An object 5 cm tall was placed in front of a spherical mirror at a 20 cm distance from the mirror. If a virtual image, 10 cm tall, was formed behind the mirror, found the focal length of the mirror and the type of mirror.

The Focal length of the mirror is

- 40 cm .

Given,
Distance of the object from the mirror,

= - 20 cm

; The height of the object,

h = + 5 cm,

and height of the virtual and erect image,

h ’ = + 10 cm

m = \frac{h^{'}}{h} = - \frac{v}{u},

hence

\frac{+ 10}{+ 5} = - \frac{v}{- 20}

\Rightarrow v = + 40 cm

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

= \frac{1}{+ 40} + \frac{1}{- 20}

= - \frac{1}{40}

f = - 40 cm

Thus, the mirror is a concave mirror with a focal length

40 cm

, and the image is formed

40 cm

behind the mirror.

An object $5 cm$ tall was placed in front of a spherical mirror at a $20 cm$ distance from the mirror. If a virtual image, $10 cm$ tall, was formed behind the mirror, found the focal length of the mirror and the type of mirror.