NCERT Solutions for Class 10 Science Chapter 10 - The Human Eye and The Colorful World

Topic 10.2 - Defects of vision and their correction

Q. 1. What is meant by the power of accommodation of the eye?

Answer:

The power of accommodation is the ability of an eye to focus near and far objects clearly and make the image on the retina by adjusting its focal length. However, the focal length of the eye lens cannot be decreased below a certain minimum limit.

Q.2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type the corrective lens used to restore proper vision?

Answer:

The correction for a myopic eye should be the concave lens (negative power) to restore proper vision.

Q.3. What is the far point and near point of the human eye with normal vision?

Answer:

The far point of the human eye with normal vision is at infinity and the near point is 25cm distance from the eye.

Q.4. A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Answer:

Since the student has difficulty reading the blackboard while sitting in the last row, he is most likely to suffer from myopia (short-sightedness), i.e., he can see objects clearly but is unable to see the far objects properly.

Hence, it can be corrected by placing the spectacles with concave lenses of appropriate power.

The Human Eye and The Colorful World Exercise Solutions

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

(a) presbyopia
(b) Accommodation
(c) near-sightedness
(d) far-sightedness

Answer:

It is due to the accommodation of the eye. The human eye can adjust its focal length according to the different distances. Therefore, the option (b) is correct.

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

(a) cornea

(b) iris

(c) pupil

(d) retina

Answer:

The human eye forms the image of an object at its retina. Therefore, the correct option is (d) Retina

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

(a) 25 m

(b) 2.5 cm

(c) 25 cm

(d) 2.5 m

Answer:

The least distance of distinct vision for a young adult with normal vision is about (c) 25 cm

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

(a) pupil

(b) retina

(c) ciliary muscles

(d) iris

Answer:

The change in focal length of an eye-lens is caused by the action of the :(c) ciliary muscles

Q.5. 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 (i) distant vision, and (ii) near vision?

Answer:

For the distant vision:

The power of the lens, given $P=-5.5\mathrm{D}$

$\mathrm{Power}(\text{ in }D)=\frac{1}{f(m)}$

Hence, the focal length will be:

$f=\frac{100}{-5.5}=-18.2\mathrm{cm}$

Therefore, the focal length of the lens required for correcting distant vision will be: -18.2 cm (Here, the negative sign of focal length tells us that it is a concave lens).

For near vision:
The power of the lens, given $P=+1.5\mathrm{D}$

$\mathrm{Power}(\text{ in }D)=\frac{1}{f(m)}$

Hence the focal length will be:

$f=\frac{100}{+1.5}=66.66\mathrm{cm}$

Therefore, the focal length of the lens required for correcting near vision will be: +66.7 cm (Here, the positive sign of focal length tells us that it is a convex lens).

Q.6. 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?

Answer:

The myopic person is suffering from near-sightedness which can be corrected by putting the concave lens in spectacles.

Given the far point up to which a myopic person can see is 80cm in front of the eye.

Here, this person can see the distant object (kept at infinity) clearly if the image of this distant object is formed at his far point.

So, we have

The object distance, $u=\mathrm{\infty }$ (in finity $)$
The image distance, $v=-80\mathrm{cm}$ (far point, in front of the lens) and the focal length, $f=$ to find.

Hence substituting the values in the lens formula: we obtain

$\begin{array}{rl}& \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ & \Rightarrow \frac{1}{-80}-\frac{1}{\mathrm{\infty }}=\frac{1}{f}\\ & \Rightarrow \frac{1}{f}=\frac{1}{-80}\end{array}$

$\Rightarrow f=-80\mathrm{cm}$
Therefore, the focal length of the required concave lens will be 80 cm
Now, Power,

$\text{ Power }=\frac{1}{f(m)}=\frac{1}{-0.8}=-1.25\mathrm{D}$

Hence the power of the concave lens required is -1.25 D

Q.7. 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.

Answer:

Given the near point for the hypermetropic eye is 1m.

Assume the near point of the normal eye is 25 cm.

So,

The object distance will be, $u=-25\mathrm{cm}$ (Normal near the point)
The image distance, $v=-1m$ (Near the point of this defective eye) or -100 cm
Then the focal length can be found from the lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Substituting the values in the equation, we obtain

$\begin{array}{rl}& \Rightarrow \frac{1}{-100}-\frac{1}{-25}=\frac{1}{f}\\ & \Rightarrow \frac{-1+4}{100}=\frac{1}{f}\\ & \Rightarrow \frac{3}{100}=\frac{1}{f}\\ & \Rightarrow f=\frac{100}{3}=33.3\mathrm{cm}\text{ or }0.333\mathrm{m}\end{array}$

Hence the power of the lens will be:

$P=\frac{1}{f(m)}=\frac{1}{+0.333\mathrm{m}}=+3.0\mathrm{D}$

Thus, the power of the convex lens required will be +0.3 D

Q.8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer:

A normal eye is not able to see clearly the objects placed closer than 25 cm because the ciliary muscles have a limit of contraction and relaxation, to see closer objects, ciliary muscles, must decrease the focal length of the eye lens. However, the focal length of the eye lens cannot be decreased below a certain minimum limit. Due to this, our eyes are not able to see objects clearly when placed closer than 25cm.

Q.9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer:

There is no change to the image distance in the eye when we increase the distance of an object from the eye. The image is always formed in the retina of our eye.

Q.10. Why do stars twinkle?

Answer:

Stars twinkle because of turbulence in the atmosphere of the Earth.

The light from the star is refracted in different directions because of the variable refractive index. This seems to us like the position of stars to be changed slightly and also brightness and position, Hence, the twinkling of stars happens.

Q.11. Explain why the planets do not twinkle.

Answer:

Planets do not twinkle the way stars do, the reason is that stars are so far away that they appear to be a point source of light on the sky, while planets have finite size and the size of a planet "averages out" the turbulent effects of the atmosphere.

Therefore, we see a relatively stable image of the eye. Hence, twinkling is not observed on planets.

Q.12. Why does the Sun appear reddish early in the morning?

Answer:

At sunrise and sunset, the sun is closer to the horizon. The sunlight near the horizon passes through denser layers of the air and covers a larger distance before reaching our eyes. Most of the blue light gets scattered. The light that reaches our eyes is of longer wavelengths, mainly orange and red. That is why the sun appears red at sunrise and at sunset.

Q.13. Why does the sky appear dark instead of blue to an astronaut?

Answer:

The dark colour is nothing but the absence of light.

The sky appears dark instead of blue to an astronaut because there is no atmosphere in outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reaches the eyes of the astronauts the sky appears black to them.