NCERT Exemplar Solutions Class 12 Physics Ray Optics and Optical Instruments — Multiple Choice Questions Q1–16

NCERT Exemplar Problems, Class 12 Physics — Ray Optics and Optical Instruments, MCQ I and MCQ II. All 16 questions solved.

The book numbers these questions 9.1 to 9.16; here they are Questions 1 to 16. Questions 1 to 11 are MCQ I (exactly one option is correct) and Questions 12 to 16 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Refraction and dispersion. Snell’s law $\mu_1\sin i = \mu_2\sin r$, with $\mu$ larger for blue than for red, fixes bending, speed and the critical angle.
• Mirror and lens formulae. $1/v + 1/u = 1/f$ for mirrors and $1/v – 1/u = 1/f$ for lenses, together with the lens-maker’s formula $1/f = (\mu-1)(1/R_1 – 1/R_2)$.
• Total internal reflection. It needs light to go from a denser to a rarer medium at an angle beyond $\sin^{-1}(1/\mu)$; a gradual change of refractive index acts like a curved version of it.
• Instruments. Angular magnification, not linear size, is what a magnifier or telescope improves.

Key insight. Most of these are decided by one inequality: which colour has the larger $\mu$, which angle exceeds the critical angle, or where the object sits relative to $f$. Questions 10 and 14 carry printed key errors.

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

Question 1

A ray of light incident at an angle $\theta$ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is $5^\circ$ and the prism is made of a material of refractive index $1.5$, the angle of incidence is

Solution. Emerging normally means the angle of refraction at the second face is $r_2 = 0$. For a prism $r_1 + r_2 = A$, so $r_1 = A = 5^\circ$. Snell’s law at the first face gives

$$\sin i = 1.5\sin 5^\circ \approx 1.5\times 0.0872 = 0.1307, \qquad i \approx 7.5^\circ.$$

For small angles the same result follows from $i \approx \mu r_1 = 1.5\times 5^\circ = 7.5^\circ$.

Question 2

A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

Solution. At normal incidence no colour is bent, so all colours travel the same thickness $t$ of glass. The time taken is $t/v = t\mu/c$. In glass $\mu$ is smallest for red, so red has the greatest speed, takes the least time and emerges first. Violet, with the largest $\mu$, is slowest and emerges last.

Question 3

An object approaches a convergent lens from the left of the lens with a uniform speed $5$ m/s and stops at the focus. The image

Solution. Take the object distance as $u$ (magnitude) and the lens formula $\dfrac1v = \dfrac1f – \dfrac1u$ in magnitudes for a real image, so

$$v = \frac{uf}{u-f}.$$

As the object comes in from far away, $u$ falls towards $f$ and $v$ rises from about $f$ towards infinity: the image moves away from the lens. Its speed is

$$\frac{dv}{dt} = -\frac{f^2}{(u-f)^2}\frac{du}{dt},$$

and since $du/dt = -5$ m/s is constant, the image speed is $5f^2/(u-f)^2$, which keeps growing as $u\to f$. A speed that changes at a changing rate means non-uniform acceleration, so (a) and (b) fail, and (d) is wrong because the image moves away, not towards the lens.

Question 4

A passenger in an aeroplane shall

Solution. A rainbow is the set of directions at about $42^\circ$ (primary) and $51^\circ$ (secondary) from the line joining the observer to the point opposite the Sun, the antisolar point. On the ground the horizon cuts off the lower part of each circle, leaving arcs. From an aeroplane there are water drops all around and below, so nothing blocks the lower part and the whole circles can be seen, the secondary one outside the primary, both centred on the antisolar point. So (a) and (d) are false, and (c) describes the view from the ground rather than from the plane.

Question 5

You are given four sources of light each one providing a light of a single colour — red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is $90^\circ$. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

Solution. Refraction at $90^\circ$ means the yellow beam is incident exactly at the critical angle, so light goes from a denser to a rarer medium and $\sin i_c = \mu_2/\mu_1$. The refractive index of the denser medium increases from red to blue, so the critical angle decreases from red to blue.

• Blue has a larger $\mu_1$, hence a smaller critical angle than yellow. The same angle of incidence now exceeds it, so blue undergoes total internal reflection. (c) is true.
• Red has a larger critical angle, so the angle of incidence is below it and red is refracted. It bends away from the normal, as light entering a rarer medium always does. So (a) and (b) are false.
• Green lies between yellow and blue in $\mu$ but is not specified: it has a smaller critical angle than yellow, so it also undergoes TIR; it cannot be refracted, so (d) is false.

Question 6

The radius of curvature of the curved surface of a plano-convex lens is $20$ cm. If the refractive index of the material of the lens be $1.5$, it will

Solution. By the lens-maker’s formula with one flat face ($R_2\to\infty$),

$$\frac1f = (\mu-1)\frac1R = 0.5\times\frac1{20\ \text{cm}} \quad\Rightarrow\quad f = +40\ \text{cm}.$$

The focal length is positive, and reversing the lens swaps $R_1$ and $R_2$ with a sign change on both, leaving $1/R_1 – 1/R_2$ unchanged. So the lens converges light whichever face the object is on.

