CBSE Solutions for Class 10 Physics

Select CBSE Solutions for class 10 Subject & Chapters Wise :

Why does the sun appear reddish early in the morning ?

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Answer :

The light coming from the sun passes through various denser layers of air in the earth’s atmosphere before reaching our eyes near the horizon. Most of the part of blue light and light of small wavelength gets scattered by dust particles near the horizon. So, the light reaching our eyes is of large wavelength. Due to this the sun appears reddish at the time of sunrise and sunset.

Explain why the planets do not twinkle ?

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Answer :

The planets are much nearer to the earth than stars and because of this they can be considered as large source of light. If a planet is considered to be a collection of a very large number of point sources of light, then the average value of change in the amount of light entering the eye from all point size light sources is zero. Due to this the effect of twinkling is nullified.

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

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Answer :

The eye lens of a normal eye forms the images of objects at various distances on the same retina. Therefore, the image distance in the eye remains the same.

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

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Answer :

At distance less than 25 cm, the ciliary muscles cannot bulge the eye lens any more, the object cannot be focused on the retina and it appears blurred to the eye, as shown in the given figure.

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 ?

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Answer :

The remedial lens should make the objects at infinity appear at the far point.
Therefore, for object at infinity, u = ∞
Far point distance of the defected eye, ν = – 80 cm

Negative sign shows that the remedial lens is a concave lens.

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 ?

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Answer :

(i) Power of distant viewing part of the lens, P1 = -5.5 D
Focal length of this part, f1 = 1p1 = 1−5.5 m = -0.182 m = -18.2 cm

(ii) For near vision,

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?

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Answer :

The child is suffering from myopia. The child should use concave lens of suitable focal length

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

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Answer :

For a human eye with normal vision the far point is at infinity and near point is 25 cm from the eye.

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 ?
 

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Answer :

Concave lens.

What is meant by power of accommodation of the eye ?

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Answer :

The power of accommodation of the eye is the maximum variation of its power for focusing on near and far (distant) objects.

Why does the sky appear dark instead of blue to an astronaut

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Answer :

The sky appears dark instead of blue to an astronaut, as scattering of light does not take place outside the earth’s atmosphere.

Why does the Sun appear reddish early in the morning?

 

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Answer :

White light coming from the sun has to travel more distance in the atmosphere before reaching the observer. During this, the scattering of all colored lights except the light corresponding to red color takes place and so, only the red colored light reaches the observer. Therefore, the sun appears reddish at sunrise and sunset.

Explain why the planets do not twinkle?

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Answer :

Unlike stars, planets don’t twinkle. Stars are so distant that they appear as pinpoints of light in the night sky, even when viewed through a telescope. Since all the light is coming from a single point, its path is highly susceptible to atmospheric interference (i.e. their light is easily diffracted).

Why do stars twinkle?

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Answer :

The twinkling of a star is due to atmospheric refraction of starlight. The starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index.

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

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Answer :

The image is formed on the retina even on increasing the distance of an object from the eye. The eye lens becomes thinner and its focal length increases as the object is moved away from the eye.

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

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Answer :

A normal eye is not able to see the objects placed closer than 25 cm clearly because the ciliary muscles of the eyes are unable to contract beyond a certain limit.

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?

 

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Answer :

The student is suffering from short-sightedness or myopia. Myopia can be corrected by the use of concave or diverging lens of an appropriate power.

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

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Answer :

The minimum distance of the object from the eye, which can be seen distinctly without strain is called the near point of the eye. For a normal person’s eye, this distance is 25 cm.

The far point of the eye is the maximum distance to which the eye can see objects clearly. The far point of a normal person’s eye is infinity.

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

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Answer :

An individual with a myopic eye should use a concave lens of focal length 1.2 m so that he or she can restore proper vision.

What is meant by power of accommodation of the eye?

 

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Answer :

The ability of the lens of the eye to adjust its focal length to clearly focus rays coming from distant as well from a near objects on the retina, is known as the power of accommodation of the eye.

The phenomenon of scattering of light by colloidal particles gives rise to …………..

 

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Answer :

Tyndall effect

The splitting of light into its component colours is called …………..

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Answer :

Dispersion

A person with ………….. can see nearby objects clearly but cannot see distant objects clearly.

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Answer :

Myopia

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

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Answer :

Accomodation

. ………….. regulates and controls the amount of light entering the eye.

 
 
 
 
 
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Answer :

Pupil

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?

