What will be the focal length of an object 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high?

The focal length of an object

3 cm

high is placed at a distance of

8 cm

from a

Concave Mirror

which produces a virtual image

4.5 cm

high will be

24 cm .

Given,
Distance of the object from the mirror,

u = - 8 cm

Height of the object

{, h}_{0} = 3 cm

Height of the image,

h_{i} = 4.5 cm

Since ho and hi are given

m = \frac{h_{i}}{h_{o}}

\frac{h_{i}}{h_{o}} = - \frac{v}{u}

\frac{4.5}{3} = - \frac{v}{(- 8)}

3 v = 4.5 \times 8

v = + 12 cm

Now using the mirror formula;

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

\frac{1}{f} = - \frac{1}{8} + (\frac{1}{12})

\frac{1}{f} = \frac{1}{24}

f = 24 cm

What will be the focal length of an object $3 cm$ high is placed at a distance of $8 cm$ from a concave mirror which produces a virtual image $4.5 cm$ high?