannot interfere and fills the minima.
• Adding intensities where amplitudes must be added in Question 21. Add the signed amplitudes first, then square.
• Using the wrong sign for the refracted angle in Question 22. With $n<0$ the refracted ray is on the same side of the normal as the incident ray.

Practise next

• Exemplar Wave Optics: Multiple Choice Questions — the first half of this unit: Brewster’s law, single-slit diffraction and the coherence conditions for fringes.
• NCERT textbook Chapter 10 — Wave Optics for Huygens’ principle, Young’s experiment, diffraction and polarisation.
• A single slit of width $0.20$ mm is illuminated by light of wavelength $600$ nm. A screen is $1.0$ m away. What is the width of the central bright fringe?

• Soap film of refractive index $1.33$ in air is lit at normal incidence by light of wavelength $600$ nm. What is the smallest thickness at which the reflected light is strongest?