NCERT Exemplar Solutions Class 10 Science The Human Eye and the Colourful World — Multiple Choice Questions Q1–14

NCERT Exemplar Problems, Class 10 Science — The Human Eye and the Colourful World, Multiple Choice Questions. All 14 questions solved.

This chapter joins two subjects that at first look unrelated — how the eye focuses, and why the sky is blue — and the exemplar’s fourteen multiple-choice questions divide almost evenly between them. Four ideas carry the lot.

• Accommodation. The eye focuses by changing the focal length of its own lens: ciliary muscles contract to make the lens thicker and shorter-focus for near objects, and relax to make it thinner and longer-focus for distant ones.
• The two defects. Myopia (short sight) — the far point has come closer, distant objects blur, corrected by a concave lens. Hypermetropia (long sight) — the near point has receded, nearby objects blur, corrected by a convex lens.
• Dispersion. A prism separates white light because the glass refracts each colour by a different amount; violet is deviated most, red least, so violet emerges nearest the base.
• Scattering. Fine particles scatter short wavelengths much more strongly than long ones, which is why the sky is blue, the setting sun is red and danger signals are red.

Key insight. Almost every scattering question here is answered by the same sentence: blue scatters most, red scatters least. Blue scattered out of the direct beam and into your eye from all over the sky makes the sky blue; the same blue removed from a long path through the atmosphere leaves the sunset red. One fact, two opposite-looking results.

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

Question 1

A person cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using a lens of power

Solution. Not seeing distant objects clearly is myopia, and the person’s far point is 2 m rather than infinity.

The corrective lens must take an object at infinity and form its image at the far point, 2 m away, so that the eye can then focus it. Applying the lens formula with $u = -\infty$ and $v = -2\ \mathrm{m}$ (the image is virtual, on the same side as the object):

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-2} – 0 = -\frac{1}{2}$$

$$f = -2\ \mathrm{m} \qquad\Rightarrow\qquad P = \frac{1}{f} = -0.5\ \mathrm{D}$$

which is option (b).

Two checks confirm it. The sign must be negative, because myopia is corrected by a diverging (concave) lens — that rules out (a) and (c) immediately. And the magnitude: the focal length of the correcting lens equals the distance of the far point, 2 m, giving $1/2 = 0.5\ \mathrm{D}$. Option (d), $-0.2\ \mathrm{D}$, would correspond to a far point of 5 m.

The shortcut worth remembering is that for myopia, $f$ = the far point distance (negative), because the lens must bring infinity to the far point.

Question 2

A student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book. Which of the following statements is correct?

Solution. Read the symptoms carefully, because the two halves point in opposite directions.

He can read the blackboard, which is far away. So his distant vision is fine and his far point is normal — that disposes of (c) and (d), both of which claim the far point has changed.

He cannot read his textbook, which is close. So his near vision has failed: the closest point he can focus on is further away than it should be. The near point has receded, which is option (a).

This is hypermetropia, long sight. A normal near point is about 25 cm; in hypermetropia it might be 50 cm or a metre, so a book held at reading distance is inside it and cannot be brought into focus.

Option (b) has the near point coming closer, which would be an improvement rather than a defect — a nearer near point means being able to focus on things closer still.

The correction is a convex lens, which adds the converging power the eye is short of. Note the contrast with Question 1: there the far point had moved in and a concave lens was needed; here the near point has moved out and a convex lens is needed.

Question 3

A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident on the prism as shown in Figure 11.1. In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky?

Solution. Work backwards from what is wanted.

The colour of the sky is blue, and the question asks for the arrangement in which blue is the third colour from the top.

The spectrum runs VIBGYOR — violet, indigo, blue, green, yellow, orange, red — and blue is the third colour counting from the violet end. So the required order from the top is violet, indigo, blue: violet must be at the top.

Now the physics. In a prism, violet is deviated most and red least, and all the colours are bent towards the base. So violet always emerges nearest the base of the prism.

For violet to be at the top, the base must be at the top.

Checking the four orientations: in (i) the base BC is at the bottom, so violet would be at the bottom and red at the top — the order from the top would be ROYGBIV, making blue the fifth colour. In (iii) and (iv) the base is vertical, so the spectrum spreads sideways rather than up and down and no colour is meaningfully “third from the top”.

