NCERT Exemplar Solutions Class 10 Science The Human Eye and the Colourful World — Short and Long Answer Questions Q15–30

NCERT Exemplar Problems, Class 10 Science — The Human Eye and the Colourful World, Short Answer Questions 15 to 24 and Long Answer Questions 25 to 30. All 16 questions solved.

Nine of these sixteen questions ask for a diagram, and the diagrams fall into just two families — the eye, and the prism. Getting each family right once is most of the work.

For the eye, every defect diagram needs the same three elements: where the rays actually converge, where the retina is, and how the corrective lens moves the first onto the second.

	Rays converge	Corrective lens	What it does
Myopia	in front of the retina	concave	diverges the light a little first, pushing the focus back
Hypermetropia	behind the retina	convex	converges it a little first, pulling the focus forward

For the prism, the single fact that generates every answer is that violet is deviated most and red least, and all colours bend towards the base.

Key insight. Dispersion is not the prism “adding” colour — the colours were in the white light all along. The prism separates them only because the glass has a slightly different refractive index for each wavelength, so each is bent through a different angle. That is why two prisms placed oppositely, in Question 19, put the white light back together again: the second undoes exactly what the first did.

Short Answer Questions

Question 15

Draw ray diagrams each showing (i) myopic eye and (ii) hypermetropic eye.

Solution. In both defects the eye lens is working, but the image lands in the wrong place relative to the retina.

(i) The myopic eye. Parallel rays from a distant object are brought to a focus in front of the retina, and by the time the light reaches the retina it has begun to diverge again, so the image there is blurred. Either the eyeball is too long from front to back, or the lens is too strongly converging. The far point has moved in from infinity to some finite distance.

(ii) The hypermetropic eye. Rays from a nearby object are still converging when they reach the retina — their meeting point lies behind it, which is why the diagram shows it with dashed lines: no image actually forms there. Either the eyeball is too short, or the lens is too weakly converging. The near point has receded beyond the usual 25 cm.

The two defects are therefore opposite errors of the same kind, and the corrections in Question 26 are correspondingly opposite.

Question 16

A student sitting at the back of the classroom cannot read clearly the letters written on the blackboard. What advice will a doctor give to her? Draw ray diagram for the correction of this defect.

Solution. She cannot see a distant object — the blackboard — clearly. That is myopia, short-sightedness, and the doctor will advise her to wear spectacles fitted with a concave (diverging) lens of suitable power.

How the correction works. The myopic eye converges light too strongly, so its focus falls short of the retina. The concave lens placed in front diverges the incoming parallel rays slightly before they reach the eye. The eye then has to work on light that is already spreading, and its focus is pushed back — onto the retina, where a sharp image forms.

Equivalently, and this is the way to calculate the power: the concave lens forms a virtual image of the distant object at the eye’s far point, and the eye can focus on anything at its far point. So the required focal length is the far point distance, negative — which is exactly the calculation in Question 1 of the multiple-choice band and in Question 18 below.

Question 17

How are we able to see nearby and also the distant objects clearly?

Solution. Because the eye can change the focal length of its own lens. This ability is called the power of accommodation.

The distance from the lens to the retina is fixed — the image distance cannot change. But the object distance varies enormously, from 25 cm to infinity. The only way to keep the image sharp on the retina throughout that range is to alter the focal length, and the ciliary muscles do it by changing the shape of the flexible crystalline lens.

For distant objects, the ciliary muscles relax. The suspensory ligaments pull the lens thin and flat, so its focal length is long, and the nearly parallel rays are brought to a focus on the retina.

For nearby objects, the ciliary muscles contract. The tension is released and the lens bulges thicker and more curved, so its focal length shortens and the strongly diverging rays from the near object are still converged onto the retina.

There is a limit to how much the lens can thicken, and that limit sets the near point — about 25 cm for a normal eye. There is no corresponding limit at the far end, so the far point for a normal eye is infinity.

This is the fact behind Question 13 of the multiple-choice band, and its failure with age — the lens stiffening so it can no longer bulge — is presbyopia.

