Chapter 10 · The Human Eye and the Colourful World

NCERT Class 10 Science Chapter 10 Exercises

All 12 questions of NCERT Class 10 Science Chapter 10 Exercises (The Human Eye and the Colourful World), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   The ​human ​eye can focus on ​objects at different distances by adjusting the focal length of the ​eye lens. This ​is ​due ​to

   1. (a)presbyopia.
   2. (b)accommodation.
   3. (c)near-sightedness.
   4. (d)far-sightedness.
   Show answer

   (b) accommodation.

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2. Q2
   The ​human eye forms the ​image ​of an ​object at ​its

   1. (a)cornea.
   2. (b)iris.
   3. (c)pupil.
   4. (d)retina.
   Show answer

   (d) retina.

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3. Q3
   The ​least ​distance of distinct vision for ​a ​young ​adult with ​normal vision is ​about

   1. (a)$25$ m.
   2. (b)$2.5$ cm.
   3. (c)$25$ cm.
   4. (d)$2.5$ m.
   Show answer

   (c) $25$ cm.

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4. Q4
   The change in focal length ​of an eye ​lens ​is caused by the ​action of the

   1. (a)pupil.
   2. (b)retina.
   3. (c)ciliary muscles.
   4. (d)iris.
   Show answer

   (c) ciliary muscles.

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5. Q5
   A person needs a lens of power $-5.5$ dioptres for correcting his distant vision. ​For ​correcting ​his near vision he needs a ​lens of power $+1.5$ dioptre. What is the focal length of the ​lens required for correcting

   1. (i)​distant vision, and
   2. (ii)​near vision?
   Show answer

   The focal lengths are $-0.18$ m for distant vision (concave lens) and $+0.67$ m for near vision (convex lens).

   (i) For distant vision, $f_1=\frac{1}{P_1}=\frac{1}{-5.5}=-0.18$ m (about $-18$ cm), a concave lens; (ii) For near vision, $f_2=\frac{1}{P_2}=\frac{1}{+1.5}=+0.67$ m (about $+67$ cm), a convex lens. Both values are rounded to two decimal places, and the focal lengths are in metres because power is measured in dioptres, where $1$ D $=1$ m${}^{-1}$.

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6. Q6
   The ​far point ​of a ​myopic person is $80$ ​cm in front of ​the eye. What is the ​nature and power of the ​lens required to correct the problem?
   Show answer

   A concave (diverging) lens of power $-1.25$ D is required.

   A concave (diverging) lens of power $-1.25$ D is required. The lens must form, at the eye's far point $80$ cm away, the image of an object at infinity, so $u=-\infty$ and $v=-80$ cm. The lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ then gives $\frac{1}{f}=-\frac{1}{80}$, that is $f=-80$ cm $=-0.8$ m, and $P=\frac{1}{f}=\frac{1}{-0.8}=-1.25$ D. The negative sign shows that the lens is concave.

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7. Q7
   Make ​a diagram to show how hypermetropia ​is corrected. ​The near point of a hypermetropic eye is $1$ m. What is ​the power of the lens required to ​correct ​this defect? Assume that the near point of the ​normal eye is $25$ cm.
   Show answer

   A convex lens of power $+3.0$ D corrects the defect.

   A convex (converging) lens of power $+3.0$ D is required. The lens must form, at the eye's near point of $100$ cm, a virtual image of an object placed at the normal near point of $25$ cm, so $u=-25$ cm and $v=-100$ cm. The lens formula gives $\frac{1}{f}=\frac{1}{-100}-\frac{1}{-25}=\frac{-1+4}{100}=\frac{3}{100}$, so $f=+\frac{100}{3}=+33.3$ cm $=+0.333$ m and $P=\frac{1}{f}=+3.0$ D. The image is erect and four times as tall as the object, and the diagram above shows the construction: the ray parallel to the axis refracted through $F^{\prime }$, the ray through the optical centre undeviated, and their backward extensions meeting at the $100$ cm mark.

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8. Q8
   Why ​is a normal ​eye ​not ​able to see ​clearly ​the objects placed ​closer than $25$ cm?
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   The lens cannot shorten its focal length below $2.27$ cm, so for objects nearer than $25$ cm the image falls behind the retina.