Only (ii) has the base BC uppermost, putting violet at the top and blue third down.

Question 4

At noon the sun appears white as

Solution. At noon the Sun is overhead, so its light travels through the shortest possible path in the atmosphere to reach us — straight down through the layers rather than slanting through them.

A short path means few scattering encounters, so very little light of any colour is scattered out of the beam. All seven colours arrive in nearly their original proportions, and light with all colours present in balance is white.

That is option (a).

Option (b) is self-defeating: if every colour were scattered away, none would reach us and the Sun would look dark.

Option (c) is a true statement — blue is always scattered most — but it does not explain whiteness. It explains why the sky around the Sun is blue, and it explains the red sunset, when the long slanting path removes enough blue to leave the direct beam reddened. At noon the path is too short for that removal to matter.

Option (d) reverses the physics: red is scattered least, which is exactly why it survives the long path at sunset.

The comparison to hold is between noon and sunset, and the only thing that changes is the length of the path through the atmosphere.

Question 5

Which of the following phenomena of light are involved in the formation of a rainbow?

Solution. Follow sunlight through a single raindrop, which acts as a tiny prism.

1. Refraction as the light enters the drop, bending at the air–water surface.
2. Dispersion at the same moment — each colour is refracted by a different amount, so the white light begins to separate.
3. Internal reflection at the back surface of the drop, sending the light back towards the observer.
4. Refraction again, and further dispersion, as it leaves the drop.

So the three phenomena are refraction, dispersion and internal reflection — option (c).

The distinction between (b) and (c) is the one being tested, and it is worth getting right. Total internal reflection happens only when light strikes the boundary at an angle greater than the critical angle, and then all of it is reflected. Inside a raindrop the light meets the back surface at an angle that is generally less than the critical angle, so only part of it is reflected and the rest escapes out of the back. That is ordinary internal reflection, not total — which is also why a rainbow is faint rather than dazzling.

Option (a) omits the internal reflection that turns the light back. Option (d) brings in scattering, which is the mechanism behind the blue sky, not the rainbow.

Question 6

Twinkling of stars is due to atmospheric

Solution. The atmosphere is not uniform. Its density falls with height, and it is constantly stirred by winds and convection, so light from a star passes through many layers of slightly different refractive index that are continually shifting.

Each layer bends the light a little — atmospheric refraction — and because the layers move, the amount of bending changes from moment to moment. The star’s apparent position therefore wavers slightly, and the amount of its light entering the eye fluctuates. That flickering is what we see as twinkling. Option (b).

The reason it happens to stars and not to planets is worth adding, since Question 30 asks for it. A star is so far away that it is effectively a point source, and the whole of its light is affected together. A planet is much nearer and appears as a small extended source — a collection of many points, each twinkling independently and at random. Their variations average out, so a planet shines steadily.

The other options name real phenomena attached to the wrong effect: dispersion by water droplets makes rainbows, scattering by particles makes the sky blue, and clouds do not internally reflect starlight in any relevant way.

Question 7

The clear sky appears blue because

Solution. The molecules of the air are much smaller than the wavelength of visible light, and such particles scatter short wavelengths far more strongly than long ones — the scattered intensity falls off very steeply as the wavelength increases.

Violet and blue have the shortest wavelengths of the visible spectrum, so they are scattered most. That scattered light comes to us from every direction in the sky, not just from the direction of the Sun, and it is what we see when we look away from the Sun. Hence the blue. Option (c).

Option (d) is the exact reverse and would make the sky orange-red.

Option (a) confuses scattering with absorption. Absorbed light is gone — converted to heat — and could not reach our eyes at all; if blue were absorbed the sky would look yellowish. Scattered light is merely redirected, and it is precisely that redirection into our line of sight that makes the sky luminous.

Option (b) is a true statement about the ozone layer but has nothing to do with the colour we see, since ultraviolet is invisible.

A fair question is why the sky is not violet, given that violet scatters even more than blue. Two reasons: sunlight contains less violet to begin with, and the human eye is considerably less sensitive to violet than to blue.

Question 8

Which of the following statements is correct regarding the propagation of light of different colours of white light in air?

Solution. The decisive words are in air.