Question 18

A person needs a lens of power –4.5 D for correction of her vision. (a) What kind of defect in vision is she suffering from? (b) What is the focal length of the corrective lens? (c) What is the nature of the corrective lens?

Solution.

(a) The power is negative, and a negative power means a diverging lens. A diverging lens is prescribed to correct myopia (short-sightedness) — the defect in which distant objects cannot be seen clearly because the eye focuses them in front of the retina.

(b) From $P = 1/f$ with $f$ in metres,

$$f = \frac{1}{P} = \frac{1}{-4.5} = -0.22\ \mathrm{m} = -22\ \mathrm{cm}$$

(c) A negative focal length means the lens is concave (diverging).

All three parts follow from the single minus sign, which is worth noticing: the sign of the power tells you the nature of the lens, and the nature of the lens tells you the defect.

The focal length also tells us something about the person’s eyesight. Since a myopic correction has $f$ equal to the far point distance, her far point is about 22 cm — she can see clearly only out to about 22 cm without glasses, which is a considerable degree of short sight.

Question 19

How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.

Solution. Place the two prisms so that the second is inverted with respect to the first — the base of one uppermost and the base of the other lowermost — with their refracting faces parallel.

Why it works. The first prism disperses the white light, bending violet most and red least, so a spectrum emerges from it.

The second prism is identical but turned the other way up, so it deviates light in the opposite direction and by exactly the same amount for each colour. Every colour is therefore bent back by precisely the amount the first prism bent it, and the seven beams converge again into a single beam. When all the colours are recombined in their original proportions the result is white light.

This is Newton’s classic experiment, and its importance is what it proves: the prism does not create colour. If the colours were manufactured by the glass, passing the spectrum through a second piece of glass would produce still more colour, not less. That the light comes out white again shows the colours were present in the white light all along, and the prism merely sorted them.

Question 20

Draw a ray diagram showing the dispersion through a prism when a narrow beam of white light is incident on one of its refracting surfaces. Also indicate the order of the colours of the spectrum obtained.

Solution.

What happens. The white beam is refracted as it enters the first face and again as it leaves the second. Glass has a slightly different refractive index for each wavelength — larger for short wavelengths — so each colour is bent through a slightly different angle, and the beam fans out into a band of colours. The band is the spectrum, and the splitting is called dispersion.

The order of the colours, from least deviated to most, is

$$\textbf{R} \ \textbf{O} \ \textbf{Y} \ \textbf{G} \ \textbf{B} \ \textbf{I} \ \textbf{V}$$

red, orange, yellow, green, blue, indigo, violet — conventionally remembered as VIBGYOR read the other way.

Two points fix the diagram and are the ones examiners look for. Red is deviated least and violet most, because violet has the shortest wavelength and the largest refractive index. And all the colours are bent towards the base of the prism, so with the base at the bottom, violet comes out lowest and red highest. That is the fact Question 3 of the multiple-choice band turns on.

Question 21

Is the position of a star as seen by us its true position? Justify your answer.

Solution. No. A star appears slightly higher in the sky than it really is.

The reason is atmospheric refraction. The atmosphere is not uniform: it is denser near the ground and thins out with height, so its refractive index decreases gradually with altitude. Starlight coming in from space therefore passes continuously from rarer layers into denser ones as it descends.

Light entering a denser medium bends towards the normal, so the ray is bent progressively downwards along its path — it follows a gentle curve rather than a straight line. But the eye assumes light has travelled in a straight line, and traces the ray back along its final direction. That backward line points to a position slightly above the star’s true position.

The effect is small but real, and it has two familiar consequences worth adding. The Sun is visible for about two minutes before actual sunrise and two minutes after actual sunset, because its light is bent around the curve of the atmosphere while the Sun itself is still below the horizon. And because the layers are constantly moving, the apparent position wavers — which is the twinkling of Questions 6 and 30.

Question 22

Why do we see a rainbow in the sky only after rainfall?

Solution. Because a rainbow needs water droplets suspended in the air, and those are present only during or just after rain.