Question 7

The phenomena involved in the reflection of radiowaves by ionosphere is similar to

Solution. The ionosphere has no sharp surface. Its electron density, and with it the refractive index for radiowaves, changes gradually with height, so a wave travelling upward is bent more and more until it turns back down. In a mirage, light from the sky passes through hot air near the ground whose refractive index also varies gradually; it bends progressively and is turned upward by what is effectively total internal reflection. Both involve a gradient of refractive index, not a mirror surface (a), not dispersion (c), and not scattering (d).

Question 8

The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (Fig. 9.1). Which of the four rays correctly shows the direction of reflected ray?

The figure shows a concave mirror with centre of curvature C and focus F on the principal axis. The incident ray PQ starts at P below the axis, passes through F and strikes the mirror at Q above the axis. Ray 1 leaves Q upward to the left, ray 2 horizontally to the left, ray 3 down to the left through the region of C, and ray 4 horizontally to the right.

Solution. A ray that passes through the focus before striking a concave mirror emerges parallel to the principal axis. Since PQ passes through F, the reflected ray is horizontal. It must travel away from the mirror, which is to the left, so it is ray 2. Ray 4 is also horizontal but would head into the mirror; rays 1 and 3 do not satisfy the rule.

Question 9

The optical density of turpentine is higher than that of water while its mass density is lower. Fig. 9.2 shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Fig. 9.2, the path shown is correct?

The figure shows air above a layer of turpentine above a layer of water in a rectangular container, with four rays 1 to 4 coming down at different angles and bending at each interface.

Solution. Optical density is about refractive index, not mass density, so $\mu_\text{air} < \mu_\text{water} < \mu_\text{turpentine}$. Going from air into turpentine the ray enters a denser medium and bends towards the normal. Going from turpentine into water it enters a rarer medium and bends away from the normal. So the correct ray must be steeper (closer to the vertical) inside the turpentine than in the air, and then less steep again in the water. Measuring the figure, only ray 2 does both; the others fail one of the two bends.

Question 10

A car is moving with at a constant speed of $60\ \text{km h}^{-1}$ on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of $100$ m and is approaching with a speed of $5\ \text{km h}^{-1}$. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every $2$ s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement(s) is/are correct?

Solution. The rear car approaches the leading car at $5\ \text{km h}^{-1}$ relative to it, so its speed over the road is $60 + 5 = 65\ \text{km h}^{-1}$: (a) is true. The stem says “statement(s) is/are correct”, so more than one option can hold.

The rear-view mirror is plane. A plane mirror has $v_i = v_o$ for the image speed relative to the mirror, so the image appears to approach at a constant $5\ \text{km h}^{-1}$ however close the car gets. (c) is false.

The side mirror is convex. For a convex mirror differentiating $\dfrac1v+\dfrac1u=\dfrac1f$ with respect to time gives the image speed

$$v_i = \left(\frac{f}{u-f}\right)^2 v_o ,$$

with $u$ the object distance and $f$ the magnitude of the (negative) focal length used with the sign convention. As the car gets nearer, $u$ decreases and the factor grows, so the image appears to speed up. (d) is true. The image speed in the convex mirror is much less than $5\ \text{km h}^{-1}$ at large distances, and it is not constant, so (b) is false.

Printed key error. The key prints (d) only. But (a) is a plain statement of the data, $60+5 = 65\ \text{km h}^{-1}$, and the question asks which statement(s) are correct, so it cannot be left out. The correct answer is (a), (d).

Question 11

There are certain material developed in laboratories which have a negative refractive index (Fig. 9.3). A ray incident from air into such a medium (medium 2) shall follow a path given by

The figure shows four diagrams (a) to (d) of a ray incident from medium 1 (above) at angle $i$ to the normal. In (a) the ray goes into medium 2 on the same side of the normal as the incident ray, bending towards the normal. In (b) it goes into medium 2 on the opposite side of the normal, bent away from the normal. In (c) it is reflected. In (d) it goes straight down along the normal.

Solution. Snell’s law $\sin i = \mu_2\sin r$ with $\mu_2<0$ gives $\sin r<0$, so $r$ is negative: the refracted ray lies on the same side of the normal as the incident ray, not across it as for an ordinary medium. Its magnitude obeys $|\sin r| = \sin i/|\mu_2|$. Diagram (a) shows the ray inside medium 2 on the incident side of the normal. (b) is the ordinary behaviour for a positive index, (c) is only reflection, and (d) would need $r = 0$ for all $i$.

Question 12

Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because

Solution. The apparent depth of a point is $d/\mu$ only for near-normal viewing. For an eye near the edge, rays from points close to the edge leave the surface at a small angle to the normal, while rays from distant points reach the surface obliquely and leave at a grazing angle.

