NCERT Exemplar Solutions Class 10 Science Light Reflection and Refraction — Multiple Choice Questions Q1–10

NCERT Exemplar Problems, Class 10 Science — Light: Reflection and Refraction, Multiple Choice Questions 1 to 10. All 10 questions solved.

This unit’s multiple-choice band runs to nineteen questions and is split here at ten; Questions 11 to 19 continue into mirror selection, ray-diagram checking and the magic-mirror question.

Almost everything on this page comes from three relations and one convention.

• Mirror formula $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$, with magnification $m = \dfrac{h’}{h} = -\dfrac{v}{u}$.
• Lens formula $\dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u}$, with $m = \dfrac{h’}{h} = \dfrac{v}{u}$, and power $P = \dfrac{1}{f}$ with $f$ in metres.
• Snell’s law, ${}_1n_2 = \dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}$.

The convention is the one that causes most of the lost marks: distances are measured from the pole or optical centre, against the direction of the incident light being negative. So an object in front of a mirror always has $u$ negative, a real image in front of a mirror has $v$ negative, and a concave mirror has $f$ negative.

Key insight. Half of these questions never need a formula at all — they need you to know what each device does: a concave mirror and a convex lens converge, a convex mirror and a concave lens diverge, a glass slab displaces without deviating, and a plane mirror does neither. Identify the device’s behaviour first and the option usually falls out.

Choose the correct answer from the given four options in the following questions:

Question 1

Which of the following can make a parallel beam of light when light from a point source is incident on it?

Solution. A point source sends light out in all directions. To turn that diverging light into a parallel beam you need a device that converges — and the trick is to put the source at the focus, so that rays which would have converged to the focus instead leave parallel.

The two converging devices in the syllabus are the concave mirror and the convex lens, which is option (a). A torch works exactly this way: a small bulb sits at the focus of a concave reflector.

Convex mirrors and concave lenses are diverging devices. Light from a point source is already diverging, and they can only make it diverge more — never parallel. That disposes of (b), and of (d), which pairs a converging mirror with a diverging lens.

Option (c) is a different idea altogether. Two plane mirrors at right angles send a ray back parallel to itself, which is Question 28 in the short-answer band — but that is about one ray returning along a parallel path, not about turning a diverging beam into a parallel one. Plane mirrors do not converge light at all.

Question 2

A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is

Solution. The image is formed in front of the mirror, so it is real, and a real image in a mirror is inverted. Set the signs before calculating:

$$h = +10\ \mathrm{mm}, \qquad h’ = -5\ \mathrm{mm}, \qquad v = -30\ \mathrm{cm}$$

From the magnification,

$$m = \frac{h’}{h} = \frac{-5}{10} = -\frac{1}{2} = -\frac{v}{u}$$

$$\Rightarrow\ \frac{1}{2} = \frac{v}{u} \ \Rightarrow\ u = 2v = -60\ \mathrm{cm}$$

Now the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-60} = \frac{-2 – 1}{60} = -\frac{3}{60} = -\frac{1}{20}$$

$$f = -20\ \mathrm{cm}$$

which is option (b). The negative sign is expected — a concave mirror always has a negative focal length in this convention — and every option is negative, so the sign is not what is being tested; the arithmetic is.

Note the sanity check the numbers offer: the object is at $60\ \mathrm{cm} = 3f$ and the image at $30\ \mathrm{cm} = 1.5f$, both beyond the centre of curvature and between C and F respectively, which is exactly where a real, inverted, diminished image should be.

Question 3

Under which of the following conditions a concave mirror can form an image larger than the actual object?

Solution. Run through the four positions and ask what size of image each gives.

(a) At C, the radius of curvature. The image is real, inverted and the same size as the object. Not larger.

(d) Beyond C. The image is real, inverted and diminished. Not larger.

(c) Between F and C. The image is real, inverted and magnified — it forms beyond C. This is the key’s answer, and it is correct.

(b) Nearer than F. The image is virtual, erect and magnified, formed behind the mirror. This is how a concave mirror is used as a shaving or make-up mirror.

So (b) and (c) are both larger than the object, and the question as printed has two correct options. The distinction the exemplar seems to intend is that (c) gives a real magnified image — one you could catch on a screen — while (b) gives a virtual one; but the question says only “an image larger than the actual object”, and a virtual image is still an image.

If this wording appeared in a board paper, (c) is the safer answer, since it is the one the key prints. But it is worth knowing that a concave mirror magnifies in two separate positions, because that fact is what Question 30 of the long-answer band asks you to draw.

