# Light

## NCERT Exercise

### Part 2

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:** 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 behind the mirror 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.

Question 13: 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.

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:** 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.

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 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.

Question 16: 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.

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:** 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.