NCERT Solutions Class 10 Science Chapter 11 Electricity

Q.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.

Ans-

Answer: The correct option is (b).
Explanation: Power of accommodation is the ability of eye lens to change its focal length so that the image of an object can be focused on the retina.

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

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

Ans-

The correct option is (d).
Explanation: In the case of human eye, image is formed at retina.

Q.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.

Ans-

The correct option is (c).
Explanation: Least distance of distinct vision is the smallest distance at which the human eye can see the objects clearly without any strain. For a normal eye, it is 25 cm.

Q.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.

Ans-

The correct option is (c).
Explanation: The curvature of the eye lens can be changed by the relaxation or contraction of ciliary muscles. The change in curvature of the eye lens changes the focal length of the eyes. Thus, the change in focal length of an eye lens is caused by the action of ciliary muscles.

Q.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?

Ans-

\begin{array}{l}\left(\text{i}\right)\text{Power}\text{of}\text{the lens used for correcting distance vision= -5.5}\\ \text{Focal length}\left(\text{f}\right)\text{of the required lens}\\ \text{f =}\frac{\text{1}}{\text{P}}\\ \text{f}\text{=}\frac{\text{1}}{\text{-5}\text{.5}}\text{=-0}\text{.181}\text{m}\\ \text{Hence, the focal length of the lens for correcting distant vision = – 0.181 m.}\\ \left(\text{ii}\right)\text{}\text{Power}\text{}\text{of the lends used for correcting near vision = +1}\text{.5}\text{D}\\ \text{Focal length (f) of the required lens,}\\ \text{f =}\frac{\text{1}}{\text{P}}\\ \text{f}\text{=}\frac{\text{1}}{\text{1}\text{.5 D}}\text{=}\text{+}\text{0}\text{.667}\text{m}\\ \text{Hence, the focal length of the lens for correcting near vision is 0}\text{.667}\text{m}\text{.}\end{array}

Q.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?

Ans-

\begin{array}{l}\text{Given,}\text{u}\text{=}\text{¥}\\ \text{v}\text{=}\text{-80}\text{cm}\\ \text{On}\text{using lens formula,}\\ \frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \text{–}\frac{\text{1}}{\text{80}}\text{–}\frac{\text{1}}{\text{¥}}\text{=}\frac{\text{1}}{\text{f}}\\ \text{or,}\text{f}\text{=}\text{-80}\text{cm}\text{=}\text{-0}\text{.8}\text{m}\\ \text{Now,}\text{P}\text{=}\frac{\text{1}}{\text{f}\left(\text{metres}\right)}\\ \text{P}\text{=}\frac{\text{1}}{\text{-0}\text{.8}}\text{=-1}\text{.25}\text{D}\text{.}\\ \text{A}\text{concave lens of power -1}\text{.25 D is required by the person to correct the defect}\text{.}\end{array}

Q.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.

Ans-

It is a visual defect in which an eye cannot see the nearby objects clearly. The image of the object, in this case, is formed beyond the retina and hence person experiences difficulty in understanding the object.

This defect of vision can be corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina. The convex lens actually creates a virtual image of a nearby object (o’) at the near point of vision (o) of the person suffering from hypermetropia.
A person can clearly see the object kept at 25 cm, if the image of the object is formed at his near point, which is given as 1 m.

\begin{array}{l}\text{Given,}\text{u}\text{=}\text{-25}\text{cm}\\ \text{v}\text{=}\text{-1}\text{m=-10}\text{cm}\\ \text{On}\text{usig}\text{lens}\text{formula,}\\ \frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \frac{\text{1}}{\text{-100}}\text{–}\frac{\text{1}}{\text{-25}}\text{=}\frac{\text{1}}{\text{f}}\\ \text{or,}\text{f=}\frac{\text{100}}{\text{3}}\text{cm}\text{=33}\text{.3}\text{cm}\text{=}\text{0}\text{.33}\text{m}\\ \text{Now,}\text{P}\text{=}\frac{\text{1}}{\text{f}\left(\text{metres}\right)}\\ \therefore \text{P=}\frac{\text{1}}{\text{0}\text{.33m}}\text{=+3D}\text{.}\\ \text{Hence,}\text{a}\text{convex}\text{lens}\text{of}\text{power}\text{+3D}\text{is}\text{required}\text{to}\text{correct}\text{the}\text{defect}\text{.}\end{array}

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

Ans-

As the ciliary muscles of eyes are incapable to contract beyond a certain limit hence, a normal eye cannot see clearly the objects kept closer than 25 cm. If an object is kept at a distance less than 25 cm from the eye, then the object seems blurred and produces strain in the eyes.

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

Ans-

For an eye, the image distance always remains constant as the size of eyes cannot increase or decrease. The increase in the object distance is balanced by the change in the focal length of the eye lens. The eye changes its focal length focal length in such a way that the image is always formed at retina.

Q.10 Why do stars twinkle?

Ans-

Twinkling of stars is due to the atmospheric refraction of light. As stars are too far from the earth therefore, they can be considered as point sources of light. When the light coming from stars passes through the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star light refracted by the atmosphere, it comes more towards us. It appears brighter when it comes less towards us.

Q.11 Explain why the planets do not twinkle.

Ans-

Planets do not twinkle like stars because planets are much closer to the earth and are thus seen as extended objects. If we consider planet as a collection of large number of point-sized sources of light, sources of light the variation in amount of light entering our eye from all the individual point sized sources will average out to zero, thereby nullifying the twinkling effect.

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

Ans-

During sunrise, the light coming from the Sun has to travel a larger distance in the earth’s atmosphere before entering to our eyes. During this, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. As blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour reaches to our eyes after the atmospheric scattering of light. Hence, the Sun seems reddish while sunrise.

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

Ans-

There is no atmosphere in the outer space. Due to this, the sunlight does not scatter and hence, no scattered light reaches to the eyes of astronauts in the space and hence, the sky appears black to them.