The Human Eye

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 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 object. By contraction of ciliary muscles thickness of eye lens increases which decrease the focal length and by relaxation of ciliary musclses thickness of eye lens decreases which increase the focal length.

Question: 2. A person with a myopic eye cannot see objects beyond 1.2m distinctly. What should be the type of the 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 human eye: The farthest point up to which the eye can see objects clearly is called the far point of the eye. Far point of a normal eye is between 25cm to infinity.

Near point of human eye: The minimum distance, at which objects can be seen most distinctly without any strain, is called the near point of human eye. Near point of human eye is known as 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 eye because of which a person cannot see a distant object clearly but can see the object placed nearby is called myopia or near sightedness.

Thus, this child is suffering from myopia. This can be corrected by using of a concave lens or diverge 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 object. By contraction of ciliary muscles thickness of eye lens increases which decrease the focal length and by relaxation of ciliary musclses thickness of eye lens decreases which increase 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. Retina is composed of enormous number of light sensitive cells. Lens forms image of the visual world on retina through cornea. Light sensitive cells of retina get activated upon illumination and generate electric signals. Optic nerves send these electric signals to the brain, where these signals 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. Least distance of distinct vision is known as near point of the eye also. Least distance of distinct vision for a young adult with 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 eye lens can be modified according to the distance of object by ciliary muscles with which lens is attached. By changing the curvature focal length of eye lens is changed.

Contraction of ciliary muscles make eye lens thicker which decreases the curvature of eye lens consequently focal length decreases. This enables us to see nearby objects clearly.

Relaxation of ciliary muscles make eye lens thin which increases the curvature of eye lens which increases its focal length. Increase in focal length of eye lens enables us to see object at distance clearly.

The ability of 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 power of 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 power of 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 image is formed at 80cm.

Thus, `v =- 80` cm `=- 0.8` m

And `u = oo`

We know that `1/v - 1/u=1/f`

`:. 1/(-0.8) - 1/oo = 1/f`

`=>1/f = 1/(-0.8) - 0`

[∵ `1/oo = 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 in negative thus, 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`

`:. 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 lens of eye is adjusted accordingly to see an object using ciliary muscles, this ability of eye is called power of accommodation.

But, focal length of eye lens can be decreased to only a certain extent. Eye lens has a minimum limit to decrease its focal length. This limit is 25cm from eyes.

Thus, a normal eye does 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 normal eye, image is always formed at retina instead of distance of 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 retina depends on the nature of defects. But in such case also after correction of vision using suitable power of lens, image is formed always on retina instead of the distance of object.

If image will not form at retina people will 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 light coming from stars enter in the atmosphere, it get refracted in zig zag pattern because of uneven refractive index of atmosphere. And light coming from stars enters in our eyes in same zig-zag pattern. This gives the appearance of stars sometime more bright and sometime 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 atmosphere of earth because of uneven refractive index of atmosphere. But, light coming from one edge of planet gets refracted in one side while light coming from other end of planet gets refracted in other side. This refraction of light in both sides nullifies the variation of refraction of rays of light coming from 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 horizon which is at more distance than in noon. Sunlight coming from horizon has to pass through thicker layer of atmosphere and is dispersed through respectively larger size of particles suspended in atmosphere. Larger suspended particles in atmosphere scatter red and orange color of light having longer wavelength more strongly than those color having shorter wavelength, such as blue light.

Since, red and orange colors of light scattered strong when sunlight is coming from horizon, thus Sun appear reddish early in the morning.

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 take place in space. While in the atmosphere of earth fine particles suspended in atmosphere scatter blue light more strongly and sky appear blue on a clear day.

Thus, sky appear dark instead of blue to an astronaut.