NCERT Solution Class 10 science chapter 10 The Human Eye and the Colourful World - Param Himalaya

NCERT Solution Class 10 science chapter 10 The Human Eye and the Colourful World .

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

Solution : The ability of the eye lens to adjust it's focal length is called power of accommodation.

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

Solution : here distance of far point , v = 1.2 m

For viewing distant objects , u = infinity

Focal length of corrective len,

\[\frac{1}{f} = \frac{1}{v}-\frac{1}{u}\]

\[\frac{1}{f}=\frac{1}{(-1.2)}-\frac{1}{\alpha}\]

\[\frac{1}{f}=-\frac{10}{12}-0\]

\[Power\;of\; lens,P=\frac{1}{f}\]

\[=-\frac{10}{12}=-.83D\]

\[P=-.83D (concave\;Lens)\]

Since the power is negative , therefore , the lens must be concave.

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

Solution : For human eye with normal vision , far point  is at infinity and near point is at 25 cm from the eye.

Question 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?

Solution : As the child has difficulty in reading the blackboard , he is suffering from myopia or short sightedness. To correct this defect , he has to use spectacles with concave lens of suitable focal length.

Exercises :

Question 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.

Solution : The property of the eye to adjust the focal length of eye lens is called accommodation. Option (b) Accommodation 

Question 2. The human eye forms the image of an object at its (a) cornea. (b) Iris. (c) Pupil. (d) Retina.

Solution : The human eye forms the image of an object at its retina. Option (d) is correct.

Question 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.

Solution : 

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

Question 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.

Solution : The action of ciliary muscles holding the eye lens changes the focal length of eye lens enabling the eye to focus the image of objects at varying distances . Option (c) ciliary muscles. Is correct.

Question 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?

Solution : 

(i) For distant vision , given f=? , P  = -5.5 D

Using the relation

P= 1/f  or f = 1/p

f  =100 / (-5.5) = -18.2 cm

(ii) For near vision ,  f = ?

P = +1.5 D

Using the relation P = 1/f or f = 1/P

f = 1/1.5 = 0.667m

The focal length of the lens for correcting near vision is 0.667 m.

Question 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?

Solution : The person is suffering from an eye defect called myopia. 

In this defect, the image is formed in front of the retina. 

Hence, a concave lens is used to correct this defect of vision. 

Object distance, u = infinity = Image distance,

v = −80 cm 

Focal length = f 

According to the lens formula,

The far point of a myopic person is 80 cm in front of the eye

Question 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.

Solution : A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.

The convex lens actually creates a virtual image of a nearby object (N’ in the figure) at the near point of vision (N) of the person suffering from hypermetropia. 

hypermetropia

The given person will be able to clearly see the object kept at 25 cm (near point of the normal eye), if the image of the object is formed at his near point, which is given as 1 m.

Object distance, u = −25 cm 

Image distance, v = −1 m = −100 cm

Focal length, f 

Using the lens formula,

. The near point of a hypermetropic eye is 1 m.


A convex lens of power +3.0 D is required to correct the defect.

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

Solution : A normal eye is unable to clearly see the objects placed closer than 25 cm because the ciliary muscles of eyes are unable to contract beyond a certain limit. If the object is placed at a distance less than 25 cm from the eye, then the object appears blurred and produces strain in the eyes.

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

Solution : Since the size of eyes cannot increase or decrease, the image distance remains constant. When we increase the distance of an object from the eye, the image distance in the eye does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed at the retina of the eye.

Question 10. Why do stars twinkle?

Solution : Stars emit their own light and they twinkle due to the atmospheric refraction of light. Stars are very far away from the earth. Hence, they are considered as point sources of light. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star light refracted by the atmosphere comes more towards us, it appears brighter than when it comes less towards us. Therefore, it appears as if the stars are twinkling at night.

Question 11. Explain why the planets do not twinkle.

Solution : Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to earth. Planets can be considered as a collection of a large number of point-size sources of light. The different parts of these planets produce either brighter or dimmer effect in such a way that the average of brighter and dimmer effect is zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.

Question 12 : Why does the sky appear dark instead of blue to an astronaut?

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


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