Describe the nature of image formed when the object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm.

Given:

Object distance, $u$ = $-$20 m     (negative, due to sign convention of mirror)

Focal length, $f$ = $-$10 m            (negative, due to sign convention of mirror)

To find: Image distance, $v$.

Solution:

From the mirror formula, we know that-

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

$\Rightarrow \frac{1}{v}+\frac{1}{(-20)}=\frac{1}{(-10)}$

$\Rightarrow \frac{1}{v}-\frac{1}{20}=-\frac{1}{10}$

$\Rightarrow \frac{1}{v}=-\frac{1}{10}+\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{-2+1}{20}$

$\Rightarrow \frac{1}{v}=\frac{-1}{20}$

$\Rightarrow v=-20m$

Hence, the image distance is -20m.

Here, the negative sign represents that the image is formed on the left side of the mirror. Thus, it is a real and inverted image, which is formed at a distance of 20 m from the mirror.