   Because the eye has run out of accommodation. The retina is fixed about $2.5$ cm behind the eye lens, so the image distance cannot change; the only way to focus a nearer object is to shorten the focal length of the lens, and the ciliary muscles do that by making the lens more curved. From $\frac{1}{f}=\frac{1}{2.5}+\frac{1}{d}$, an object at $25$ cm needs $f=2.27$ cm, which is the shortest focal length a normal eye lens can reach. An object at $10$ cm would need $f=2.00$ cm, which the lens cannot supply, so with $f$ stuck at $2.27$ cm the image is formed $2.94$ cm behind the lens — behind the retina. The rays are still converging as they cross the retina, so a blurred patch falls on it and the object cannot be seen clearly. That is why $25$ cm is called the least distance of distinct vision.

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9. Q9
   What happens to the ​image distance in the ​eye when ​we increase ​the distance of ​an object from ​the ​eye?
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   The image distance stays constant at about 2.5 cm because the retina is fixed; the eye lens changes its focal length instead.

   The image distance in the eye does not change: it stays at about $2.5$ cm, the fixed distance from the eye lens to the retina. Because the retina cannot move, the image must always be formed on it, so in $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ the quantity $v$ is held constant and the **focal length** of the eye lens changes instead. As the object is moved further away, the ciliary muscles relax, the eye lens becomes thinner and less curved, and its focal length increases — from about $2.27$ cm for an object at $25$ cm to about $2.5$ cm for a very distant one. This adjustment is called accommodation. What does get smaller as the object recedes is the **size** of the image on the retina, not its distance.

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10. Q10
    Why do stars twinkle?
    Show answer

    Atmospheric refraction by moving air layers keeps changing the light from a point-like star, so it seems brighter and fainter by turns.

    Stars twinkle because of atmospheric refraction. The earth's atmosphere is made of layers of air whose density increases towards the ground, so light from a star is refracted continuously as it comes down, bending towards the normal and following a curved path; this is also why a star seen near the horizon appears a little higher in the sky than it really is. The layers of air are constantly moving and changing density, so the amount of bending keeps changing, and with it the path the starlight takes and the amount of it that enters the eye. Because a star is so far away that it acts as a single point source, all its light shifts together, so the star seems brighter at one moment and fainter the next. This continual variation in brightness is the twinkling of stars.

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11. Q11
    Explain ​why ​the planets do ​not twinkle.
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    A planet is a disc of many point sources whose independent flickers average out, so its light stays steady.

    Planets do not twinkle because they are extended sources of light, not point sources. A planet is millions of times nearer to us than any star, so it subtends a measurable angle at the eye and is seen as a small disc, which can be treated as a very large number of point sources lying side by side. The atmosphere refracts the light from each of those points along a path that keeps changing, exactly as it does for a star, so each point does flicker. But the flickers are independent of one another, so at any instant the points that are momentarily brighter are balanced by those that are momentarily dimmer. The total light the eye receives from the whole disc therefore stays very nearly constant, and the planet shines steadily. A star, being effectively a single point source, has nothing to average its fluctuations against, so it twinkles.

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12. Q12
    Why does ​the sky ​appear dark ​instead ​of ​blue ​to ​an ​astronaut?
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    There is no atmosphere above the astronaut to scatter sunlight, so no scattered blue light reaches the eye and the sky looks dark.

    Because there is no atmosphere above the astronaut to scatter sunlight. The sky looks blue from the ground only because sunlight passing through the air is scattered by its molecules, and since light of shorter wavelength is scattered much more strongly than light of longer wavelength, it is mainly blue light that is thrown sideways and reaches the eye from every direction of the sky. An astronaut is above practically all the air, so there are no molecules to scatter the sunlight and no scattered light arrives from the empty directions. Light reaches the astronaut only directly from the Sun and from lit bodies such as the earth, the moon and the stars, so the rest of the sky appears completely dark. There is no shortage of sunlight up there — it is stronger than at the ground — only a shortage of anything to scatter it.

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