What will be the size of the image when an object of 5 cm in size is placed at a distance of 20 cm from a converging mirror of a focal length of 15 cm?

The size of the image will be

15 cm

when an object of

5 cm

in size is placed at a distance of

20 cm

from a converging mirror of a focal length of

15 cm

.
Given,

f =

-15 cm

u = - 20 cm

h' = 5 cm

Substituting the given values in mirror formula we get,

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

\Rightarrow \frac{1}{(- 15)} = \frac{1}{v} + \frac{1}{- 20}

\Rightarrow \frac{1}{v} = \frac{3 - 4}{60}

\Rightarrow \frac{1}{v} = - \frac{1}{60}

\Rightarrow v = - 60 cm

Now, substituting the value of

v

u

and

h ’

in magnification formula, we get

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

\Rightarrow \frac{h}{5} = - \frac{(- 60)}{(- 20)}

\Rightarrow h = - 3 \times 5

\Rightarrow h = - 15 cm

So, the height of the image will be

15 cm

and the negative sign shows that the image formed will be below the principal axis.

What will be the size of the image when an object of $5 cm$ in size is placed at a distance of $20 cm$ from a converging mirror of a focal length of $15 cm$ ?