NCERT Solutions Science Chapter 10 Human Eye and Colourful World for Class 10th
Chapter 10 of NCERT Science for Class 10, titled “Human Eye and Colourful World,” discusses the structure and functioning of the human eye, the nature of light, and how we perceive colors. It explains the anatomy of the eye, including parts like the cornea, lens, retina, and the role of the iris in controlling light entry. The chapter also covers topics like refraction, the formation of images, and common vision defects such as myopia and hyperopia, along with their corrections.
Additionally, it touches on the dispersion of light and the formation of a rainbow, illustrating how light can separate into different colors. Overall, the chapter emphasizes the importance of the human eye in experiencing the vibrant world around us and the science behind our visual perception.
NCERT Solutions Science Chapter 10 Human Eye and Colourful World
Page No: 190
1. What is meant by power of accommodation of the eye?
Answer-
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.
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-
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.
3. What is the far point and near point of the human eye with normal vision?
Answer-
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.
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-
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.
Page No: 197
Exercise
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
Due to accommodation the human eye can focus objects at different distances by adjusting the focal length of the eye lens.
Page No: 198
2. The human eye forms an image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina
Answer –
(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.
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
25 cm is the least distance of distinct vision for a young adult with normal vision.
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
The action of the ciliary muscles changes the focal length of an eye lens
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
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.
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-
The individual is suffering from myopia. In this defect, the image is formed in front of the retina. Therefore, 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,
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.
Solution:
(i) The near point N of hyper metropic 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:
At distance less than 25 cm, the ciliary muscles cannot bulge the eye lens any more, the object cannot be focused on the retina and it appears blurred to the eye, as shown in the given figure.
Question 9
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
Answer:
The eye lens of a normal eye forms the images of objects at various distances on the same retina. Therefore, the image distance in the eye remains the same.
Question 10
Why do stars twinkle ?
Answer:
Stars appear to twinkle due to atmospheric refraction. The light of star after the entry of light in earth’s atmosphere undergoes refraction continuously till it reaches the surface of the earth. Stars are far away. So, they are the point source of light. As the path of light coming from stars keep changing, thus the apparent position of stars keep changing and amount of light from stars entering the eye keeps twinkling. Due to which a star sometimes appear bright and sometimes dim, which is the effect of twinkling.
Question 11
Explain why the planets do not twinkle ?
Answer:
The planets are much nearer to the earth than stars and because of this they can be considered as large source of light. If a planet is considered to be a collection of a very large number of point sources of light, then the average value of change in the amount of light entering the eye from all point size light sources is zero. Due to this the effect of twinkling is nullified.
Question 12
Why does the sun appear reddish early in the morning ?
Answer:
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:
As an astronaut moves away from the atmosphere of earth, the atmosphere becomes thin. Due to the absence of molecules (or dust particles) in air, the scattering of light does not take place. Thus, sky appears dark in the absence of scattering.