# Question #08234

Jan 30, 2017

We know the mirror formula as below

$\textcolor{b l u e}{\frac{1}{v} + \frac{1}{u} = \frac{2}{r}}$

where

$u \to \text{object distance} = - 20 c m$

$v \to \text{image distance}$

$r \to \text{radius of curvature}$

Here it is also given that size of the virtual image seen behind the mirror is twice that of object.This means the magnification $m = - 2$

Again $m = \frac{v}{u} \implies - 2 = \frac{v}{u}$

So $v = - 2 u = - 2 \times \left(- 20\right) c m = 40 c m$

Inserting these values in mirror equation we get

$\textcolor{g r e e n}{\frac{1}{v} + \frac{1}{u} = \frac{2}{r}}$

$\implies \frac{1}{40} + \frac{1}{-} 20 = \frac{2}{r}$

$\implies \frac{1 - 2}{40} = \frac{2}{r}$

$\implies r = - 80 c m$