• (a) True. Points near the edge are seen at nearly their normal apparent depth, whereas oblique viewing makes points farther away appear nearer the surface. The object is seen distorted in depth.
• (b) True. For an eye near the edge, distant points are seen at grazing emergence, so the image is compressed and subtends a smaller angle than the object would in air.
• (c) True. A ray from a far point that reaches the surface at more than the critical angle ($\approx 49^\circ$ for water) is totally internally reflected and never reaches the eye, so some points are not visible.
• (d) False. A plane water surface has no focusing power; it does not magnify.

Question 13

A rectangular block of glass ABCD has a refractive index $1.6$. A pin is placed midway on the face AB (Fig. 9.4). When observed from the face AD, the pin shall

Solution. In the figure A is the top left corner, B top right, C bottom right and D bottom left; the pin sits on the top face AB, and the face AD is on the left. Light from the pin enters the glass at the face AB, with $\sin i = 1.6\sin r$. The critical angle for glass in air is

$$\sin c = \frac1{1.6},\qquad c = 38.7^\circ.$$

Even for $i = 90^\circ$ the angle of refraction satisfies $\sin r\le 1/1.6$, so $r\le c = 38.7^\circ$ at the face AB. Face AD is perpendicular to AB, so the angle of incidence of the ray at AD is $90^\circ – r \ge 51.3^\circ$. That is greater than $c$, so every ray from the pin is totally internally reflected at AD. No light leaves through AD and the pin cannot be seen from that face.

Question 14

Between the primary and secondary rainbows, there is a dark band known as Alexandar’s dark band. This is because

Solution. Rays that are reflected once inside a raindrop (primary bow) leave at angles of at most about $42^\circ$ from the direction opposite the Sun, with the brightest light piled up at the edge. Rays reflected twice (secondary bow) leave at about $51^\circ$ or more. Geometrically, no ray with either path comes out at an angle in the gap between these values, so there is simply no light scattered into that band by the drops. That is (b), and the angular range quoted in (d), roughly $42^\circ$ to $51^\circ$ (the book says $50^\circ$), is a correct description of where the band lies.

Alexander’s dark band is not interference (a), since there are no waves arriving there to cancel, and absorption (c) would darken the whole sky.

Printed key error. The key prints (a), (d). Option (a) attributes the band to destructive interference, which is wrong: the dark band is purely geometric. The correct answer is (b), (d).

Question 15

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

Solution. The object is placed within the focal length of the convex lens, so the image is virtual, erect and magnified (b). Because the object is now much closer than $25$ cm, it subtends a larger angle at the eye, and that is what shows fine detail (a). The angular magnification is $m = 1 + D/f$ at the near point and $D/f$ for the image at infinity; both are finite, so (d) is false. A magnifier does not widen the field of view (c); it narrows it as the object fills more of the lens.

Question 16

An astronomical refractive telescope has an objective of focal length $20$ m and an eyepiece of focal length $2$ cm.

Solution. In normal adjustment the tube length is the sum of the focal lengths:

$$L = f_o + f_e = 20\ \text{m} + 0.02\ \text{m} = 20.02\ \text{m}.$$

The angular magnification is

$$M = \frac{f_o}{f_e} = \frac{2000\ \text{cm}}{2\ \text{cm}} = 1000.$$

A telescope with two convex lenses gives an inverted final image. A larger aperture does gather more light and so brightens the image, but it does not reduce chromatic aberration; that depends on the lens material and design, and a wider lens can even make it worse. So (d) is false.

Common mistakes

• Using the wrong colour order in Question 2. Red is the fastest in glass because it has the smallest $\mu$.
• Treating the image speed as constant in Question 3. The image speed $5f^2/(u-f)^2$ grows without bound as the object nears the focus.
• Confusing the view in Question 4. Arcs belong to the ground; from the air the circles are complete.
• Mixing up the colours in Question 5. Blue has a larger $\mu$ and so a smaller critical angle than yellow.
• Using the sign of the curved side in Question 6. The focal length is $+40$ cm whichever side the object is on.
• Reading density as optical density in Question 9. Turpentine is optically denser than water despite being lighter.
• Dropping (a) in Question 10 because the key lists only (d). The rear car’s speed is $65\ \text{km h}^{-1}$.
• Blaming interference in Question 14. Alexander’s dark band is a gap in the geometry of drop deflections, not a pattern of waves.
• Giving the tube length as $f_o\times f_e$ or the magnification as $f_o+f_e$ in Question 16. Add for length, divide for magnification.

Practise next

• Exemplar Ray Optics and Optical Instruments: Very Short, Short and Long Answer Questions — where the same ideas are applied to prisms, lenses, critical angles and gravitational lensing.
• NCERT textbook Chapter 9 — Ray Optics and Optical Instruments for the theory of refraction, lenses, prisms and instruments.
• Exemplar Electromagnetic Waves — the previous unit, which describes light as an electromagnetic wave.
• Exemplar Wave Optics — the next unit, which goes beyond rays to interference and diffraction.