# Illustrations

An object of height 4 cm is placed at 30 cm from a concave mirror of focal length 20 cm. Find the position of the image. Comment on the nature and size of the image.

Given, h_{o} = 4 cm, u = +30 cm, f = + 20 cm

To find: v, h_{i} and nature of image.

From mirror formula, we have

Magnification,

or

Thus, image is formed at a distance of 60 cm in front of the mirror. Height of the image is 8 cm. As v is positive, the image is real. Since h_{i} is negative, it is inverted.

A point source of light is kept in front of a convex mirror of radius of curvature 40 cm. The image is formed at 10 cm behind the mirror. Calculate the object distance.

Given: For a convex mirror, R = â€“40 cm.

v = â€“10 cm (image is virtual).

From mirror formula, we have

= +20 cm.

Thus, object is placed 20 cm in front of the mirror.

The refractive index of glass is 1.5. The speed of light in vacuum is 3 10^{8} m s^{â€“1}. Calculate the speed of light in glass.

Given, refractive index of glass, n_{g} = 1.5

Velocity of light in vacuum, c = 3 10^{8} m s^{â€“1}

Velocity of light in glass, v_{g} = ?

Refractive index of glass with respect to air or vacuum is

i.e., v_{g} = 2 10^{8} m s^{â€“1}.

A ray of light travels from water to glass. The angle of incidence is 40Â°. Calculate the angle of refraction and the deviation produced. Given n_{g} = 1.5, n_{w} = 1.3.

i = 40Â°, r = ?, n_{g} = 1.5, n_{w} = 1.3

Since, light is travelling from water to glass, we have

i.e.

*n*g sin *r* = *n*_{w} sin *i*

sin r =

âˆ´ r = sin^{â€“1} (0.5570) = 33Â° 51â€²

Thus, angle of refraction is 33Â° 51â€².

Deviation, d = i â€“ r = 40Â° âˆ’ 33Â° 51â€²

âˆ´ d = 6Â° 9â€².

At what distance should an object be placed in front of a convex lens of focal length 0.4 m so that the image is thrice the size of the object?

Given f = 0.4 m, h_{i} = 3h_{o}, u = ?

We know that,

Also from lens formula, we have

i.e., object distance, u = 0.267 m.

The object must be placed at 0.267 m from the lens.

The eye lens has a focal length of 2.5 cm. Find the range of accommodation while changing the focus from infinity to D = 25 cm.

For the object at infinity, u_{1} = âˆž,

v_{1} = f_{1} = 2.5 cm

For the object at 25 cm, u_{2} = 25 cm,

v_{2} = 2.5 cm, f_{2} = ?

= 44 dioptre.

Thus, the change in power, i.e., the range of accommodation is p_{2} âˆ’ p_{1} = 4 dioptre.