In vacuum, and to a very good approximation in air, all colours of light travel at the same speed, $3 \times 10^8\ \mathrm{m\,s^{-1}}$. Air’s refractive index is so close to 1, and varies so little across the spectrum, that no separation of colours occurs. Option (c).

This is why white light travelling through air stays white, and why the light from a distant star arrives with all its colours together rather than spread out in time.

It is in a denser medium such as glass or water that the colours part company. There the refractive index depends on wavelength — it is larger for violet than for red — so violet travels slowest and red fastest. That difference is what makes dispersion possible, and it is the reason a prism works while a stretch of air does not.

So option (a) is a true statement about glass, not about air, and it is the trap in this question. Option (b) gets the order right for a dense medium — blue is slower than green there, so it is wrong even then. Option (d) invents an averaging rule that holds in no medium.

Question 9

The danger signals installed at the top of tall buildings are red in colour. These can be easily seen from a distance because among all other colours, the red light

Solution. Red has the longest wavelength in the visible spectrum, and scattering is far weaker for long wavelengths than for short ones.

So red light is scattered the least by the smoke, fog and dust in the air. It therefore travels furthest through murky air without being diverted out of the beam, and a red signal stays visible from a distance when a blue one would have been scattered away. Option (b).

Option (a) is the reverse and would make red the worst choice. Option (c) confuses scattering with absorption again — and if red were strongly absorbed it would be a poor signal colour for that reason too.

Option (d) is the same mistake as Question 8: in air all colours travel at the same speed, and in any case the speed of light has nothing to do with how far a signal can be seen.

The same physics explains several everyday facts worth grouping together: red is used for traffic stop lights and brake lights, the setting Sun looks red, and red is chosen for fog lamps. In each case the reason is that long wavelengths survive a long or dusty path.

Question 10

Which of the following phenomena contributes significantly to the reddish appearance of the sun at sunrise or sunset?

Solution. At sunrise and sunset the Sun is near the horizon, so its light reaches us through a much longer path in the atmosphere than at noon — several times longer.

Over that long path, scattering removes most of the short-wavelength light. Blue and violet are scattered out of the beam in every direction and never reach the eye; what survives the journey is chiefly the red and orange, which scatter least. The Sun therefore appears reddish. Option (b).

Compare Question 4 directly: the same mechanism gives a white Sun at noon and a red one at sunset, and the only variable is the length of the atmospheric path. Short path, little scattering, all colours survive, white. Long path, much scattering, blue removed, red.

The other options do not apply. Dispersion separates colours by refraction in a prism or raindrop and would give a spectrum, not a uniformly reddened disc. Total internal reflection requires a denser-to-rarer boundary and a large angle, which does not arise here. Reflection from the earth plays no part in the colour of the direct sunlight.

Question 11

The bluish colour of water in deep sea is due to

Solution. Water molecules and the fine particles suspended in sea water scatter light in the same way the air does — preferentially at short wavelengths. Blue light is scattered most, and much of it is scattered back upwards out of the water towards an observer. That scattered blue is the colour we see. Option (c).

Option (b) is the plausible-sounding one and is worth ruling out carefully. Reflection of the sky does contribute to the appearance of a water surface, particularly at a glancing angle. But it cannot be the explanation the question wants, because the deep sea looks blue even on an overcast day, and looks blue when viewed from directly above where reflection is weakest. The colour comes from the water itself.

Option (a) fails on the colours: algae and water plants are green or brown, and where they are abundant the water looks greenish rather than blue.

Option (d) has the wrong mechanism. Water does absorb light — and it absorbs red more strongly than blue, which also favours blue — but absorption alone removes light rather than sending any back to the eye. Something must scatter light upwards for the sea to look coloured at all.

Question 12

When light rays enter the eye, most of the refraction occurs at the

Solution. This one runs against intuition, which is why it is asked.

Bending at a surface depends on the difference in refractive index across it. The largest such difference anywhere in the eye is at the very front, where light passes from air ($n = 1.00$) into the cornea ($n \approx 1.38$). That single surface does most of the focusing — roughly two-thirds to three-quarters of the eye’s total converging power. Option (b).

