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.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 2 of 9

1. Q21 Exemplar · Multiple Choice Questions
   Which of ​the following ​phenomena of ​light are ​involved in ​the formation ​of a ​rainbow?

   1. (a)Reflection, refraction and dispersion
   2. (b)Refraction, dispersion and total internal reflection
   3. (c)Refraction, dispersion and internal reflection
   4. (d)Dispersion, scattering and total internal reflection
   Show answer

   (c) Refraction, dispersion and internal reflection

   Step-by-step solution →
2. Q22 Exemplar · Multiple Choice Questions
   Twinkling ​of stars ​is due ​to ​atmospheric

   1. (a)dispersion of light by water droplets
   2. (b)refraction of light by different layers of varying refractive indices
   3. (c)scattering of light by dust particles
   4. (d)internal reflection of light by clouds
   Show answer

   (b) refraction of light by different layers of varying refractive indices

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3. Q23 Exemplar · Multiple Choice Questions
   The clear ​sky ​appears ​blue because

   1. (a)blue light gets absorbed in the atmosphere
   2. (b)ultraviolet radiations are absorbed in the atmosphere
   3. (c)violet and blue lights get scattered more than lights of all other colours by the atmosphere
   4. (d)light of all other colours is scattered more than the violet and blue colour lights by the atmosphere
   Show answer

   (c) violet and blue lights get scattered more than lights of all other colours by the atmosphere

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4. Q24 Exemplar · Multiple Choice Questions
   Which ​of the following statements is ​correct regarding ​the propagation ​of light ​of different colours ​of white ​light ​in air?

   1. (a)Red light moves fastest
   2. (b)Blue light moves faster than green light
   3. (c)All the colours of the white light move with the same speed
   4. (d)Yellow light moves with the mean speed as that of the red and the violet light
   Show answer

   (c) All the colours of the white light move with the same speed

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5. Q25 Exemplar · Multiple Choice Questions
   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

   1. (a)is scattered the most by smoke or fog
   2. (b)is scattered the least by smoke or fog
   3. (c)is absorbed the most by smoke or fog
   4. (d)moves fastest in air
   Show answer

   (b) is scattered the least by smoke or fog

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6. Q26 Exemplar · Multiple Choice Questions
   Which of ​the following phenomena contributes ​significantly to the ​reddish appearance ​of the sun at sunrise ​or ​sunset?

   1. (a)Dispersion of light
   2. (b)Scattering of light
   3. (c)Total internal reflection of light
   4. (d)Reflection of light from the earth
   Show answer

   (b) Scattering of light

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7. Q27 Exemplar · Multiple Choice Questions
   The bluish ​colour ​of ​water ​in ​deep sea is ​due ​to

   1. (a)the presence of algae and other plants found in water
   2. (b)reflection of sky in water
   3. (c)scattering of light
   4. (d)absorption of light by the sea
   Show answer

   (c) scattering of light

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8. Q28 Exemplar · Multiple Choice Questions
   When light ​rays ​enter the eye, most of ​the refraction ​occurs ​at ​the

   1. (a)crystalline lens
   2. (b)outer surface of the cornea
   3. (c)iris
   4. (d)pupil
   Show answer

   (b) outer surface of the cornea

   Step-by-step solution →
9. Q29 Exemplar · Multiple Choice Questions
   The ​focal ​length of the ​eye lens ​increases when ​eye muscles

   1. (a)are relaxed and lens becomes thinner
   2. (b)contract and lens becomes thicker
   3. (c)are relaxed and lens becomes thicker
   4. (d)contract and lens becomes thinner
   Show answer

   (a) are relaxed and lens becomes thinner

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10. Q30 Exemplar · Multiple Choice Questions
    Which of ​the ​following ​statement is ​correct?

    1. (a)A person with myopia can see distant objects clearly
    2. (b)A person with hypermetropia can see nearby objects clearly
    3. (c)A person with myopia can see nearby objects clearly
    4. (d)A person with hypermetropia cannot see distant objects clearly
    Show answer

    (c) A person with myopia can see nearby objects clearly

    Step-by-step solution →
11. Q31 Exemplar · Short Answer Questions
    Draw ​ray diagrams ​each ​showing ​

    1. (i)myopic eye ​and
    2. (ii)hypermetropic ​eye.
    Show answer

    In a myopic eye the image forms in front of the retina; in a hypermetropic eye it forms behind it.

