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NCERT Solution for Class 10 Science Chapter 11 : Human Eye and Colourful World

Human Eye and Colourful World
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Chapter 11 of NCERT Class 10 Science is Human Eye and Colourful World. In this chapter, you will learn about the primary structure of the human eye and functions of each part of it. The chapter has 13 questions in its exercise which are going to help you understand specific topics like refraction through a prism, scattering of light, and refraction through the atmosphere.

NCERT Solutions for Class 10 Science Chapter 11 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 10 Science Chapter 11 are designed as per the CBSE Class 10th Science Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 10 Science Chapter 11.

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NCERT Solution for Class 10 Science Chapter 11 Human Eye and Colourful World PDF

 

NCERT 10th Science Chapter11 1

Exercise: 11.1

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

 

Solution:
As soon as the ciliary muscles of our eyes are relaxed, the eye lens becomes thin, and the focal length of the eye increases as well, due to which distant objects are clearly visible to the eyes. Contrary to this when we try to see the nearby objects, the ciliary muscles of the eye contract in order to make the eye lens thicker thus, reducing the focal length of the eye lens, as a result, the nearby objects become visible to the eyes. Therefore, we can conclude that the human eye lens can adjust its focal length as required to view distant and nearby objects clearly on the retina. This ability to change the eye’s focal length is called the power of accommodation of the eyes.

 

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?

 

Solution:
If a person can see nearby objects clearly but have a problem in seeing objects beyond 1.2 m. This type of condition occurs when the image of an object beyond 1.2 m is formed in front of the retina and not at the retina, this condition of the eye is called myopia or near-sightedness. This defect of vision can be corrected using a concave lens. The concave lens of correct optical power will bring the image of a nearby object back to the retina.

 

Question3.
What is the far point and near point of the human eye with normal vision?

 

Solution:
The near point of the human eye is defined as the minimum distance at which an object can be seen clearly by a human eye without any strain. This near point for a normal human eye is 25 cm.
The far point of the human eye is defined as the maximum distance to which the objects can be seen clearly by the human eye.
The far point is infinity, for a normal human 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 4:
A student is having difficulty in reading the text written on the blackboard while sitting in the last row. It shows that he has a problem in seeing distant objects clearly. By this information, we can state that he is suffering from myopia. This defect of the eye of the student can be corrected using a concave lens.

Exercise: 11.2

 

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

 

Solution 1:
The human eye can adjust the focal length of its eye lens accordingly to see objects situated at various distances from the eye. This property of changing the focal length is feasible due to the power of accommodation 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

 

Solution 2:
The image of an object in the human eye is formed at its retina.

 

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 3:

The minimum distance of an object to see a clear and distinct image of an object is called the least distance of distinct vision. For a young adult with normal vision, the least distance of distinct vision is equal to 25 cm.

 

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 4:
The curvature of the eye lens changes with the relaxation or contraction of ciliary muscles. This change in curvature of the eye lens results in the change of the focal length of the eyes. Hence, the focal length change of an eye lens of the human eye is caused by the action of ciliary muscles.

 

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 5:
Given:
The relation between the powers P of a lens and its focal length f is given by
P = 1/f (in meters)
(i) Power of the lens used for by the person correcting distant vision = −5.5 D
Focal length of the lens required to correct distant vision, f = 1/P
f = 1/-5.5 = -0.181 m
The focal length of the lens used by the person for correcting distant vision is −0.181 m.
(ii) Power of the lens used by the person for correcting near vision = +1.5 D
Focal length of the lens required to correct near vision, f = 1/P
f = 1/1.5 = +0.667
The focal length of the lens used by the person for correcting near vision is 0.667m.
Therefore,
Distant vision of the person= −0.181 m,
Near vision of the person = 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 correcting the problem?

 

Solution 6: Refer pdf.

 

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 7: Refer pdf.

 

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

 

Solution 9:
The image distance remains constant as the size of the eyes cannot increase or decrease. When we increase the distance of an object from the eye, this doesn’t affect the image distance of the eye. This increase in the distance of the object is compensated by changing the focal length of the eye lens. The focal length of the eyes changes accordingly to always form an image at the retina of the eye.

 

Question 10:
Why do stars twinkle?

 

Solution 10:
Stars emit their own light but due to the atmospheric refraction of light, they appears to shine. They are considered as point sources of light as stars are very far away from the earth. When the light coming from stars enters into the earth’s atmosphere, it gets refracted at different levels due to the change in density of air at different levels of the atmosphere. When the starlight refracted by the atmosphere reaches closer to us, it appears brighter than while it was far from us. Due to this effect of refraction, it appears as if the stars are twinkling at night.

 

Question 11:
Explain why the planets do not twinkle?

 

Solution 11:
Planets are relatively closer to earth as compared to stars. As they appear larger in size than that of stars, planets can be considered as a collection of a large number of point-size sources of light. The various parts of these planets generate a different intensity of light in such a manner that the average of brighter and dimmer effect is zero. Therefore, the planets don’t appear to twinkle.

 

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

 

Solution 12:
During sunrise, the Sun rays travel a greater distance in the earth’s atmosphere before they reach our eyes. During this journey, the rays of shorter wavelengths are dispersed out and only rays with longer wavelengths can reach our eyes. Since blue colour rays have a shorter wavelength it is dispersed out while red colour has a longer Wavelength, therefore, only red colour reaches our eyes after the atmospheric scattering of light, as a result, the Sun appearsreddish early in the morning.

 

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

 

Solution 13:
The sunlight doesn’t scatter as there is no atmosphere in outer space, so the sky appears dark instead of blue to an astronaut. The eyes of astronaut’s don’t receive any light as the sunlight is not scattered, due to this sky appears black to them.

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