CBSE Class X

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:

The person is suffering from an eye defect called myopia. In this defect, the image

is formed in front of the retina. Hence, 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 person to correct his defect

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:

For distant vision = −0.181 m, for near vision = 0.667 m

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

 

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

Focal length of the required lens, f = 1/P

 

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

 

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

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) The relaxation or contraction of ciliary muscles changes the curvature of the eye

lens. The change in curvature of the eye lens changes the focal length of the eyes.

Hence, the change in focal length of an eye lens is caused by the action of ciliary

muscles.

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:

(c) The least distance of distinct vision is the minimum distance of an object to

see clear and distinct image. It is 25 cm for a young adult with normal visions.

Question 2:

The human eye forms the image of an object at its

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

Answer:

(d) The human eye forms the image of an object at its retina.

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) Human eye can change the focal length of the eye lens to see the objects

situated at various distances from the eye. This is possible due to the power of

accommodation of the eye lens.

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:

A student has difficulty in reading the blackboard while sitting in the last row. It

shows that he is unable to see distant objects clearly. He is suffering from myopia.

This defect can be corrected by using a concave lens.

Question 3:

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

Answer:

The near point of the eye is the minimum distance of the object from the eye,

which can be seen distinctly without strain. For a normal human eye, this distance

is 25 cm. The far point of the eye is the maximum distance to which the eye can

see the objects clearly. The far point of the normal human eye is infinity.

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?

Answer:

The person is able to see nearby objects clearly, but he is unable to see objects

beyond 1.2 m. This happens because the image of an object beyond 1.2 m is

formed in front of the retina and not at the retina, as shown in the given figure.

 

To correct this defect of vision, he must use a concave lens. The concave lens will

bring the image back to the retina as shown in the given figure.

 

Question 1:

What is meant by power of accommodation of the eye?

Answer:

When the ciliary muscles are relaxed, the eye lens becomes thin, the focal length

increases, and the distant objects are clearly visible to the eyes. To see the nearby

objects clearly, the ciliary muscles contract making the eye lens thicker. Thus, the

focal length of the eye lens decreases and the nearby objects become visible to the

eyes.Hence, the human eye lens is able to adjust its focal length to view both

distant and nearby objects on the retina. This ability is called the power of

accommodation of the eyes.

Question 17:

A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of

the lens. Is the prescribed lens diverging or converging?

Answer:

 

A convex lens has a positive focal length. Hence, it is a convex lens or a converging

lens.

Question 16:

Find the focal length of a lens of power −2.0 D. What type of lens is this?

Answer:

 

A concave lens has a negative focal length. Hence, it is a concave lens.

Question 15:

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal

length 18 cm. At what distance from the mirror should a screen be placed, so that

a sharp focused image can be obtained? Find the size and the nature of the image.

Answer:

Object distance, u = −27 cm

Object height, h = 7 cm

Focal length, f = −18 cm

According to the mirror formula,

 

The screen should be placed at a distance of 54 cm in front of the given mirror.

 

The negative value of magnification indicates that the image formed is real.

 

The negative value of image height indicates that the image formed is inverted.

Question 14:

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex

mirror of radius of curvature 30 cm. Find the position of the image, its nature

and size.

Answer:

Object distance, u = −20 cm

Object height, h = 5 cm

Radius of curvature, R = 30 cm

Radius of curvature = 2 × Focal length

R = 2f

f = 15 cm

According to the mirror formula,

 

The positive value of v indicates that the image is formed behind the mirror.

 

The positive value of magnification indicates that the image formed is virtual.

 

The positive value of image height indicates that the image formed is erect.

Therefore, the image formed is virtual, erect, and smaller in size.

Question 13:

The magnification produced by a plane mirror is +1. What does this mean?

Answer:

Magnification produced by a mirror is given by the relation

The magnification produced by a plane mirror is +1. It shows that the image formed

by the plane mirror is of the same size as that of the object. The positive sign

shows that the image formed is virtual and erect.

Question 12:

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm.

Find the position and nature of the image.

Answer:

Focal length of convex mirror, f = +15 cm

Object distance, u = −10 cm

According to the mirror formula,

 

The positive value of v indicates that the image is formed behind the mirror.

 

The positive value of magnification indicates that the image formed is virtual and

erect.

Question 11:

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far

is the object placed from the lens? Draw the ray diagram.

Answer:

Focal length of concave lens (OF₁), f = −15 cm

Image distance, v = −10 cm

According to the lens formula,

 

The negative value of u indicates that the object is placed 30 cm in front of the lens.

This is shown in the following ray diagram.

 

Question 10:

An object 5 cm in length is held 25 cm away from a converging lens of focal length

10 cm. Draw the ray diagram and find the position, size and the nature of the image

formed.

Answer:

Object distance, u = −25 cm

Object height, h_o = 5 cm

Focal length, f = +10 cm

According to the lens formula,

 

The positive value of v shows that the image is formed at the other side of the lens.

 

The negative sign shows that the image is real and formed behind the lens.

 

The negative value of image height indicates that the image formed is inverted.

The position, size, and nature of image are shown in the following ray diagram.

 

Question 9:

One-half of a convex lens is covered with a black paper. Will this lens produce a

complete image of the object? Verify your answer experimentally. Explain your

observations.

Answer:

The convex lens will form complete image of an object, even if its one half is covered

with black paper. It can be understood by the following two cases.

Case I

When the upper half of the lens is covered

In this case, a ray of light coming from the object will be refracted by the lower half

of the lens. These rays meet at the other side of the lens to form the image of the

given object, as shown in the following figure.

 

Case II

When the lower half of the lens is covered

In this case, a ray of light coming from the object is refracted by the upper half of

the lens. These rays meet at the other side of the lens to form the image of the

given object, as shown in the following figure.

 

Question 8:

Name the type of mirror used in the following situations.

(a) Headlights of a car

(b) Side/rear-view mirror of a vehicle

(c) Solar furnace

Support your answer with reason.

Answer:

(a) Concave (b) Convex (c) Concave

Explanation

(a) Concave mirror is used in the headlights of a car. This is because concave mirrors

can produce powerful parallel beam of light when the light source is placed at their

principal focus.

(b) Convex mirror is used in side/rear view mirror of a vehicle. Convex mirrors give a

virtual, erect, and diminished image of the objects placed in front of it. Because of

this, they have a wide field of view. It enables the driver to see most of the traffic

behind him/her.

(c) Concave mirrors are convergent mirrors. That is why they are used to construct

solar furnaces. Concave mirrors converge the light incident on them at a single point

known as principal focus. Hence, they can be used to produce a large amount of heat

at that point.