KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World.

Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

KSEEB SSLC Class 10 Science Chapter 11 Intext Questions

Text Book Part II Page No. 100

Question 1.
What is meant by power of accommodation of the eye?
Answer:
When the ciliary muscles are relaxed, the eye lens becomes thin, the focal length increases and the distant objects are clearly visible to the eyes. To see the nearby objects clearly, the cilliary muscles contract making the eye lens thicker. Thus, the focal length of the eye lens decreases and the nearby objects become visible to the eyes. Hence, the human eye lens is able to adjust its focal length to view both distant and nearby objects on the retina. This ability is called 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 the corrective lens used to restore proper vision?
Answer:
A person with a myopic eye cannot see objects beyond 1.2 m distinctly because, In a myopic eye, the image of a distant object is formed in front of the retina. This defect can be corrected by using a concave lens.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
The near point of the eye is the minimum distance of the object from the eye, which can be seen distinctly without strain. For a normal human eye, this distance is 25 cm. The far point of the eye is the maximum distance to which the eye can see the objects clearly. The far point of the normal human eye is infinity.

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:
A student has difficulty in reading the blackboard while sitting in the last row. It shows that he is unable to see distant objects clearly. He is suffering from myopia. This defect can be corrected by using a concave lens.

KSEEB SSLC Class 10 Science Chapter 11 Textbook 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) accomodation
(c) near-sightedness
(d) far-sightedness
Answer:
(b) accomodation

Question 2.
The human eye forms the image of an object at its
(a) cornea
(b) Iris
(c) pupil
(d) retina
Answer:
(d) 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
Answer:
(c) The least distance of distinct vision is the minimum distance of an object to see clear and distinct image. It is 25 cm for a young adult with normal visions.

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:
(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens 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?
Answer:
The power P of a lens of focal length f is given by the relation
MP Board Class 10th Science Solutions Chapter 11 Human Eye and Colourful World 3
(i) Power of the lens used for correcting distant vision = -5.5 D
Focal length of the required lens, f = \(\frac { 1 }{ p } \).
f = \(\frac { 1 }{ -5.5 } \) = -0.181m
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 = \(\frac { 1 }{ p } \) : f = \(\frac { 1 }{ 1.5 } \) = + 0.667 m.
The focal length of the lens for correcting near vision is 0.667m.

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:
A person is suffering from near vision. So image is formed in front of retina.
Distance of Image, V = – 80 cm
focal length = f
As per lens formula.
KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World Ex Q 6
f = 80 cm = – 0.8 m,
KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World Ex Q 6.1
∴ Power of -125 D is required to correct this problem.

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.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World Ex Q 7
Object distance, u = 25 cm
Image distance, v = 1m = -100 m
As per lens formula,
KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World Ex Q 7.1
This defect should be corrected by +3.0 power.

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

Question 9
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
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?
Answer:
The twinkling of a star is due to atmospheric refraction of starlight. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. This apparent position of the star is not stationary, but keeps on changing slightly, as the physical conditions of the earths atmosphere are not stationary since the stars are very distant, they approximate point – sized sources of light. As the path of rays of light coming from the star goes on varying slightly, the’ apparent position of the star fluctuates and the amount. Of starlight entering the eye flickers some other time, fainter, which is the twinkling effect.

Question 11.
Explain why the planets do not twinkle.
Answer:
The planets are much closer to the earth and are thus seen as extended sources. If we consider a plant as a collection of a large number of point sized sources of light, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
During sunrise, the light ray coming from the Sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. Since, blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning.

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

KSEEB SSLC Class 10 Science Chapter 11 Additional Questions and Answers

Question 1.
Draw a neat diagram showing
(a) neat point of a Hypermetropic eye
(b) Hypermetropic eye
(c) correction for Hypermetropic eye.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World Ad Q 1

Question 2.
Which causes presbyopia? Name the lens to correct this defect.
Answer:
This causes due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens.
A common type of bi-focal lense is necessary to correct this defect.

Question 3.
Who discovered spectrum of sunlight?
Answer:
Isaac Newton.

Question 4.
What is the reason for Advance sunrise and delayed sunset?
Answer:
The sun is visible to us about 2 minutes before the actual sunrise, and about 2 minutes after the actual sunset because of atmospheric refraction.

Question 5.
At noon, the sun appears white. Give reason.
Answer:
Because at noon, the sun appears white as only a little of the blue and violet colours are scattered.

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