NCERT Solutions for class 10th Science Chapter 11 The Human Eye and Colourful World

NCERT INTEXT QUESTIONS

Question 1. What is meant by power of accommodation of the eye?

Sol. Power of accommodation is the ability of the eye lens to adjust its focal length with the help of ciliary muscles, depending on the distance of the object

Note: When the muscles of the eyes are relaxed, the lens becomes thin. Hence, its focal length increases which enables us to see far objects clearly. When the eyes look at nearby objects, the ciliary muscles contract which increases the curvature of the eye lens and becomes thicker. Consequently, the focal length of the eye lens de­ creases, which enables us to see nearby objects clearly.

Question 2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

Sol. – Myopia is a vision condition in which people can see close objects clearly, but objects farther away appear blurred.

  • Myopia can be corrected by the application of a concave lens of suitable power. This concave lens will bring the image back on to the retina. Focal length (f) of the lens as person cannot see objects beyond 1.2 m distinctly= 1.2 m

Power = 1/ Focal length (f)

= 1/-1.2

= -0.83 D

Hence, the defect can be corrected by using the concave lens of power—0.83 D.

Note: Myopia may arise due to excessive curvature of the eye lens or due to elongation of the eyeball.

Question 3. What is the far point and near point of the human eye with normal vision?

Sol. – The far point of the eye is the maximum distance up to which the normal eye can see the objects clearly. The far point of the normal human eye is infinity.

•  The minimum distance, at which objects can be seen most distinctly without strain, is called the least distance of distinct vision. It is also called the near point of the eye. For a young adult with normal vision, the near point is about 25 cm.

Question 4. A student has difficulty in reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Ans. – If the student is facing difficulty in reading the blackboard while sitting in the last row then it means that he is not able to see far-off objects clearly and he is suffering from a condition called short-sightedness or myopia. It can be corrected by using a concave lens of suitable power

Exercise

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.

Sol. (b) In accomodation, the focal length of the eye lens can be adjusted according to the distance of the object. The ability of the eye to increase or decrease its focal length with the help of ciliary muscles, with respect to the distance of the object is known as power of accommodation of the eye.

Question 2. The human eye forms the image of an object at its

(a) cornea.   (b)  iris.                  (c)  pupil.                      (d)  retina.

Sol. (d) The human eye forms the image of an object at its retina. Retina is neural portion of the eye. Retina, located near the optic nerve, is a thin layer of tissue that lines the back of the eye on the inside. When light enters the eye through the cornea, then with the help of cornea and eye lens rays coming from the object bend and strike the retina. The retina contains specialised cells that respond to light and some of these cells send signals through the optic nerve to the brain for visual recognition. So, brain interprets these signals as upright images.

Note: The image formed on the retina is inverted.

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.

Sol. (c) The minimum distance between the eye lens and the object to form a clear image is known as the least distance of distinct vision. It is also called the near point of the eye. For a young adult with normal vision, the near point is about 25 cm.

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.

Sol. (c) Ciliary muscle is a circular muscles around the lens of the eye that can change the shape of the lens in order to produce a clear image. The relaxation or contraction of ciliary muscles changes the curvature of the eye lens which changes the focal length of the eyes, which enables the individual to see nearby or distant objects depending upon the distance of the object.

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?

Sol. (i)  Power oflens for correcting distant vision (P) = -5.5 dioptre
The focal length (f) of the lens required for correcting distant vision will be given as:
Power (P) of a lens= 1/ focal length (f)
= 1/-5.5
=(-)0.18m
The negative sign (-) indicates that the lens is concave.

(ii)  Power for correcting distant vision=+1.5 dioptre
The focal length of the lens required for correcting near vision will be given as:
Power (P) of a lens= 1/ focal length (f)
= 1/+1.5
=(+)0.67m
The positive sign (+) indicates that the lens used by a person convex.

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?

Sol. Far point of the myopic eye, v = -80 cm Object distance, u = oo (infinity) According to lens formula,
1/f = l/v-1/u
= 1/ -80-1/oo
=-1/80 +0
= -1/80
or f=-80 cm or-0.8 m
So, Power (P) of a lens= 1/f(in meters)
= 1/—0.8
=-1.25 Dioptres
Since the power of a lens is negative, so the nature of the lens is concave.

Question 7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is lm. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Sol. Hypermetropia is a defect of the eye in which parallel rays are focused behind the retina and a person unable to see nearby objects distinctly. This defect can be corrected by using a convex lens of appropriate power.

Hypermetropic eye and its correction,

According to lens formula, 1/f = 1/v -1/u
= 1/-100-1/-25 [Since v = lm = 100 cm and u = 25 cm from question]
=-1/100+ 1/25
=3xl00/100 f=0.33m
So, power (P) ofa lens= 1/f(in metres)
= 1/0.33
= +3.0 Dioptres required to correct this defect.

Question 8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Sol. For a young adult with normal vision, the near point is about 25 cm. If the object is placed closer than 25 cm, it appears blurred because light coming from the object appears to meet beyond retina of the eyes and ciliary muscles of the eye are not able to contract beyond this limit. This is the reason why a normal eye is unable to clearly see the objects placed closer than 25 cm.

Question 9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Sol. The focal length of the eye lens varies according to the distance of the object. When the distance of the object changes then eye lens accommodates itself to form a clear image on the retina of the eye by the action of ciliary muscles. Image distance is equal to the distance from the centre of the eye lens and retina. Thus, the image distance remains the same, only the focal length of the eye lens changes to accommodate the image on the retina.

Question 10. Why do stars twinkle?

Sol. Stars twinkle due to atmospheric refraction. When a ray of light travels from one medium to another it ‘bends’. This phenomenon is referred to as refraction. If it travels from a rarer medium to a dense medium, it bends towards the normal and if it travels from a dense medium to a rarer medium, it bends away from the normal. The speed of the light through which it travels in the atmosphere changes depending on the medium and therefore this bending occurs. The temperature and the refractive index of the atmosphere changes continuously. Therefore, apparent position of the star and the amount of light from the stars reaching to our eye, both changes continuously which leads to the twinkling of stars.

Question 11.Explain why the planets do not twinkle.

Sol. Planets can be considered as a collection of a large number of point-size sources of light. As planets are relatively closer to the Earth, due to which they appears larger in size as compare to stars. If we consider a planet as a collection ofa large number of point-sized sources oflight, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.Why does the Sun appear reddish early in the morning?

Sol. During sunrise, sunlight has to travel a greater distance before reaching to our eyes. During this journey, the shorter wavelength of light scattered and only longer wavelength of light is able to reach our eyes. Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelengths. This gives rise to the reddish appearance of the Sun in the morning.

Question 13. Why does the sky appear dark instead of blue to an astronaut?

Sol. The sky appear dark instead of blue to an astronaut because at a very high altitude, when the astronauts moves away from the atmosphere, the medium becomes rarer. Since there is no atmosphere in the space, sunlight passing through the space fails to get scattered. As the sunlight not gets scattered so no scattered light reach to the eyes of the astronauts and the sky appears them black.

Note: Scattering of light: When light rays pass through an imperfect medium i.e. atmo­ sphere containing dust particles, after striking dust particles deflected from its straight path and scattered haphazardly. The intensity of scattered light depends on the size of the particles and wavelength of the light (A).
Scattering a: 1/ lamda4
For example: Blue colour of the sky, colour of water in deep sea, the reddening of the Sun at sunrise and sunset, etc. are due to scattering of light.

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