Class 10 Science · Chapter 10

The Human Eye and the Colourful World

163 Class 10 Science questions on The Human Eye and the Colourful World from the NCERT textbook, NCERT Exemplar and CBSE board papers 2023–2026, each with its answer and a step-by-step solution.

NCERT exercises

Exercises (12) Intext Questions (Page 164) (4)

NCERT Exemplar exercises

Multiple Choice Questions (14) Short Answer Questions (10) Long Answer Questions (6)

CBSE Class 10 Science previous year papers, with answers →

Questions · page 3 of 9

Q41 Exemplar · Long Answer Questions

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

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The eye forms a real, inverted image on the retina; the ciliary muscles change the lens focal length to see near and distant objects.
The eye is a nearly spherical ball about $2.3$ cm across. Light enters through the transparent **cornea**, where most of the refraction takes place; it crosses the **aqueous humour**, passes through the **pupil** whose size is controlled by the coloured **iris**, and is finely focused by the **crystalline lens**, whose curvature is set by the **ciliary muscles**. It then crosses the **vitreous humour** and forms a real, inverted, diminished image on the **retina**, whose rods and cones send signals along the **optic nerve** to the brain; where that nerve leaves the eyeball is the **blind spot**, which has no receptors. We can see near and distant objects alike because of the power of accommodation: the retina is fixed $2.5$ cm behind the lens, so the eye varies the focal length of the lens instead. For a distant object the ciliary muscles are relaxed, the lens is thin and $f = 2.50$ cm ( $40$ D); for an object at the near point they contract, the lens thickens and $f = 2.27$ cm ( $44$ D). A normal eye therefore sees clearly from $25$ cm out to infinity.

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Q42 Exemplar · Long Answer Questions

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?

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Myopia, with the image in front of the retina, needs a concave lens; hypermetropia, with the image behind it, needs a convex lens.
A person is **myopic** when he can see nearby objects clearly but not distant ones: his far point has come in from infinity, so the image of a distant object falls in **front** of the retina (here at $2.27$ cm against a retina at $2.5$ cm). It is corrected by a **concave** (diverging) lens whose focal length equals minus the far-point distance, which for a far point of $25$ cm is $f = - 25$ cm, that is $P = - 4$ D; the lens forms a virtual image of the distant object at the far point and the eye then focuses it on the retina. A person is **hypermetropic** when he can see distant objects clearly but not nearby ones: his near point lies beyond $25$ cm, so the image of a book at $25$ cm would fall **behind** the retina (here at $2.70$ cm). It is corrected by a **convex** (converging) lens which forms a virtual image of the book at his own near point; for a near point of $1$ m this needs $f = + 33.3$ cm, that is $P = + 3$ D.

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Q43 Exemplar · Long Answer Questions

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

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The ray bends towards the normal at the first face and away at the second, turning towards the base; $δ = i + e - A$ .
A ray $P Q$ falling on the face $A B$ of the prism $A B C$ passes from air into the denser glass at $Q$ and so bends **towards** the normal: $sin i = n sin r_{1}$ gives $r_{1} = 28. 1^{\circ}$ for $i = 4 5^{\circ}$ and $n = 1.50$ . The geometry of the prism forces $r_{1} + r_{2} = A$ , so $r_{2} = 31. 9^{\circ}$ . At $R$ on the face $A C$ the ray passes from glass into the rarer air and bends **away** from the normal: $n sin r_{2} = sin e$ gives $e = 52. 4^{\circ}$ . Both bends turn the ray in the same sense, towards the base, so the total turning is $δ = (i - r_{1}) + (e - r_{2}) = i + e - A = 4 5^{\circ} + 52. 4^{\circ} - 6 0^{\circ} = 37. 4^{\circ}$ . The **angle of deviation** is defined from this figure as the angle between the incident ray produced forward and the emergent ray produced backward, that is the angle through which the prism turns the ray away from its original direction, always towards the base.

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Q44 Exemplar · Long Answer Questions

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

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The long slant path at sunrise and sunset scatters out the blue, leaving red; at noon the short path leaves the Sun white.
The Sun sends the same white light all day; what changes is the thickness of atmosphere its light must cross. At sunrise and sunset the Sun is near the horizon, so the light travels a slant path about $2 R / H \approx 39$ times longer than the overhead path. Because scattering by air molecules varies as $1/ λ^{4}$ , and $(700/400)^{4} = 9.4$ , the short-wavelength blue and violet is scattered sideways out of the beam far more strongly than red. Over so long a path nearly all of it is removed, so the light that still travels straight to the eye is mainly red, and the Sun looks reddish. At noon the Sun is almost overhead and the light crosses the least thickness of atmosphere; comparatively little of any colour is scattered away, all the colours reach the eye in nearly their original proportions, and the Sun appears white.

