What is the focal length of a converging mirror which forms a real image of height 4 cm to that of an object of height 1cm placed at 20 cm away from the mirror?

The focal length of a converging mirror which forms a real image of height

4 cm

to that of an object of height

1 cm

placed at

20 cm

away from the mirror will be

16 cm

.
Given,
Distance of the object from the mirror,

u = - 20 cm

Height of the object

{, h}_{0} = 1 cm

Height of the image,

h_{i} = 4 cm

Since

h_{o}

and

h_{i}

is given by using the magnification formula.

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

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

\frac{- 4}{1} = - \frac{v}{(- 20)}

v = - 80 cm

Using the mirror formula;

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

\frac{1}{- 80} + \frac{1}{- 20} = \frac{1}{f}

Therefore

f = - 16 cm

.
The focal length (

f

) of the converging mirror is

16 cm

What is the focal length of a converging mirror which forms a real image of height $4 cm$ to that of an object of height $1 cm$ placed at $20 cm$ away from the mirror?