The Human Eye - Class 10th Science
NCERT Question and Answer
NCERT In Text Questions and Answer
Question (1) What is meant by power of accommodation of the eye?
Answer: The ability of the eye lens to adjust its focal length is called accommodation or power of accommodation.
Eye adjusts its focal length using ciliary muscles according to the distance of the object. By contraction of ciliary muscles thickness of the eye lens increases which decrease the focal length and by relaxation of ciliary muscles thickness of the eye lens decreases which increases the focal length.
Question (2) A person with a myopic eye cannot see objects beyond 1.2m distinctly. What should be the type of corrective lens used to restore proper vision?
Answer: In the case of myopia a person can see the object nearby clearly while cannot see a distant object clearly.
For myopic eye a diverging lens or concave lens of suitable power is used to restore proper vision.
Question (3) What is the far point and near point of the human eye with normal vision?
Answer:
Far point of the human eye: The farthest point up to which the eye can see objects clearly is called the far point of the eye. The far point of a normal eye is between 25cm to infinity.
Near the point of the human eye: The minimum distance, at which objects can be seen most distinctly without any strain, is called the near point of the human eye. The near point of the human eye is known as the least distance of distinct vision also.
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: Defect of the eye because of which a person cannot see a distant object clearly but can see the object placed nearby is called myopia or nearsightedness.
Thus, this child is suffering from myopia. This can be corrected by using a concave lens or diverging lens of suitable power.
NCERT Exercise Question and Answer
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
Explanation : The ability of eye lens to adjust its focal length is called accommodation or power of accommodation.
Eye adjusts its focal length using ciliary muscles according to the distance of the object. By contraction of ciliary muscles thickness of the eye lens increases which decrease the focal length and by relaxation of ciliary muscles thickness of the eye lens decreases which increases the focal length.
Question (2) The human eye forms the image of an object at its
(a) Cornea
(b) Iris
(c) Pupil
(d) Retina
Answer: (d) Retina
Explanation : Retina is the inner back portion of eye. The retina is composed of an enormous number of light-sensitive cells. The lens forms an image of the visual world on the retina through the cornea. Light-sensitive cells of the retina get activated upon illumination and generate electric signals. Optic nerves send these electric signals to the brain, where these signals are finally processed and interpreted to see an object as they are.
Question (3) The least distance of distinct vision for a young adult with normal vision is about
(a) 25m
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m
Answer: (c) 25 cm
Explanation: The minimum distance, at which objects can be seen most distinctly without any strain, is called the least distance of distinct vision. The least distance of distinct vision is known as the near point of the eye also. The least distance of distinct vision for a young adult with a normal eye is about 25cm.
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
Explanation :
The curvature of the eye lens can be modified according to the distance of the object by the ciliary muscles with which the lens is attached. By changing the curvature focal length of the eye lens is changed.
Contraction of ciliary muscles makes the eye lens thicker which decreases the curvature of the eye lens consequently focal length decreases. This enables us to see nearby objects clearly.
Relaxation of ciliary muscles makes the eye lens thin which increases the curvature of the eye lens which increases its focal length. An increase in the focal length of the eye lens enables us to see objects at distance clearly.
The ability of the eye lens to adjust its focal length is called accommodation or power of accommodation.
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:
Given, power of lens for distant vision (P) = –5.5 dioptres
Power of lens for near vision (P) = +1.5 dioptre
(i) Focal length for distant vision
We know that power of lens P = 1/f
Where P is the power of the lens and f is the focal length
∴ –5.5= 1/f
⇒ f = 1/–5.5 = –0.18 m
Thus, focal length of the lens for distant vision = – 0.18 m or – 18 cm
(ii) Focal length for near vision
We know that power of lens P = 1/f
Where P is the power of the lens and f is the focal length
∴ 1.5 = 1/f
⇒ f = 1/1.5 = 0.66 m = 66 cm
Thus, focal length of lens for near vision = 0.66 m or 66 cm
Question (6) The far point of a myopic person is 80cm in from of the eye. What is the nature and power of the lens required to correct the problem?
Answer: Far point of a normal eye is between 25cm to infinity.
