Light Reflection and Refraction class 10 science NCERT Exercise Solution Part 2

# Light

## NCERT Solution 2

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?

Answer: Power P = -2.0D
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.