Each droplet behaves like a tiny prism. Sunlight entering it is refracted and dispersed into its colours; the separated colours are then internally reflected at the back of the drop; and they are refracted and dispersed again on the way out. The result is that each drop sends a fan of colours back towards the observer, and the drops at different angles send different colours, building the arc we see.

All four steps need the drop. With no droplets there is nothing to disperse the light, and the sky stays its ordinary colour.

Two further conditions are worth stating, because they explain why a rainbow does not appear after every shower. The Sun must be shining — there must be direct sunlight to disperse. And the observer must have the Sun behind them and the rain in front, because the light returns from the drops roughly back towards the direction it came from. That is why rainbows are seen in the eastern sky in the evening and the western sky in the morning.

The same phenomenon appears wherever droplets and sunlight meet — in the spray of a waterfall, or from a garden hose held with the Sun behind you.

Question 23

Why is the colour of the clear sky blue?

Solution. Because of the scattering of sunlight by the molecules of the air, and because that scattering is far stronger for short wavelengths than for long ones.

The molecules of nitrogen and oxygen are very much smaller than the wavelength of visible light. Particles of that size scatter light in a way that depends steeply on wavelength: the shorter the wavelength, the more strongly it is scattered.

Blue and violet have the shortest wavelengths in the visible spectrum, so they are scattered far more than red, orange and yellow. That scattered blue light is redirected all over the sky and reaches our eyes from every direction, not merely from the Sun’s direction. When we look at any part of the sky away from the Sun, what we see is this scattered light — and it is blue.

Two supplementary points complete the answer:

The sky is blue rather than violet, even though violet scatters more still, because sunlight contains less violet and the eye is much less sensitive to it.

And to an astronaut above the atmosphere the sky is black, because there are no molecules to scatter the sunlight. That observation is the clearest proof that the colour comes from the air and not from the Sun.

Question 24

What is the difference in colours of the Sun observed during sunrise/sunset and noon? Give explanation for each.

Solution. At sunrise and sunset the Sun appears reddish; at noon it appears white. The single quantity that changes is the length of the path its light takes through the atmosphere, and both results follow from scattering.

At sunrise and sunset the Sun is near the horizon, so its light travels a long, slanting path through the atmosphere — several times longer than the vertical path. Over that distance, scattering removes most of the short-wavelength light: blue and violet are scattered out of the beam in all directions and never reach the observer. What survives the journey is chiefly the long wavelengths, red and orange, which scatter least. The Sun therefore looks reddish, and so does the sky around it.

At noon the Sun is overhead and its light takes the shortest path through the atmosphere. Comparatively little light of any colour is scattered out of the beam, so all seven colours arrive in nearly their original proportions — and light containing all colours in balance is white.

The point worth emphasising is that this is one mechanism producing two opposite appearances. Nothing about the Sun changes between noon and sunset; only the thickness of air its light must cross.

Long Answer Questions

Question 25

Explain the structure and functioning of Human eye. How are we able to see nearby as well as distant objects?

Solution.

The structure, following the light inwards:

• Cornea — the transparent bulge at the front. Because light passes from air into it, this is where most of the refraction happens, as Question 12 established.
• Aqueous humour — the clear fluid just behind the cornea.
• Iris — the coloured muscular diaphragm. It controls the size of the pupil and therefore how much light enters.
• Pupil — the central opening in the iris. It widens in dim light and narrows in bright light.
• Crystalline lens — a flexible, transparent lens held by ciliary muscles. It provides the fine, adjustable focusing.
• Vitreous humour — the jelly filling the main chamber and keeping the eyeball in shape.
• Retina — the light-sensitive screen at the back, carrying rods (which respond to dim light) and cones (which respond to colour).
• Optic nerve — carries the signals from the retina to the brain.

The functioning. Light is refracted at the cornea and again by the lens, and a real, inverted, diminished image is formed on the retina. The light-sensitive cells convert it into electrical signals, the optic nerve carries these to the brain, and the brain interprets them — including turning the inverted image the right way up.