Here given far of the myopic person = 80cm.
This means the person can see the object if the image is formed at 80cm.
Thus, v = – 80 cm = – 0.8 m
And u = ∞
We know that 1/v – 1/u = 1/f
∴ 1/–0.8 – 1/∞ = 1/f
⇒ 1/f = 1/–0.8 – 0
[∵ 1/∞ = 0]
⇒ 1/f = 1/–0.8
⇒ f = – 0.8 m
Now we know that, P = 1/f
Where, P = power of lens, and f = focal length
∴ P = 1/–0.8
⇒ P = – 1.25 dioptre
Since, here focal length is negative thus, the lens is of concave nature.
And power of lens = – 1.25 dioptre.
Question (7) Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25cm.
Answer :
Here given, distance of image (v) = – 1 m
Distance of object (u) = – 25 cm = – 0.25m
[∵ near point of the normal eye = 25cm]
We know that 1/v – 1/u = 1/f
&there4 1/– 1 – 1/–0.25 = 1/f
⇒ – 1 + 4 = 1/f
⇒ 1/f = 3 m
⇒ f = 1/3 m
Since, focal length is positive, thus lens is convex.
Now, Now we know that, P = 1/f
Where, P = power of lens, and f = focal length (in meter)
∴ P = 1/1/3
⇒ P = + 3 dioptre
Thus, a convex lens of +3 dioptre is required to correct this defect.
Question (8) Why is a normal eye not able to see clearly the objects placed closer than 25cm?
Answer: Focal length of the lens of the eye is adjusted accordingly to see an object using ciliary muscles, this ability of the eye is called the power of accommodation.
But, the focal length of the eye lens can be decreased to only a certain extent. The eye lens has a minimum limit to decrease its focal length. This limit is 25cm from the eyes.
Thus, a normal eye is not able to see clearly the objects placed closer than 25cm.
Question (9) What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer: In the case of a normal eye, the image is always formed at the retina instead of the distance of the object. Thus, there is no effect on image distance when we increase the distance of an object.
However, in the case of eye defects image is formed either before or after the retina depending on the nature of the defects. But in such cases also after correction of vision using suitable power of the lens, the image is formed always on the retina instead of the distance of the object.
If the image will not form at the retina people will be unable to see the object clearly.
Question (10) Why do stars twinkle?
Answer: Stars are too far from us because of that they appear as tiny sources of light. When the light coming from stars enters the atmosphere, it gets refracted in a zig-zag pattern because of the uneven refractive index of the atmosphere. And the light coming from stars enters our eyes in the same zig-zag pattern. This gives the appearance of stars sometimes more bright and sometimes less bright, which gives the twinkling effect.
Thus because of uneven refraction of light coming from starts, they twinkle.
Question (11) Explain why the planets do not twinkle.
Answer: Planets are closer than stars to us and appear bigger than stars. Light coming from planets also gets unevenly refracted by the atmosphere of the earth because of the uneven refractive index of the atmosphere. But, light coming from one edge of the planet gets refracted on one side while light coming from another end of the planet gets refracted on another side. This refraction of light on both sides nullifies the variation of refraction of rays of light coming from the planet and we see planets do not twinkle.
Question (12) Why does the Sun appear reddish early in the morning?
Answer: Early in the morning Sun is at the horizon which is at more distance than at noon. Sunlight coming from the horizon has to pass through a thicker layer of the atmosphere and is dispersed through the respectively larger size of particles suspended in the atmosphere. Larger suspended particles in the atmosphere scatter red and orange colors of light having longer wavelengths more strongly than those colors having a shorter wavelength, such as blue light.
Since, red and orange colors of light are scattered strongly when sunlight is coming from the horizon, thus Sun appears reddish early in the morning.
Sun appears reddish while sunset in the evening because of the same reason.Question (13) Why does the sky appear dark instead of blue to an astronaut?
Answer: There is no atmosphere in the space. Thus, no scattering of light takes place in space. While in the atmosphere of earth fine particles suspended in the atmosphere scatter blue light more strongly and the sky appears blue on a clear day.
Thus, the sky appears dark instead of blue to an astronaut.