Question 4

Figure 10.1 shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is

Solution. Two things have to be read off the figure before Snell’s law can be applied, and both are easy to get wrong.

Which angle is which. Angles in optics are always measured from the normal, not from the surface. The figure marks $30^\circ$ from the surface in medium A, and its complement $60^\circ$ from the normal — which is the one to use. In medium B the angle from the normal is $45^\circ$.

Which is the incidence. The question says the light travels from A to B. So the ray starts in A, and the angle of incidence is the one in A:

$$i = 60^\circ \ (\text{in A}), \qquad r = 45^\circ \ (\text{in B})$$

Now Snell’s law:

$${}_A n_B = \frac{\sin i}{\sin r} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{\sqrt{2}/2} = \frac{\sqrt{3}}{\sqrt{2}}$$

which is option (a), and it is about $1.22$.

Two checks confirm it. The answer is greater than 1, which it must be: the ray bends towards the normal on entering B ($60^\circ$ down to $45^\circ$), so B is the denser medium. And option (b) is the same expression inverted — it is ${}_B n_A$, the answer to the question asked the other way round, which is the commonest mistake here.

Question 5

A light ray enters from medium A to medium B as shown in Figure 10.2. The refractive index of medium B relative to A will be

Solution. No numbers are given, so the answer has to come from the direction of bending alone — which is what the question is testing.

Compare the two segments against the normal, the vertical line. In medium A the ray runs at a shallow angle to the boundary, so it makes a large angle with the normal. In medium B it is much steeper, so it makes a smaller angle with the normal.

The ray has therefore bent towards the normal on crossing into B. Light bends towards the normal when it enters a denser medium — one in which it travels more slowly — so B is denser than A, and

$${}_A n_B = \frac{\sin i}{\sin r} > 1$$

since $i$ (in A) is larger than $r$ (in B). Option (a).

Option (c) would mean the two media were optically identical and the ray went straight on, which the figure plainly contradicts. Option (d) is impossible: a refractive index of zero would mean light travelled infinitely fast in B. Option (b) would be right if the ray had bent away from the normal, and it is the answer to this figure viewed upside down.

The rule to carry: towards the normal means entering a denser medium; away from the normal means entering a rarer one.

Question 6

Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the Figure 10.3. Which of the following could be inside the box?

Solution. Read what the figure actually shows: each beam comes out parallel to the direction it went in, but shifted sideways. Both beams are displaced by the same amount and in the same direction, and neither is bent.

That combination — no deviation, but lateral displacement — is the signature of a rectangular glass slab. A ray entering a slab bends towards the normal at the first face and away from it by exactly the same amount at the second, because the two faces are parallel; the two bends cancel in direction and leave only a sideways shift. This is Question 21 in the short-answer band.

The other three all change the direction of the light, which the figure shows is not happening.

A convex lens would converge the two beams towards each other, so they would cross. A concave lens would spread them apart. A prism has non-parallel faces, so the two refractions add instead of cancelling, and the emergent ray is deviated towards the base — and it would also disperse white light into colours.

The test to remember: parallel in, parallel out, but displaced means a slab.

Question 7

A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the Figure 10.4. Which of the following could be inside the box?

Solution. The rays go in parallel and come out diverging from a single point. The whole answer is in recognising that the point they diverge from is a real crossing point — the beam has been brought to a focus and has then spread out again beyond it.

Only a converging device does that, and the converging lens is the convex lens. Parallel rays entering a convex lens meet at its focus; past the focus they continue and separate, which is the fan the figure shows.

A concave lens would also produce a diverging fan, and this is the option worth thinking about rather than dismissing. But its rays diverge as though coming from a virtual focus on the incoming side — they never actually meet — so the fan would start spreading immediately at the lens, not from a point beyond it. The figure’s rays clearly cross first.

A rectangular glass slab leaves parallel rays parallel, as Question 6 has just established. A prism deviates all the rays through the same angle, so they would emerge still parallel to one another, merely turned towards the base.

Question 8

Which of the following statements is true?

Solution. Every option states the same focal length, $0.25\ \mathrm{m}$, so two things are being tested: the magnitude of the power and its sign.

The magnitude comes from the definition, with $f$ in metres:

$$P = \frac{1}{f} = \frac{1}{0.25} = 4\ \mathrm{D}$$

so the number is 4 in every case, and the sign decides everything.

A convex lens is converging, so its focal length is positive and its power is positive. A concave lens is diverging, with negative focal length and negative power. Since all four options quote $f = +0.25\ \mathrm{m}$ — a positive focal length — the lens must be convex, and its power must be $+4\ \mathrm{D}$.

