Class 10 - Physics - Human Eye and Colourful World

Question1.

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:

Correct option: - (b) accommodation

The ability of the eye lens to adjust its focal length is called accommodation.

Question2.

The human eye forms the image of an object at it’s:-

(a) cornea. (b) Iris. (c) Pupil. (d) Retina.

Answer:

Correct option: - (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:

Correct option: - (c) 25cm.

To see an object comfortably and distinctly, we must hold it at about 25 cm from the eyes.

The minimum distance, at which objects can be seen most distinctly without strain, is called the least distance of distinct vision.

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:

Correct option: - (c) ciliary muscles

The eye lens is composed of a fibrous, jelly-like material.

Ciliary muscles change the curvature of the eye lens and that changes the focal length.

It helps to see distant and nearby objects clearly.

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:-

  • Correcting distant vision;

Power P = -5.5D

Using P = (1/f) where f= focal length of the lens.

=> f = (1/-5.5) = -0.181m

The focal length of the lens for correcting distant vision =-0.181m.

  • Correcting near vision

Power P = + 1.5D

Using P = (1/f) where f= focal length of the lens.

=> f = (1/1.5) = +0.667m

The focal length of the lens for correcting distant vision =+0.667m.

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:

For a myopic eye, concave lens should be used to correct the defect;

Object distance u = -infinity

Far point of the defective eye v = -80cm

Using (1/f) = (1/v) – (1/u)

(1/f)  = 1/ (-80) + (1/-infinity)

(1/f)  = - (1/80) + (-0)

f = -80 cm

f = -0.8 m

Power = (1/f)

= 1/ (-0.8)

= -1.25 D

The nature of lens required to see the distant objects clearly is convex lens is power -1.25 D.

                                                                                                                                                                            

Question7.

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropia 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:

It is corrected by using convex lens of suitable focal length.

Near point hypermetropia eye v = -1m = -100cm

Object distance u = -25cm

According to formula, (1/v) – (1/u) = (1/f)

 Near point of hypermetropia eye:-

(1/f) = - (1/100) – (1/-25) = (-1/100) + (1/25)

(1/f) = (-1+4)/ (100) = (3/100)

f = (100/3)

Physics Class 10 Human Eye Defect Of Hypermetropic1

  • Hypermetropia eye:-

f = (100/3)

P = (1/f) = (3 x 100)/ (100) = 3D

Physics Class 10 Human Eye Defect of Hypermetropic

  • Correction of a hypermetropia eye

 Physics Class 10 Human Eye Hypermetropia                                                                                                                                                                          

Question 8.

Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer:

The rays coming from the objects which are at a distance less than 25cm are not properly focussed at retina because ciliary muscles cannot be contracted beyond this limit.

The focal length of the eye cannot be reduced below 25cm.

The minimum distance, at which objects can be seen most distinctly without strain, is called the least distance of distinct vision.

Therefore normal eye cannot see the 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:

The image distance in the eye, is fixed and cannot be changed. The ability of the eye lens to adjust its focal length is called accommodation.

Due to this eye can increase or decrease focal length of the lens in order to see either close or distant objects.

In order to see nearby objects the ciliary muscles contract and lens become thick and focal length is decreased which helps to see nearby objects.

In order to see distant objects the ciliary muscles expand and the lens become thin and focal length increases which help to see distant objects.

 Question 10.

Why do stars twinkle?

Answer:

Stars twinkle due to atmospheric refraction of starlight. The light from the star on entering earth’s atmosphere undergoes refraction continuously before it reaches the earth.

Refraction occurs and refractive index changes.

As the path of rays of light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of starlight entering the eye flickers.

Therefore sometimes star appears brighter and sometimes twinkle.

Physics Class 10 Human Eye Twinkling of Star

Question 11.

Explain why the planets do not twinkle.

Answer:

Planets, do not twinkle as they are closer to the Earth than those distant stars, so planets appear larger in comparison.

Due to the planets’ closeness to Earth, the light coming from them does not bend much due to Earth’s atmosphere.

Therefore, the light coming from our solar system’s planets does not appear to twinkle.

 Question 12.

Why does the Sun appear reddish early in the morning?

Answer:

In the early morning, as the Sun rises above horizon, light from the sun has to cover a larger distance in the denser medium of the

earth’s atmosphere before reaching our eyes. The blue light due to shorter wavelength gets scattered in the half way

and not visible in the sky. Only the red light which has larger wavelength is able to reach our eyes due to less scattering.

 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 as there is no atmosphere.

There is no outer space that can scatter the sunlight. As the sunlight is not scattered therefore no light reaches the eyes of the astronaut.

Question 9.

What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer:

The image distance in the eye, is fixed and cannot be changed. The ability of the eye lens to adjust its focal length is called accommodation.

Due to this eye can increase or decrease focal length of the lens in order to see either close or distant objects.

In order to see nearby objects the ciliary muscles contract and lens become thick and focal length is decreased which helps to see nearby objects.

In order to see distant objects the ciliary muscles expand and the lens become thin and focal length increases which help to see distant objects.

Question 10.

Why do stars twinkle?

Answer:

Stars twinkle due to atmospheric refraction of starlight. The light from the star on entering earth’s atmosphere

undergoes refraction continuously before it reaches the earth. Refraction occurs and refractive index changes.

As the path of rays of light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of starlight entering the eye flickers.

Therefore sometimes star appears brighter and sometimes twinkle.

Physics Class 10 Human Eye Syy

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