NCERT Exemplar Solutions Class 12 Physics Wave Optics — Multiple Choice Questions Q1–9

NCERT Exemplar Problems, Class 12 Physics — Wave Optics, MCQ I and MCQ II. All 9 questions solved.

The book numbers these questions 10.1 to 10.9; here they are Questions 1 to 9. Questions 1 to 5 are MCQ I (exactly one option is correct) and Questions 6 to 9 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Polarisation by reflection. At Brewster’s angle the reflected light is entirely polarised perpendicular to the plane of incidence, so the transmitted light is only partially polarised.
• Size against wavelength. Diffraction through a slit or hole of width $a$ is strong when $a$ is comparable to $\lambda$ and negligible only when $a \gg \lambda$.
• Coherence. A steady interference pattern needs two sources of the same wavelength with a constant phase difference.
• Phase on reflection. Reflection from a denser medium adds a phase of $\pi$; reflection from a rarer one adds none.

Key insight. Several of these are decided by one comparison: the width of the opening against $\lambda\approx 5\times10^{3}$ Å. Questions 2 and 7 turn on it, and the printed key gets both wrong.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in Fig. 10.1. A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.

The figure shows a ray striking the top of a rectangular glass slab, a reflected ray leaving at the same angle, and a refracted ray passing through the slab and emerging at P.

Solution. The incident light is unpolarised: it contains vibrations both in the plane of incidence (p) and perpendicular to it (s). At Brewster’s angle none of the p component is reflected, so the reflected beam is purely s. The refracted beam therefore loses part of its s component and keeps all of its p component. It is partially polarised, a mixture of a strong p part and a weakened s part. Passing through the slab and out of the second surface does not change this.

A polaroid in a partially polarised beam gives a maximum when its axis is along the p direction and a minimum when it is along the s direction. These are $90^\circ$ apart, so in one full turn there are two minima, $180^\circ$ apart. The minimum is not zero, because some s light is still present. So (a) is false, (b) is false because the beam is not unpolarised, and (d) is false because there are two minima in a rotation, not four.

Question 2

Consider sunlight incident on a slit of width $10^4$ Å. The image seen through the slit shall

Solution. The slit width is $a = 10^4$ Å $= 1\ \mu\text{m}$, while visible light has $\lambda$ between about $4\times10^3$ Å and $7\times10^3$ Å. So $a$ is only about two wavelengths. A sharp geometric image needs $a \gg \lambda$; here diffraction is strong. The first minimum of a single slit lies at

$$\sin\theta = \frac{\lambda}{a},$$

which is about $0.4$ for blue and $0.7$ for red. The minima therefore fall at clearly different angles for different colours. At the centre ($\theta = 0$) every colour has its maximum, so the centre is bright and white. Away from the centre the colours separate, because each has its own minima and secondary maxima, giving coloured regions. That is (c).

Option (a) would need a slit much wider than $\lambda$. Option (b) is wrong because the zeros come at different places for different colours, so the intensity does not fall to zero for all colours together. Option (d) leaves out the colour.

Printed key error. The key gives (a), a “fine sharp slit”. That would be right only for $a \gg \lambda$, which $10^4$ Å is not.

Question 3

Consider a ray of light incident from air onto a slab of glass (refractive index $n$) of width $d$, at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Solution. Let the refracted ray make angle $r$ with the normal, so $\sin\theta = n\sin r$. The ray reflected at the bottom surface travels down and back up the slab. Allowing for the part of the path it saves in air, its optical path difference from the top reflection works out to

$$\Delta = 2d\cos r, \qquad \cos r = \sqrt{1-\frac{\sin^2\theta}{n^2}}.$$

The options use $\lambda$ as the wavelength in the glass, as the form of the answer implies, so the phase from this path is $\dfrac{2\pi}{\lambda}\cdot 2d\cos r = \dfrac{4\pi d}{\lambda}\left(1-\dfrac{\sin^2\theta}{n^2}\right)^{1/2}$.

Now the reflections. The top-surface reflection is from air to glass (rarer to denser), which adds a phase of $\pi$ to that ray. The bottom-surface reflection is from glass to air (denser to rarer), which adds none. The phase difference between the two reflected rays therefore carries one extra $\pi$ from the top reflection, and the expression in (a) is the result. Options (b), (c) and (d) have no $\pi$, a $\pi/2$ that no reflection produces, and $2\pi$ (equivalent to nothing), respectively.

Question 4

In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

Solution. Interference needs two coherent waves, which means the same frequency and a constant phase difference. Red and blue light have different frequencies, so their phase difference changes continuously and rapidly with time. At any point on the screen the two intensities simply add, with no steady bright and dark pattern. The screen shows a uniformly lit region, tinted by both colours. Each slit passes only one colour, so there is no red pattern or blue pattern of its own either: each colour comes through a single slit, which gives a diffraction pattern, not two-slit fringes. So (a), (b) and (d) are all false.

Question 5

Figure 10.2 shows a standard two slit arrangement with slits $S_1$, $S_2$. $P_1$, $P_2$ are the two minima points on either side of $P$. At $P_2$ on the screen, there is a hole and behind $P_2$ is a second 2-slit arrangement with slits $S_3$, $S_4$ and a second screen behind them.

The figure shows a source S, two slits $S_1$ and $S_2$, a first screen on which $P$ is the centre with the minima $P_1$ above and $P_2$ below it. At $P_2$ a small hole opens onto a second pair of slits $S_3$, $S_4$ and a second screen.

