Chapter 11 · The Human Eye and the Colourful World

NCERT Exemplar Class 10 Science Chapter 11 Short Answer Questions

All 10 questions of NCERT Exemplar Class 10 Science Chapter 11 Short Answer Questions (The Human Eye and the Colourful World), in the book's order, each with its answer and a step-by-step solution.

1. Q15
   Draw ​ray diagrams ​each ​showing ​

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

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2. Q16
   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.
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   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.

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3. Q17
   How ​are we ​able to see nearby ​and also ​the ​distant objects clearly?
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   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.

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4. Q18
   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?
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   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.

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5. Q19
   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.
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   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.

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6. Q20
   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.
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   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.

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7. Q21
   Is the position ​of a star ​as seen by us its ​true position? Justify ​your ​answer.
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   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.

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8. Q22
   Why ​do ​we ​see a rainbow in ​the ​sky ​only after ​rainfall?
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   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.

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9. Q23
   Why ​is ​the ​colour ​of the ​clear ​sky blue?
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   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.

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10. Q24
    What ​is the ​difference in ​colours of the Sun observed ​during sunrise/sunset ​and noon? ​Give explanation ​for each.
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    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.

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