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# 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.