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Answer :

The power (P) of a lens of focal length f is given by the relation

Power (P) = 1/f

(i) Power of the lens (used for correcting distant vision) = – 5.5 D

Focal length of the lens (f) = 1/Pf= 1/-5.5 = -0.181 m

The focal length of the lens (for correcting distant vision) is – 0.181 m.

(ii) Power of the lens (used for correcting near vision) = +1.5 D

Focal length of the required lens (f) = 1/P

= 1/1.5 = +0.667 m

The focal length of the lens (for correcting near vision) is 0.667 m.

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?

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Answer :

The individual is suffering from myopia. In this defect, the image is formed in front of the retina. Therefore, a concave lens is used to correct this defect of vision.

Object distance (u) = infinity = ∞

Image distance (v) = – 80 cm

Focal length = f

According to the lens formula,

A concave lens of power – 1.25 D is required by the individual to correct his defect.

Why does the sun appear reddish early in the morning?  [2011,2012]

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Answer :

At the time of shnrise, the rays of the sun have to travel a larger atmospheric distance. As the wavelength of red colour is the largest of all the colours of sunlight, most of the blue colour and other colours are scattered away. Only red . colour which is,,least scattered, enters into our eyes. Hence, the sun appears reddish at the time of sunrise.
 

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. [2011]

 

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Answer :

The required diagram is shown below:

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye ¡s 1 m. What ¡s the power of a lens required to correct this defect? Assume that near point of the normal eye is 25 cm.

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Answer :

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?

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Answer :

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?

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Answer :

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?

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Answer :

The far point for myopic eye is 1.2m.

Why do stars twinkle ?

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Answer :

Stars appear to twinkle due to atmospheric refraction. The light of star after the entry of light in earth’s atmosphere undergoes refraction continuously till it reaches the surface of the earth. Stars are far away. So, they are the point source of light. As the path of light coming from stars keep changing, thus the apparent position of stars keep changing and amount of light from stars entering the eye keeps twinkling. Due to which a star sometimes appear bright and sometimes dim, which is the effect of twinkling.

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.

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Answer :

An individual suffering from hypermetropia can see distinct objects clearly but he or she will face difficulty in clearly seeing objects nearby. This happens because the eye lens focuses the incoming divergent rays beyond the retina. This is corrected by using a convex lens. A convex lens of a suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.

The convex lens creates a virtual image of a nearby object (N’ in the above figure) at the near point of vision (N) of the individual suffering from hypermetropia.

The given individual will be able to clearly see the object kept at 25 cm (near point of the normal eye), if the image of the object is formed at his near point, which is given as 1 m.

Object distance, u= – 25 cm

Image distance, v= – 1 m = – 100 m

Focal length, f

Using the lens formula,

A convex lens of power +3.0 D is required to correct the defect.

Match the following columns

Column I

Column II

(a) Rainbow formation

(i) A person cannot see nearby objects clearly

(b) Twinkling of stars

(ii) Regulates and controls amount of light entering the eye

(c) Blue colour of sky

(iii) Scattering of light

(d) Myopia

(iv) Dispersion

(e) Hypermetropia

(v) Infinity

(f) Least distance of distinct vision

(vi) Membrane which controls the size of pupil

(g) Far point of eye

(vii) Atmospheric refraction

(h) Iris

(viii) Twenty five centimeters

(i) Pupil

(ix) A person cannot see distant objects distinctly

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Answer :

Column I

Column II

(a) Rainbow formation

(iv) Dispersion

(b) Twinkling of stars

(vii) Atmospheric refraction

(c) Blue colour of sky

(iii) Scattering of light

(d) Myopia

(ix) A person cannot see distant objects distinctly

(e) Hypermetropia

(i) A person cannot see nearby objects clearly

(f) Least distance of distinct vision

(viii) Twenty five centimeters

(g) Far point of eye

(v) Infinity

(h) Iris

(vi) Membrane which controls the size of pupil

(i) Pupil

(ii) Regulates and controls amount of light entering the eye

a. Lens used for correction of myopia
 

Spectrum
 

b. The phenomenon of scattering of light by colloidal particles .

Infinity

 

c. . The band of coloured components of a light beam
 

Pupil

d. Part of eye which regulates and controls the amount of light entering the eye
 

Concave lens
 

e. Lens used in correction of presbyopia

Red

f. Lens used in correction of hypermetropia

Convex lens
 

g. Colour used for danger signal
 

Tyndall effect
 

h. Farthest point for a normal eye

 Bifocal lens
 

 

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Answer :

a. Concave lens
b. Tyndall effect
c. Spectrum
d. Pupil
e. Bifocal lens
f. Convex lens
g. Red
h. Infinity

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The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.

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