    (i) In a myopic eye the parallel rays from a distant object are brought to a focus at $I$ **in front** of the retina (at $2.27$ cm from the lens against a retina at $2.5$ cm), so the distant object looks blurred and the far point lies at a finite distance instead of at infinity. (ii) In a hypermetropic eye the rays from an object at $25$ cm would meet at $I$ **behind** the retina (at $2.70$ cm), so a near object can never be seen sharply and the near point lies beyond $25$ cm.

    Step-by-step solution →
12. Q32 Exemplar · Short Answer Questions
    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.
    Show answer

    She has myopia; the doctor will advise a concave lens, for example $P=-4$ D for a far point of 25 cm.

    She is suffering from myopia (short-sightedness): her far point has come in from infinity, so the image of the distant blackboard falls in front of her retina. The doctor will advise an eye test and spectacles carrying a concave (diverging) lens of focal length equal to minus the distance of her far point. For a far point of $25$ cm this is $f=-25$ cm $=-0.25$ m, that is a power of $P=-4$ D. The lens forms a virtual image of the blackboard at her far point and the eye then focuses it on the retina, as the ray diagram shows.

    Step-by-step solution →
13. Q33 Exemplar · Short Answer Questions
    How ​are we ​able to see nearby ​and also ​the ​distant objects clearly?
    Show answer

    Accommodation: the ciliary muscles change the focal length of the eye lens, so near and distant objects both focus on the retina.

    Because of the power of accommodation of the eye. The retina stays a fixed $2.5$ cm behind the eye lens, so the ciliary muscles change the curvature, and hence the focal length, of the lens itself. For a distant object the muscles are relaxed, the lens is thin and $f$ is largest, $2.5$ cm ($40$ D), and the parallel rays are focused on the retina. For an object at $25$ cm the muscles contract, the lens becomes thicker and $f$ falls to $2.27$ cm ($44$ D), so the diverging rays are still brought to a focus on the retina. A normal eye therefore sees everything clearly from its near point at $25$ cm to its far point at infinity.

    Step-by-step solution →
14. Q34 Exemplar · Short Answer Questions
    A ​person needs a lens of ​power $-4.5$ D for correction of her vision.

    1. (a)What kind of defect in vision is she suffering from?
    2. (b)What is the ​focal ​length of the corrective ​lens?
    3. (c)What is the nature ​of the corrective ​lens?
    Show answer

    She has myopia (short-sightedness); the corrective lens has $f=-22.2$ cm and is concave (diverging).

    (a) She is suffering from myopia (short-sightedness). (b) $f=\frac{1}{P}=\frac{1}{-4.5\text{D}}=-0.222$ m $=-22.2$ cm. (c) The corrective lens is a concave (diverging) lens.

    Step-by-step solution →
15. Q35 Exemplar · Short Answer Questions
    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.
    Show answer

    The second prism, identical and inverted, undoes the deviation and dispersion of the first, so white light emerges parallel to the incident beam.

    Place the second prism inverted with respect to the first, that is turned through $180^{\circ }$ so that its base is on the opposite side, with its refracting faces parallel to the corresponding faces of the first prism. The first prism disperses the white light into its spectrum; because the second prism is identical and inverted, it deviates every colour through exactly the angle the first gave it but in the opposite sense. The net deviation and the net dispersion are therefore both zero, the colours recombine, and white light emerges parallel to the incident beam. The diagram above shows the arrangement.

    Step-by-step solution →
16. Q36 Exemplar · Short Answer Questions
    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.
    Show answer

    Violet is deviated most and red least, so with the base down the spectrum runs red at the top to violet at the bottom.

    The ray diagram is drawn above. White light incident on the face $AB$ at $45^{\circ }$ is refracted towards the normal on entering, refracted away from the normal on leaving the face $AC$, and so is deviated towards the base of the prism. Because the refractive index of the glass is largest for violet and smallest for red, violet is deviated most ($\delta =40.4^{\circ }$) and red least ($\delta =38.6^{\circ }$), an angular dispersion of $1.8^{\circ }$. With the base of the prism at the bottom, the order of the spectrum read from the **top** downwards is red, orange, yellow, green, blue, indigo, violet, that is VIBGYOR read from the bottom up.

