Class 10 Physics

# Light: NCERT Exercise Solution

We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer: When object is between principal focus and pole of a concave mirror, an erect, enlarged and virtual image is formed. So, we need to keep the object at a distance which is less than 15 cm (the given focal length).

Which one of the following materials cannot be used to make a lens?

• Water
• Glass
• Plastic
• Clay

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

• Between the principal focus and the centre of curvature
• At the centre of curvature
• Beyond the centre of curvature
• Between the pole of the mirror and its principal focus.

Answer: (d) Between the pole of the mirror and its principal focus

Where should an object be placed in front of a convex lens to get a real image of the size of the object?

• At the principal focus of the lens
• At twice the focal length
• At infinity
• Between the optical centre of the lens and its principal focus.

Answer: (b) At twice of the focal length

A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

• both concave
• both convex
• the mirror is concave and the lens is convex
• the mirror is convex, but the lens is concave

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

• plane
• concave
• convex
• either plane or convex

Answer: (d) either plane or convex

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

• A convex lens of focal length 50 cm
• A concave lens of focal length 50 cm
• A convex lens of focal length 5 cm
• A concave lens of focal length 5 cm

Answer: (c) A convex lens of focal length 5 cm

Name the type of mirror used in the following situations.

Answer: When a source of light is kept at focus of a concave mirror, the reflected rays form a parallel beam of light and go to infinity. Hence, concave mirror is used as reflector in the headlights. This helps in getting a parallel beam of light.
• Side/rear-view mirror of a vehicle.

Answer: A convex mirror can show image of a wider area, because of its wide field of view. This enables the driver to see more of traffic coming from behind. Hence, convex mirror is used as rear-view mirror in vehicles.
• Solar furnace.

Answer: The light rays coming from infinity converge at the focus after reflection from a concave mirror. Hence, a concave mirror is used in solar furnace because it helps in concentrating the solar energy at a point.

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: Covering the half portion of a convex lens will not affect the image making ability of the lens. The following two figures illustrate this.

Condition 1: When upper half of the lens is covered:

Condition 2: When the lower half of the lens is covered

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: Given, height of object = 5cm

Position of object, u = - 25cm

Focal length of the lens, f = 10 cm

Hence, position of image, v =?

We know that;

1/v-1/u=1/f

Or, 1/v+(1)/(25)=(1)/(10)

Or, 1/v=(1)/(10)-(1)/(25)

Or, 1/v=(5-2)/(50)

Or, 1/v=(3)/(50)

Or, v=(50)/(3)=16.66 cm

Thus, distance of image is 16.66 cm on the opposite side of lens.

Now, magnification = v/u

Or, m=(16.66 cm)/(-25 cm)=-0.66

m=(h_i)/(h_o)

Or, -0.66=(h_i)/(5 cm)

Or, h_i=-0.66xx5 cm=-3.3 cm

The negative sign of height of image shows that an inverted image is formed.

Thus, position of image = At 16.66 cm on opposite side of lens

Size of image = - 3.3 cm at the opposite side of lens

Nature of image – Real and inverted.

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: Given, focal length, f = - 15 cm
Distance of image, v = - 10 cm
Distance of object, v =?
Object distance can be calculated using the lens formula

1/v-1/u=1/f

Or, (1)/(-10)-1/u=(1)/(-15)

Or, -1/u=-(1)/(15)+(1)/(10)

Or, -1/u=(-2+3)/(30)=(1)/(30)

Or, u=-30 cm

Negative sign shows that object is at 30cm in front of the lens.

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: Given, distance of object, u = -10cm
Focal length, f = 15cm
Distance of image, v =?; which can be calculated as follows:

1/v-1/u=1/f

Or, 1/v-(1)/(10)=(1)/(15)

Or, 1/v=(1)/(15)+(1)/(10)

Or, 1/v=(5)/(30)

Or, v=6 cm

The positive sign of image shows that image is formed at the other side of lens at 6 cm

Magnification can be calculated as follows:
m = v/u
Or, m = (-6  cm)/(-10  cm) = 3/5  cm = 0.6  cm

The positive sign of magnification shows that image is erect. Image is erect and virtual and is formed 6 cm behind the mirror.

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

Answer: The magnification of +1 means that the image size is same as object size.

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: Given, distance of object, u = - 20 cm
Height of object = 5.0 cm
Radius of curvature, R = 30cm.
Hence, focal length = R/2 = 30/2 cm = 15 cm
Distance of image, v =?; which can be calculated as follows:

1/v-1/u=1/f

Or, 1/v-(1)/(20)=(1)/(15)

Or, 1/v=(1)/(15)+(1)/(20)

Or, 1/v=(4+3)/(60)

Or, v=8.57 cm

m=-(8.57 cm)/(-20 cm)=0.428

m=(h_i)/(h_o)

Or, 0.428=(h_i)/(5 )

Or, h_i=0.428xx5 cm=2.14 cm

The positive sign of height of image is erect. Therefore, position of image 8.57 cm behind the mirror.

Nature of image: erect and virtual.

Size of image: 2.14 cm, this means image is smaller than object.

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 focussed image can be obtained? Find the size and the nature of the image.

Answer: Given, size of object, h = 7.0 cm
Distance of object, u = - 27 cm
Focal length, f = - 18 cm
Distance of image, v =?; which can be calculated as follows:

1/v-1/u=1/f

Or, 1/v-(1)/(27)=-(1)/(18)

Or, 1/v=-(1)/(18)+(1)/(27)

Or, 1/v=(-3+2)/(54)

Or, 1/v=-(1)/(54)

Or, v=-54 cm

m=-v/u

=(-54)/(-27)=-2

Height of image can be calculated as follows:

m=(h_i)/(h_o)

Or, -2=(h_i)/(7)

Or, h_i=-14 cm

Negative sign of height of image shows, that image in inverted. Thus, screen should be placed at 54 cm in front of mirror.

The size of image = 14cm

Nature of image – real and inverted.

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

Since, P = 1/f
Hence, -2.0 D = 1/f
Or, f = (1)/(-2D) = -0.5  m

Since focal length is negative hence, it is a concave lens.

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: Given, power = +1.5 D
Since, P = 1/f
Hence, 1.5 D = 1/f
Or, f = (1)/(1.5 D) = 0.66  m

The positive sign shows it is a converging lens.