NCERT Solutions Class 10 for Science Chapter 11 Human Eye annd Colorful World : In this post, we will share with you all the detailed NCERT Solutions of Class 10 Science Chapter 11 Human Eye and Colorful World. This will contain both in-text and back-exercise questions for Science and Social Science, and all exercise questions for Mathematics. For all school and board level examinations, doing all the NCERT Questions is a must.

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Class 10 Science Chapter 11 Human Eye and Colorful World

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In-Text Questions (Page 190)

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

Answer 1:
The ability of the lens of the eye to adjust its focal length to clearly focus rays coming from distant as well from a near objects on the retina, is known as the power of accommodation of the eye.

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

Answer 2:
An individual with a myopic eye should use a concave lens of focal length 1.2 m so that he or she can restore proper vision.

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

Answer 3:
The minimum distance of the object from the eye, which can be seen distinctly without strain is called the near point of the eye. For a normal person’s eye, this distance is 25 cm.
The far point of the eye is the maximum distance to which the eye can see objects clearly. The far point of a normal person’s eye is infinity.

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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?

Answer 4:
The student is suffering from short-sightedness or myopia. Myopia can be corrected by the use of concave or diverging lens of an appropriate power.

Exercise Questions (Page 197-198)

Question 1:
The human eye can focus 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 1:
(b) accommodation
Due to accommodation the human eye can focus objects at different distances by adjusting the focal length of the eye lens.

Question 2:
The human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina

Answer 2:
(d) retina
The retina is the layer of nerve cells lining the back wall inside the eye. This layer senses light and sends signals to the brain so you can see.

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

Answer 3:
(c) 25 cm
25 cm is the least distance of distinct vision for a young adult with normal vision.

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

Answer 4:
(c) ciliary muscles
The action of the ciliary muscles changes the focal length of an eye lens

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?

Answer 5:
The power (P) of a lens of focal length f is given by the relation
Power (P) = 1/f

(i) Power of the lens (used for correcting distant vision) = – 5.5 D
Focal length of the lens (f) = 1/P
f = 1/-5.5
f = -0.181 m
The focal length of the lens (for correcting distant vision) is – 0.181 m.

(ii) Power of the lens (used for correcting near vision) = +1.5 D
Focal length of the required lens (f) = 1/P
f = 1/1.5 = +0.667 m
The focal length of the lens (for correcting near vision) is 0.667 m.

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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?

Answer 6:
The remedial lens should make the objects at infinity appear at the far point.
Therefore, for object at infinity, u = ∞
Far point distance of the defected eye, ν = – 80 cm

Negative sign shows that the remedial lens is a concave lens

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 the defect ? Assume that the near point of the normal eye is 25 cm.

Answer 7:
(i) The near point N of hypermetropic eye is farther away from the normal near point N.

(ii) In a hypermetropic eye, the image of nearby object lying at normal near point N (at 25 cm) is formed behind the retina.

(iii) Correction of hypermetropia : The convex lens forms a virtual image of the object (lying at normal near point N) at the near point N’ of this eye.

The object placed at 25 cm from the correcting lens must produce a virtual image at 1 m or 100 cm.
Therefore, u = – 25 cm, ν = 100 cm

The positive sign shows that it is a convex lens

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

Answer 8:
A normal eye is not able to see the objects placed closer than 25 cm clearly because the ciliary muscles of the eyes are unable to contract beyond a certain limit.

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

Answer 9:
The image is formed on the retina even on increasing the distance of an object from the eye. The eye lens becomes thinner and its focal length increases as the object is moved away from the eye.

Question 10:
Why do stars twinkle?

Answer 10:
The twinkling of a star is due to atmospheric refraction of starlight. The starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index.

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Question 11:
Explain why the planets do not twinkle?

Answer 11:
Unlike stars, planets don’t twinkle. Stars are so distant that they appear as pinpoints of light in the night sky, even when viewed through a telescope. Since all the light is coming from a single point, its path is highly susceptible to atmospheric interference (i.e. their light is easily diffracted).

Question 12:
Why does the Sun appear reddish early in the morning?

Answer 12:
The light coming from the sun passes through various denser layers of air in the earth’s atmosphere before reaching our eyes near the horizon. Most of the part of blue light and light of small wavelength gets scattered by dust particles near the horizon. So, the light reaching our eyes is of large wavelength. Due to this the sun appears reddish at the time of sunrise and sunset.

Question 13:
Why does the sky appear dark instead of blue to an astronaut ?

Answer 13:
The sky appears dark instead of blue to an astronaut, as scattering of light does not take place outside the earth’s atmosphere.

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For More Content related to Class 10 –

• Get NCERT Solutions of all Science Chapters of Class 10
• Get NCERT Solutions of all Mathematics Chapters of Class 10
• Get NCERT Solutions of all Social Science Chapters of Class 10
• Get Notes, PYQs, Free E-Books, and other material for Class 10
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