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Q45 Exemplar · Long Answer Questions

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

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White light splits into seven colours because glass has a different refractive index for each, so violet is deviated most and red least.
Dispersion is the splitting of white light into its seven constituent colours. It happens because white light is a mixture and the refractive index of the glass is different for each colour, being largest for violet ( $n = 1.532$ ) and smallest for red ( $n = 1.513$ ). Since a prism deviates a ray by $δ = i + e - A$ and both $r_{1}$ and $e$ depend on $n$ , each colour is turned through a different angle. Tracing a beam incident at $4 5^{\circ}$ on a $6 0^{\circ}$ prism gives $δ_{red} = 38. 6^{\circ}$ and $δ_{violet} = 40. 4^{\circ}$ , an angular dispersion of $1. 8^{\circ}$ . Violet, being deviated most, emerges nearest the base and red, deviated least, furthest from it, so with the base down the band on the screen runs red, orange, yellow, green, blue, indigo, violet from the top downwards. That band is the spectrum, and Newton's inverted second prism recombines it into white light, proving the prism only separates colours that were already present.

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Q46 Exemplar · Long Answer Questions

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

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Air density increases downwards, so light bends along a curved path; stars twinkle as point sources, while a planet's many points average out.
The atmosphere is not uniform: its density, and hence its refractive index, increases steadily downwards. Applying Snell's law layer by layer gives $n sin z =$ constant, so as light descends into denser air it bends **towards** the vertical and travels along a curved path; in the worked model the zenith angle falls from $6 2^{\circ}$ to $5 0^{\circ}$ . We see a body along the tangent to that path at the eye, so objects near the horizon appear raised, and the sun is seen about two minutes before actual sunrise and two minutes after actual sunset. **Stars twinkle** because a star is so distant that it is a point source: the layers of air are constantly moving, so the bending keeps changing, the apparent position of that point shifts and the amount of starlight entering the eye keeps varying. **Planets do not twinkle** because a planet is far nearer and is seen as a small disc, that is as a large collection of point sources. Each point flickers independently, and the total light entering the eye is their sum, so the variations average out and the planet shines steadily.

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Q47 CBSE 2026
The pair of eye parts responsible for admitting different amount of light in to the eyes is
1. (a)Iris and pupil
2. (b)Ciliary muscles and pupil
3. (c)Retina and Iris
4. (d)Ciliary muscles and cornea
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(a) Iris and pupil

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Q48 CBSE 2026
The vision defect which arises due to gradual weakening of the ciliary muscles and diminishing flexibility of eye lens is :
1. (a)Myopia
2. (b)Hypermetropia
3. (c)Presbyopia
4. (d)Myopia and hypermetropia both
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(c) Presbyopia

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Q49 CBSE 2026
When you look at an object very close to your eyes, the :
1. (a)Ciliary muscles of your eye contract and the eye lens becomes thick.
2. (b)Ciliary muscles of your eye get relaxed and the eye lens becomes thick.
3. (c)Ciliary muscles of your eye contract and the eye lens becomes thin.
4. (d)Ciliary muscles of your eye get relaxed and the eye lens becomes thin.
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(a) Ciliary muscles of your eye contract and the eye lens becomes thick.

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Q50 CBSE 2026
A thin beam of white light falls on a face of a glass prism and a spectrum is obtained on a screen. Which of the following options is the correct sequence of colours from top to bottom on the screen ?
1. (a)Red, Green, Yellow, Violet, Blue
2. (b)Red, Yellow, Green, Blue, Violet
3. (c)Violet, Blue, Green, Yellow, Red
4. (d)Violet, Red, Green, Yellow, Blue
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(b) Red, Yellow, Green, Blue, Violet

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Q51 CBSE 2026
Observe the given ray diagram showing refraction through a prism and choose the correct option for colour of rays 1 and 2 :
1. (a)Colour of ray 1: Blue; Colour of ray 2: Red
2. (b)Colour of ray 1: Green; Colour of ray 2: Yellow
3. (c)Colour of ray 1: Red; Colour of ray 2: Violet
4. (d)Colour of ray 1: Violet; Colour of ray 2: Red
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(c) Colour of ray 1: Red; Colour of ray 2: Violet

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Q52 CBSE 2026
Vision defect far-sightedness is found in the eye defects named as :
1. (a)Myopia and cataract
2. (b)Myopia and hypermetropia
3. (c)Presbyopia and hypermetropia
4. (d)Presbyopia and myopia
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(c) Presbyopia and hypermetropia

