Human Eye and Colourful World – NCERT Solutions for Class 10 Science Chapter 11
NCERT Solutions for Class 10 Science Chapter 11 – The Human Eye And Colorful World, contains solutions to various questions in Exercises for Chapter 11. The Human Eye And Colorful World Class 10 NCERT Solutions have been explained in a simple and easy-to-understand manner. NCERT Solutions for Class 10 help to check the concept you have learned from detailed classroom sessions and the application of your knowledge.
| Category | NCERT Solutions for Class 10 |
| Subject | Science |
| Chapter | Chapter 11 – The Human Eye And Colorful World |
Download NCERT Solutions for Class 10 Science Chapter 11 – The Human Eye And Colorful World
NCERT Solutions for Class 10 Science Chapter 11 – The Human Eye And Colorful World – NCERT Exercises
The Human Eye And Colorful World – Intext Questions – Page 190
1. What is meant by the power of accommodation of the eye?
Answer:
The ability of the eye to adjust its focal length with the help of ciliary muscles to focus on the rays coming from a distant object or from a nearby object on the retina in order to see it clearly is called 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:
The person is not able to see objects beyond 1.2m distinctly because the image is made in front of the retina and not on the retina. To rectify this condition, the person must wear a concave lens to push the image back to the retina.
3. What is the far point and near the point of the human eye with normal vision?
Answer:
The minimum distance of the object from the eye that can be seen distinctly without feeling any strain on the eyes is called the near point of the eye. For a person with normal vision, 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 for a person with normal vision 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:
A student who has difficulty reading the blackboard while sitting in the last row must be suffering from myopia. It is a condition in which a person is not able to see distant objects. This condition can be corrected by using a concave lens.
NCERT Solutions for Class 10 Science Chapter 11 – NCERT Exercise – Page 197
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) accommodation
(c) near-sightedness
(d) far-sightedness
Answer:
(b) accommodation
2. The human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina.
Answer:
(d) retina
(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
(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:
(b) Ciliary muscles
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:
Given: Power of lens needed for correcting distant vision: –5.5 D
Power of lens needed for correcting near vision: +1.5 D.
We know The power (P) of any lens of focal length f is given by:
=>Power (P)Â = 1/f
(i) Focal length of lens needed = 1/f
= 1/ – 5.5
= – 0.181 m
Hence, the focal length of the lens for correcting distant vision is – 0.181 m.
(ii) Focal length of lens needed = 1/f
= 1/ +1.5
= + 0.667 m
Hence, 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:
Since the person is myopic, the image is formed in front of the retina. To correct this condition, we must use a concave lens.
We are considering the far point, so the object is placed at infinity.
=> u = ∞
Image distance, v = 80 cm
We know according to the lens formula,
– 1 / 80 – 1/ ∞ = 1/f
=> 1/f = – 1 / 80
f = – 80 cm
f = – 0.8 cm
According to power and focal length relation:
Power (P) = 1/f (in m)
P = 1/ -0.8
P = – 1. 25 D
Hence, the person requires a concave lens of power -1.25 D.
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:
Hypermetropia is a condition in which a person can see distant objects clearly but has difficulty in seeing nearby objects. The image of the nearby object in this case is formed beyond the retina. This condition can be corrected by wearing a convex lens of suitable power. We can see the image formation in the diagram given below.
A virtual image of a nearby object (N’ in the above figure) is formed at the near point of vision (N) of the individual suffering from hypermetropia. This helps in bringing the image on the retina.
Given: Near point of a hypermetropic eye is 1 m.
Object distance, u = – 25 cm
Image distance, v = – 1 m = – 100 cm
Using the lens formula:
1/f = – 1/ 100 – 1/ – 25
1/f = 1/ 25 – 1/ 100
1/f = 3/ 100
f = 33.3 cm = 0.33 m
Also, Power (P) = 1/f (in m)
P = 1/ 0.33 = +3 D
Hence, the person requires a concave lens of +3D.
8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The ciliary muscles that are responsible for helping the eye adjust the focus for making clear images, cannot contract over a certain limit, so we cannot clearly see the objects placed closer than 25 cm.
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 object distance, the image distance from the retina increases too. To focus this image on the retina, the lens of the eye becomes thinner and its focal length increases.
10. Why do stars twinkle?
Answer:
Stars can be seen as point sources of light in the night sky. When the light coming from these sources enters the atmosphere, it gets refracted multiple times because of varying levels of air density in different layers of the atmosphere. This gives an illusion that the stars are twinkling.
11. Explain why the planets do not twinkle.
Answer:
Planets appear much larger than stars. They can be seen as a mass of multiple point sources of light. When the light comes from these multiple sources, each source gives light of varying intensity. These intensities cancel each other and the light is easily diffracted. Hence, the planets do not twinkle.
12. Why does the Sun appear reddish early in the morning?
Answer:
Early in the morning, the rays traveling from the sun travel to greater lengths as compared to the rest of the day. Because of this, the light rays with shorter wavelengths such as Blue rays, scatter around in the atmosphere and don’t reach our eyes. The rays with longer wavelengths like red rays travel to our eyes making the sky seem reddish.
13. Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears dark to an astronaut because there is no atmosphere in the space that can scatter the sunlight. Since no scattering of light occurs, no blue light reaches its eyes and the sky appears black.
Topics Covered in The Human Eye And Colorful World Class 10 Science Chapter 11
- The Human Eye
- Defects of Vision and their Correction
- Refraction of Light Through A Prism
- Dispersion of White Light By A Glass Prism
- Atmospheric Refraction
- Scattering Of Light
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