Seeing near and far. The lens-to-retina distance is fixed, so the eye focuses by changing the focal length of its lens — the power of accommodation of Question 17. Ciliary muscles relax for distant objects, making the lens thin and long-focus; they contract for nearby objects, making it thick and short-focus. This keeps the image sharp on the retina from the near point, about 25 cm, out to infinity.

Question 26

When do we consider a person to be myopic or hypermetropic? Explain using diagrams how the defects associated with myopic and hypermetropic eye can be corrected?

Solution.

A person is myopic when they can see nearby objects clearly but not distant ones. The far point has moved in from infinity to some finite distance, because the eye converges light too strongly — either the eyeball is too long or the lens too curved — and the image of a distant object falls in front of the retina.

A person is hypermetropic when they can see distant objects clearly but not nearby ones. The near point has receded beyond the usual 25 cm, because the eye converges light too weakly — the eyeball too short or the lens too flat — and the image of a near object would fall behind the retina.

Correcting myopia — a concave (diverging) lens. It spreads the incoming parallel rays slightly before they enter the eye, so the eye’s over-strong convergence now brings them to a focus further back, exactly on the retina. Its focal length equals the far point distance, negative; that is the calculation of Question 1 and Question 18.

Correcting hypermetropia — a convex (converging) lens. It converges the rays from a near object a little before they enter the eye, making up the convergence the eye lacks, so the focus moves forward onto the retina. Its power is chosen so that an object at the normal near point of 25 cm produces a virtual image at the person’s own, more distant, near point.

The symmetry is worth stating plainly: myopia over-converges and is corrected by diverging; hypermetropia under-converges and is corrected by converging.

Question 27

Explain the refraction of light through a triangular glass prism using a labelled ray diagram. Hence define the angle of deviation.

Solution.

The path of the ray. A ray strikes the first refracting face at an angle of incidence $i$. Glass is denser than air, so the ray bends towards the normal, entering at the angle of refraction $r_1$.

It travels in a straight line through the glass and meets the second refracting face from the inside at $r_2$. Passing now from denser to rarer, it bends away from the normal and emerges at the angle of emergence $e$.

The crucial difference from a glass slab is that the two faces of a prism are not parallel. In a slab the two bends are equal and opposite and cancel; here they are both in the same rotational sense and add. The emergent ray is therefore not parallel to the incident ray — it has been turned towards the base of the prism.

The angle of deviation is the angle through which that turning amounts to:

The angle of deviation $D$ is the angle between the direction of the incident ray produced forward and the direction of the emergent ray.

It is shown dashed in the diagram, where the incident ray is extended forward and the emergent ray backward until they meet.

Two standard results are worth adding. The angles obey $A + D = i + e$ and $r_1 + r_2 = A$, where $A$ is the angle of the prism. And as $i$ is increased from small values, $D$ first decreases to a minimum — the angle of minimum deviation, where $i = e$ and the ray passes symmetrically through the prism — and then increases again.

Question 28

How can we explain the reddish appearance of sun at sunrise or sunset? Why does it not appear red at noon?

Solution. Both halves are answered by scattering, and the only thing that differs between them is the length of the path the light takes through the atmosphere.

At sunrise and sunset the Sun is close to the horizon, so its light reaches us along a long, slanting path — passing through several times as much air as it does at noon. Scattering by air molecules is far stronger for short wavelengths, so over that long path the blue and violet are almost entirely scattered out of the direct beam and dispersed across the sky. Only the longest wavelengths, red and orange, which scatter least, survive the journey to the observer. The Sun therefore appears reddish.

At noon the Sun is overhead and its light takes the shortest possible path through the atmosphere. Far less light of any colour is scattered out of the beam, so all seven colours arrive in nearly their original proportions — and all the colours together make white.

So the Sun does not “turn red”; the atmosphere removes the blue when there is enough of it in the way, and leaves it alone when there is not.

The same reasoning accounts for the reddish sky near the horizon at those times: the blue scattered out of the direct beam has to go somewhere, and the light reaching us from that part of the sky has itself been reddened.

Question 29

Explain the phenomenon of dispersion of white light through a glass prism, using suitable ray diagram.