The crystalline lens sits behind the cornea, surrounded by fluids of refractive index close to its own, so the difference across its surfaces is small and it bends light comparatively little. Its importance is not the amount of refraction but that it is the only adjustable part: the ciliary muscles change its shape, and that fine adjustment is accommodation, the subject of Question 13.

Options (c) and (d) do not refract light at all. The iris is the coloured muscular diaphragm that controls how much light enters, and the pupil is simply the hole in it — an aperture, not a lens. A hole cannot refract anything.

So the division of labour is: the cornea does the bulk focusing, fixed; the lens does the fine focusing, variable; the iris and pupil control the brightness.

Question 13

The focal length of the eye lens increases when eye muscles

Solution. Two things must be paired correctly: what the muscles do, and what happens to the lens.

A thicker, more curved lens has a shorter focal length — more curvature means more bending. So a longer focal length requires a thinner, flatter lens. That eliminates (b) and (c), which both pair an increase in focal length with a thicker lens.

Now which muscle state gives a thin lens? The ciliary muscles relax when viewing distant objects; the suspensory ligaments then pull the lens flat and thin, giving it a long focal length. They contract for near objects, releasing that tension and letting the lens bulge thicker and shorter-focus.

So an increased focal length goes with relaxed muscles and a thinner lens — option (a).

The point worth carrying is that the relaxed state is the distant-vision state. Looking at faraway things rests the eye; sustained close work keeps the ciliary muscles contracted, which is why it causes eye strain.

Object	Ciliary muscles	Lens	Focal length
Distant	relaxed	thin, flat	long
Near	contracted	thick, curved	short

Question 14

Which of the following statement is correct?

Solution. Set out what each defect does before testing the statements.

Myopia — short sight. The far point has moved in from infinity. Distant objects blur; nearby objects are seen clearly.

Hypermetropia — long sight. The near point has moved out from 25 cm. Nearby objects blur; distant objects are seen clearly.

Now the four options:

• (a) Myopia and distant objects clear — false, that is exactly what myopia prevents.
• (b) Hypermetropia and nearby objects clear — false, the near point has receded so close work is the problem.
• (c) Myopia and nearby objects clear — true. A short-sighted person can read comfortably without glasses.
• (d) Hypermetropia and distant objects not clear — false, distant vision is the part that still works.

So (c). Notice that (a) and (c) are opposite claims about myopia, and (b) and (d) are opposite claims about hypermetropia, so exactly one from each pair is true — and only (c) is offered among the true ones.

This is the fact behind Question 2, where the student can read the blackboard but not his book: distant vision intact, near vision failing, therefore hypermetropia.

Common mistakes

• Confusing scattering with absorption, in Questions 7 and 11. Scattered light is redirected and still reaches the eye from a new direction — that is why the sky glows blue. Absorbed light is gone.
• Applying “red travels fastest” to air, in Questions 8 and 9. All colours travel at the same speed in air; they differ only in a dense medium such as glass, which is what makes dispersion possible.
• Choosing total internal reflection for the rainbow, in Question 5. The light meets the back of a raindrop below the critical angle, so the reflection is partial — ordinary internal reflection, not total.
• Mixing up which point has moved, in Questions 2 and 14. Myopia moves the far point in; hypermetropia moves the near point out. Reading the symptom first — which distance is blurred — settles it.
• Assuming the eye lens does most of the focusing, in Question 12. The cornea does, because the air-to-cornea boundary is the biggest jump in refractive index. The lens is the adjustable part, not the powerful one.
• Pairing a thicker lens with a longer focal length, in Question 13. More curvature means more bending and therefore a shorter focal length.
• Forgetting that violet sits nearest the base, in Question 3. Violet is deviated most, so the base position fixes which end of the spectrum is on top.
• Giving the sign of the corrective lens by guesswork, in Question 1. Myopia takes a negative power, and its magnitude is the reciprocal of the far point distance in metres.

Practise next

• Exemplar Short and Long Answer Questions — where the eye defects have to be drawn as ray diagrams and the dispersion explained rather than recognised.
• NCERT textbook Chapter 10 — The Human Eye and the Colourful World for the end-of-chapter exercise on accommodation, defects of vision and atmospheric refraction.
• Exemplar Light — Reflection and Refraction for the lens formula and power calculations that Question 1 above depends on.
• Textbook activities for the prism and dispersion experiments behind Questions 3 and 5.