RBSE Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

Rajasthan Board RBSE Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World Textbook Exercise Questions and Answers.

RBSE Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

RBSE Class 10 Science Chapter 11 Human Eye and Colourful World InText Questions and Answers

Page 190.

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye to focus both near and distant objects, by adjusting its focal length, is called the 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:
By using concave lens, the defect in the person can be corrected.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
Far point is at infinity and near point is at 25 cm from the eye.

Question 4.
A student has difficulty in 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:
As the student has difficulty in reading the blackboard, he is suffering from myopia. To correct this defect, he has to use spectacles with concave lens of suitable focal length.

RBSE Class 10 Science Chapter 11 Human Eye and Colourful World Textbook Questions and Answers

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:
(b) accommodation.

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) 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.
Answer:
(c) 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:
(i) Given Power of lens to clearly see distant objects
P = – 5.5 D
p = \(\frac{1}{f}\)
f = \(\frac{1}{p}\) m
f = \(\frac{1}{-5.5}\) m = -0.181m
(ii) Given Power of lens to clearly see near objects:
p = +1.5 D
P = \(\frac{1}{f}\)
f = \(\frac{1}{p}\) m
f = \(\frac{1}{1.5}\) m
f = +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?
Answer:
Object distance (u) = a
Image distance (v)= -80 cm
Focal length = f
According to lens formula:
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{80}\) – \(\frac{1}{α}\) = \(\frac{1}{f}\)
\(\frac{1}{f}\) = – \(\frac{1}{80}\)
f = -80cm = -0.8cm
Power of lens, p = \(\frac{1}{f}\)
p = \(\frac{1}{ -0.8}\)
p = -1.25 D
Therefore, to correct this problem -1.25 D power concave lens is required.

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:

The near point of human eye has shifted from 25 cm to 1 m (100) cm. Therefore Given, object distance (u) = -25 cm Image distance (v) = -1 m = -100 cm Focal length f = ?
Using the lens formula, \(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)
\(\frac{1}{f}\) =\(\frac{1}{-100}\) – \(\frac{1}{-25}\)
⇒ \(\frac{1}{f}\) = \(\frac{3}{100}\) cm or \(\frac{1}{3}\)m
Therefore, Power of the lens required:
p = \(\frac{1}{f}\) = \(\frac{1}{1/3}\) = 3
p = +3 D
Thus, 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?
Answer:
A normal eye is not able to see clearly the objects placed closer than 25 cm because the focal length of lens of eye cannot be less than the minimum range (25 cm).

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
When we increase the distance of an object from the eye, the muscles are relaxed and the lens become thin. Thus, its focal length increases which enables us to see distant objects clearly because it forms the image on the retina.

Question 10.
Why do stars twinkle?
Answer:
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 density of various layers of atmosphere changes variably due to the change in temperature. Thus the refractive index also changes. Due to the change in refractive index, the starlight continuously changes its path and the light entering the eye flickers. Thus the stars twinkle.

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 planet 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. That is why, they do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
At the time of sunset or sunrise, the sun’s rays have to pass through a larger distance in atmosphere most of the blue and other shorter wavelengths are removed by scattering. The red colour having largest wavelength is scattered least and this least scattered light reaches our eyes, therefore the sun looks reddish to us.

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 (dark) to them.