Solution. Dispersion is the splitting of white light into its constituent colours on passing through a refracting medium. The ray diagram is the one drawn in Question 20 above.

Why it happens. White light is not a single colour but a mixture of seven, each with its own wavelength. The refractive index of glass is not the same for every wavelength: it is largest for violet (the shortest wavelength) and smallest for red (the longest).

Since the amount of bending at a surface is governed by the refractive index,

$$n = \frac{\sin i}{\sin r}$$

each colour is refracted through a different angle at the same surface. Violet, with the largest $n$, is bent most; red, with the smallest, is bent least.

Why a prism and not a slab. A single refraction does separate the colours, but in a glass slab the second face undoes the first exactly, and the colours recombine on emerging. A prism’s faces are inclined to each other, so the second refraction increases the separation instead of cancelling it. By the time the light leaves the prism the colours have fanned out far enough to be seen as a band — the spectrum.

The order of the spectrum is red, orange, yellow, green, blue, indigo, violet, with red deviated least and violet most, all bent towards the base.

That the colours are in the light and not made by the glass is proved by Question 19: a second, inverted prism recombines them into white.

Question 30

How does refraction take place in the atmosphere? Why do stars twinkle but not the planets?

Solution.

Refraction in the atmosphere. The air is not uniform. Its density is greatest near the ground and falls steadily with height, so the atmosphere behaves like a stack of very many layers of gradually decreasing refractive index. Light coming in from space passes continuously from rarer layers to denser ones and so is bent towards the normal at each step, following a smooth curve instead of a straight line. This is atmospheric refraction.

Because the air is constantly disturbed by wind and by convection currents, the layers are not still — the refractive index along any given line of sight fluctuates from moment to moment.

Why stars twinkle. A star is so distant that it is effectively a point source. All the light we receive from it follows essentially one path, so when the atmosphere along that path shifts, the whole of the star’s light is deflected together. Its apparent position wavers slightly and the amount of light entering the eye rises and falls. That fluctuation is the twinkling.

Why planets do not. A planet is enormously closer, so it appears not as a point but as a small extended source — effectively a collection of very many point sources side by side. Each of those points twinkles, but independently and at random. The brightenings of some coincide with the dimmings of others, and the variations average out. The total light reaching the eye stays nearly constant, so a planet shines with a steady light.

This is a useful practical test: a steady point of light in the night sky is a planet; a flickering one is a star.

Common mistakes

• Mixing up which lens corrects which defect. Myopia over-converges, so it needs a concave lens; hypermetropia under-converges, so it needs a convex one. Questions 16, 18 and 26 all turn on this.
• Forgetting to convert the power to metres in Question 18. $f = 1/P = 1/(-4.5) = -0.22$ m. Writing $-22$ without the unit, or dividing by 4.5 cm, both go wrong.
• Saying the prism “adds” colour, in Questions 20 and 29. The colours are already in the white light; the prism separates them because the refractive index differs with wavelength. Question 19 is the proof.
• Drawing violet at the top of the spectrum. Violet is deviated most, so it emerges nearest the base. With the base at the bottom, violet is lowest.
• Explaining the blue sky by absorption, in Question 23. Absorbed light never reaches the eye. It is scattering — redirection — that fills the whole sky with blue.
• Answering Question 30 with “planets are brighter”. Brightness is not the reason. A planet is an extended source whose many points twinkle independently and average out; a star is a point source.
• Drawing the hypermetropic focus as a real point behind the eyeball, in Question 15. No image actually forms there — the rays are intercepted by the retina first — so the convergence point must be dashed.
• Confusing the angle of deviation with the angle of the prism, in Question 27. $A$ is the angle between the two refracting faces; $D$ is the angle between the incident ray produced and the emergent ray. They are related by $A + D = i + e$.

Practise next

• Exemplar Multiple Choice Questions — fourteen questions on the same material in recognition form, including the prism-orientation question that Question 20 above explains.
• 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 Questions 18 and 26 depend on.
• Textbook activities for the prism and spectrum experiments behind Questions 19, 20 and 29.