    Step-by-step solution →
17. Q37 Exemplar · Short Answer Questions
    Is the position ​of a star ​as seen by us its ​true position? Justify ​your ​answer.
    Show answer

    No; atmospheric refraction bends starlight, so the star appears slightly higher than its true position.

    No. Because of atmospheric refraction the light of a star does not reach us in a straight line. The density, and hence the refractive index, of the air increases as we come down towards the earth, so starlight entering obliquely passes continuously from rarer to denser layers and keeps bending towards the vertical normal, following a curved path. We see the star along the tangent to that path at the eye, which is a steeper direction than the true one, so the star appears slightly **higher** (nearer the zenith) than its actual position. In the worked model the zenith angle falls from $62^{\circ }$ to $50^{\circ }$. The effect vanishes overhead and is greatest near the horizon; it is also why the sun is seen about two minutes before actual sunrise and two minutes after actual sunset.

    Step-by-step solution →
18. Q38 Exemplar · Short Answer Questions
    Why ​do ​we ​see a rainbow in ​the ​sky ​only after ​rainfall?
    Show answer

    Rainbows need water droplets in the air, which exist only just after rain; each droplet disperses sunlight like a tiny prism.

    Because a rainbow needs tiny water droplets suspended in the air, and those are present only just after a shower. In each droplet sunlight is refracted and dispersed as it enters, then partly reflected at the far surface (an internal reflection: the ray meets that surface at about $40^{\circ }$, below the critical angle of about $49^{\circ }$ for water, so the reflection is not total), then refracted and dispersed again as it leaves, so the droplet acts as a tiny prism. Tracing the ray gives about $42.4^{\circ }$ for red and $40.6^{\circ }$ for violet to the incoming sunlight, so all the drops lying at that angle deliver colour to the eye and we see a coloured arc, always in the part of the sky opposite the sun, with red on the outside. Without the droplets none of this can happen, so no rainbow is seen. The same effect appears in the spray of a waterfall or a fountain, where droplets are present without any rain.

    Step-by-step solution →
19. Q39 Exemplar · Short Answer Questions
    Why ​is ​the ​colour ​of the ​clear ​sky blue?
    Show answer

    Air molecules scatter short wavelengths about nine times as strongly as long ones, so scattered blue light reaches the eye from the whole sky.

    Because of the scattering of sunlight by the molecules of the air. Sunlight is a mixture of all colours, and for particles much smaller than the wavelength of light the scattered intensity varies as $1/{\lambda }^4$. Taking red as $700$ nm and violet as $400$ nm, $(700/400{)}^4=9.4$, so the short wavelengths are scattered about nine times as strongly as the long ones. This scattered blue light reaches the eye from every direction of the sky, and since sunlight carries more blue than violet and the eye is far more sensitive to blue, the clear sky is seen as blue. If the earth had no atmosphere there would be nothing to scatter the light and the sky would look dark.

    Step-by-step solution →
20. Q40 Exemplar · Short Answer Questions
    What ​is the ​difference in ​colours of the Sun observed ​during sunrise/sunset ​and noon? ​Give explanation ​for each.
    Show answer

    At sunrise and sunset the Sun looks red because its long path scatters out the blue; at noon the short path leaves it white.

    At sunrise and sunset the Sun appears **red** or reddish-orange, while at noon it appears **white**. At sunrise and sunset the Sun is near the horizon, so its light must travel through a far greater thickness of the atmosphere, roughly $\sqrt{2R/H}\approx 39$ times the overhead path. Since scattering varies as $1/{\lambda }^4$, and blue is scattered about nine times as strongly as red, nearly all the blue and violet is scattered out of the beam on that long journey and mainly red light survives to reach the eye. At noon the Sun is almost overhead and its light crosses the least thickness of atmosphere, so little of any colour is scattered away, all the colours arrive in nearly their original proportions and the Sun looks white.

    Step-by-step solution →