That is option (a). Option (b) gives a convex lens a negative power, contradicting itself. Options (c) and (d) both call it concave, which a positive focal length rules out — and (d) is doubly inconsistent, since a concave lens would need $f = -0.25\ \mathrm{m}$ to have $-4\ \mathrm{D}$.

The unit is worth stating: one dioptre is the power of a lens of focal length one metre. Converting centimetres to metres before dividing is where this calculation usually goes wrong.

Question 9

Magnification produced by a rear view mirror fitted in vehicles

Solution. A rear view mirror is a convex mirror, and that single fact settles the question — because a convex mirror behaves the same way no matter where the object is.

Whatever the object’s distance, a convex mirror forms an image that is virtual, erect and diminished, located between the pole and the focus behind the mirror. Diminished means the magnification is always less than one.

Option (d) is the one to reject carefully, because it would be right for a concave mirror, whose magnification really does depend on the object’s position — more than one between F and C, equal to one at C, less than one beyond C. The convex mirror has no such variety, and that reliability is exactly why it is chosen for the job.

The reason a driver wants a diminished image is the accompanying gain: a convex mirror has a much wider field of view than a plane mirror of the same size, so more of the road behind is visible. The price is that everything looks smaller and therefore further away, which is why such mirrors carry the warning that objects are closer than they appear.

Question 10

Rays from Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of its image is equal to the size of the object?

Solution. The first sentence gives the focal length, though it does not say so directly. The Sun is effectively at infinity, so rays from it arrive parallel, and parallel rays converge at the focus. They converge 15 cm in front of the mirror, so

$$f = 15\ \mathrm{cm} \quad (\text{numerically}), \qquad \text{so } R = 2f = 30\ \mathrm{cm}$$

An image the same size as the object is formed by a concave mirror only when the object is at the centre of curvature, C. There the image is real, inverted and equal in size, and it forms at C as well.

So the object goes at $R = 30$ cm in front of the mirror — option (b).

The mirror formula confirms it. Put $u = -30$ and $f = -15$:

$$\frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{-15} – \frac{1}{-30} = \frac{-2 + 1}{30} = -\frac{1}{30}$$

so $v = -30\ \mathrm{cm}$, and $m = -v/u = -(-30)/(-30) = -1$. The magnification is $-1$: same size, inverted. ✓

The other options each describe a different case: at 15 cm the object is at F and no image forms (the reflected rays are parallel); between 15 and 30 cm the image is magnified; beyond 30 cm it is diminished.

Common mistakes

• Measuring angles from the surface instead of the normal, in Question 4. The figure marks $30^\circ$ from the surface; Snell’s law needs the $60^\circ$ from the normal. Always convert before substituting.
• Inverting the refractive index, in Question 4. ${}_A n_B$ puts the angle in A on top, because the light travels from A into B. Option (b) there is the correct value for ${}_B n_A$.
• Forgetting to convert to metres in Question 8. $P = 1/f$ requires $f$ in metres. A focal length in centimetres gives a power a hundred times too large.
• Assuming a rear view mirror behaves like a concave one, in Question 9. A convex mirror’s image is always virtual, erect and diminished, whatever the object distance — which is the whole reason it is used.
• Reading “rays from the Sun converge at 15 cm” as an object distance, in Question 10. The Sun is at infinity, so that 15 cm is the focal length, and the answer needs $R = 2f$.
• Treating a glass slab and a prism as interchangeable, in Questions 6 and 7. A slab’s faces are parallel, so the two refractions cancel and only a sideways shift remains; a prism’s are not, so the deviations add.
• Missing that Question 3 has two correct options. A concave mirror magnifies both when the object is nearer than F (virtual image) and when it is between F and C (real image). The key prints only the second.
• Dropping signs in Question 2. The image is in front of the mirror, so $v$ is negative and the image is inverted, making $h’$ negative. Getting $m = +1/2$ instead of $-1/2$ puts the object on the wrong side.

Practise next

• Exemplar Multiple Choice Questions Q11–19 — mirror selection, the ray-diagram checking questions and the magic mirror.
• Exemplar Short and Long Answer Questions — where the mirror and lens formulae of Questions 2 and 10 have to be applied without options to check against.
• NCERT textbook Chapter 9 — Light: Reflection and Refraction for the end-of-chapter exercise on the mirror formula, refractive index and lens power.
• Textbook activities for the experiments behind Questions 6 and 7 — the glass-slab and lens ray tracings are performed there rather than described.