Solution. The hole sits at $P_2$, a point of minimum intensity of the first arrangement. At a true minimum the two waves from $S_1$ and $S_2$ cancel exactly, but only at a single point. A real hole has a finite size, so it also admits points near $P_2$ where the cancellation is incomplete. A small amount of light therefore enters.

That light then behaves as a single new source of light. Every part of the hole is reached by the same original wavefronts, so the light leaving the hole is coherent. It illuminates $S_3$ and $S_4$ equally, which gives a regular two-slit pattern on the second screen, only fainter than usual. Option (b) would hold for a hole of zero size, but such a hole would also pass nothing, which is not the situation described. A pattern with a regular fringe system rules out (a) and (c).

Question 6

Two sources $S_1$ and $S_2$ of intensity $I_1$ and $I_2$ are placed in front of a screen [Fig. 10.3 (a)]. The pattern of intensity distribution seen in the central portion is given by Fig. 10.3 (b). In this case which of the following statements are true.

The pattern in Fig. 10.3 (b) is a row of equally spaced humps of similar height, with the intensity falling to zero between neighbouring humps and the central hump at $x=0$.

Solution. For two sources the intensity swings between $(\sqrt{I_1}+\sqrt{I_2})^2$ and $(\sqrt{I_1}-\sqrt{I_2})^2$. The minima in the figure reach zero, which needs $\sqrt{I_1}=\sqrt{I_2}$: the intensities are equal, so (a) is true.

A steady pattern, fixed in position, needs a constant phase difference between the sources, so (b) is true. It needs the same wavelength too: with different wavelengths the fringe spacing and phase would not stay fixed, so (d) is true.

(c) is not required. If the sources had a constant phase offset, the whole pattern would only slide along the screen. What matters is that the phase difference does not change with time, not that it is zero. Nothing in the figure says the central hump is a maximum caused by zero phase difference, so (c) cannot be inferred.

Question 7

Consider sunlight incident on a pinhole of width $10^3$ Å. The image of the pinhole seen on a screen shall be

Solution. The pinhole has $a = 10^3$ Å $= 0.1\ \mu\text{m}$, much smaller than any visible wavelength. For a circular hole the first dark ring needs $\sin\theta = 1.22\lambda/a$, and here this is greater than $1$ for every colour. So there is no dark ring at all: the hole acts almost like a point source and sends light over the whole forward hemisphere. Nothing sharp remains.

The geometrical image would be a small sharp spot, so the image is different from it: (b) is true. At the centre every colour is bright together, and the light spreads smoothly outward as one diffused spot, white in colour: (c) is true. There are no minima, so the ring in (a) cannot form. Option (d) needs a “sharp central white spot”, and a pinhole smaller than the wavelength cannot give one; a sharp centre needs the pinhole to be wider than $\lambda$.

Printed key error. The key gives (b), (d). It keeps (d), which describes a sharp central spot, in place of (c). For $a < \lambda$ the centre is diffuse, not sharp, so (c) and not (d) goes with (b).

Question 8

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased

Solution. The angular size of the central diffraction spot is about $\lambda/a$, which falls as the hole width $a$ grows. So the pattern shrinks, (a), and (c) is false. At the same time a larger hole lets more light through, and that light is concentrated into a smaller spot, so the intensity rises, (b), and (d) is false.

Question 9

For light diverging from a point source

Solution. All points at the same distance from a point source are reached by the light at the same instant, so the surface of constant phase is a sphere centred on the source: (a). The power is spread over a sphere of area $4\pi r^2$, so the intensity is $I = P/4\pi r^2$, falling as $1/r^2$. The wording of (b), “decreases in proportion to the distance squared”, is loose (the intensity is inversely proportional to it), but it is the intended statement, and it is the only reading under which (d) is false. A parabolic wavefront (c) is not produced by a point source.

Common mistakes

• Treating the emergent light as unpolarised in Question 1. Only the reflected beam is fully polarised; the transmitted beam is partly polarised, so there is a minimum but never darkness.
• Calling a slit of $10^4$ Å “fine and sharp” in Question 2. It is only about twice the wavelength, so diffraction is strong and colours separate.
• Dropping the $\pi$ in Question 3. Reflection at the top, rarer to denser, adds $\pi$; the path term alone is not the full phase difference.
• Expecting red and blue fringes in Question 4. Different frequencies are not coherent, so nothing steady forms.
• Declaring the second screen dark in Question 5. A hole of finite size passes light, and that light is coherent for $S_3$, $S_4$.
• Asking for equal phase in Question 6. A constant phase difference is enough; the offset only shifts the pattern.
• Keeping a sharp spot in Question 7. A hole narrower than $\lambda$ has no minima and gives only a diffuse spot.

Practise next

• Problem. Sunlight of wavelength $600$ nm falls on a single slit of width $2\ \mu\text{m}$. At what angle does the first minimum occur?

• Problem. Light in air is incident on glass of refractive index $1.5$ at Brewster’s angle. Find the angle of incidence and the angle between the reflected and refracted rays.

• Exemplar Wave Optics: Very Short, Short and Long Answer Questions — where the same ideas are applied to Huygens’ principle, resolving power, Young’s slits and thin films.
• NCERT textbook Chapter 10 — Wave Optics for the theory of interference, diffraction and polarisation.
• Exemplar Ray Optics and Optical Instruments — the previous unit, on rays, mirrors and lenses.
• Exemplar Dual Nature of Radiation and Matter — the next unit, where light also behaves as particles.