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Q53 CBSE 2026
To restore clear vision in persons whose size of the eye ball has reduced, he/she is suggested to use suitable
1. (a)Converging lens
2. (b)Diverging lens
3. (c)Bifocal lens
4. (d)Cylindrical lens
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(a) Converging lens

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Q54 CBSE 2026
Angle of deviation in refraction of light through a glass prism is the :
1. (a)Angle between the normal at the face on the emerging side of the prism and the emergent ray.
2. (b)Angle between the normal at the face on the emerging side of the prism and the incident ray.
3. (c)Angle between the incident ray and the refracted ray.
4. (d)Angle between the directions of incident ray and emergent ray.
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(d) Angle between the directions of incident ray and emergent ray.

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Q55 CBSE 2026
The part of eye which causes change in the focal length of eye lens to provide it exceptional accommodation power is :
1. (a)Aqueous humour
2. (b)Ciliary muscles
3. (c)Pupil
4. (d)Iris
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(b) Ciliary muscles

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Q56 CBSE 2026
Directions : For question number 32, two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to this question from the codes (a), (b), (c) and (d) as given below :
Assertion (A): When rays of white light pass through a prism, on emerging they give spectrum of seven colours.
Reason (R): It is due to the scattering of light that red light bends minimum and violet light bends the maximum.
1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (c)Assertion (A) is true, but Reason (R) is false.
4. (d)Assertion (A) is false, but Reason (R) is true.
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(c) Assertion (A) is true, but Reason (R) is false.

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Q57 CBSE 2026
What is meant by scattering of light ? How do our observations differ when white light (sun light) is scattered from
1. (i)very fine particles of the medium and
2. (ii)very large particles of the medium ?
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Scattering sends light off in different directions; very fine particles give blue scattered light, very large particles give white.
Scattering of light is the phenomenon in which a part of the light falling on the particles of a medium - molecules, dust, smoke or water droplets - is redirected (sent out again) in different directions, away from its original straight path. (i) Very fine particles: they scatter mainly the shorter wavelengths, so the scattered light appears **blue**. This is why the clear sky looks blue. (ii) Very large particles: they scatter all wavelengths almost equally, so the scattered light appears **white**. This is why clouds, fog and mist look white.

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Q58 CBSE 2026

How does the focal length of eye lens change as the distance of the object from the eyes is altered ? Explain.

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The focal length of the eye lens decreases for nearer objects and increases for farther ones, so the image stays on the retina.
The focal length of the eye lens decreases as the object is brought nearer and increases as the object moves farther away. The image must always form on the retina, whose distance from the lens is fixed, so the eye adjusts the lens itself (power of accommodation). For a distant object the ciliary muscles relax, the lens becomes thin and its focal length increases; for a near object the ciliary muscles contract, the lens becomes thicker and its focal length decreases, so the image of the near object still falls on the retina.

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Q59 CBSE 2026
What happens when light is scattered from
1. (i)very fine particles and
2. (ii)very large size particles of medium ? Justify your answer in each case.
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Very fine particles scatter mainly blue light; very large particles scatter all wavelengths almost equally, so the light appears white.
(i) Very fine particles (much smaller than the wavelength of light, such as air molecules) scatter the shorter wavelengths far more than the longer ones, so mainly blue light is scattered and the scattered light appears blue - the reason the clear sky is blue. Justification: red light has a wavelength about $1.8$ times that of blue, and fine particles scatter short wavelengths much more strongly. (ii) Very large particles (larger than the wavelength, such as the water droplets of clouds) scatter all wavelengths nearly equally, so the scattered light contains all the colours and appears white - the reason clouds look white. Justification: no colour is favoured, and a mixture of all colours in their original proportions is white.

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Q60 CBSE 2026
1. (i)How does the change in curvature of the eye lens helps us in the process of seeing the nearby objects clearly ?
2. (ii)State the range of the power of accommodation of a normal human eye.
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Contracting ciliary muscles thicken the lens, shortening its focal length for near objects; normal range is about 25 cm to infinity.
(i) To see a nearby object clearly the ciliary muscles contract, which increases the curvature of the eye lens so that it becomes thicker. A more curved lens has a shorter focal length (greater power), so it converges the steeply diverging light coming from the nearby object enough to form a sharp image on the retina, whose distance from the lens cannot change. This adjustment of focal length is called accommodation; for a distant object the muscles relax, the lens thins and its focal length increases. (ii) A normal human eye can see objects clearly from its near point, about $25$ cm, to its far point at infinity; over this range the power of its lens changes by about $4$ D.

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