Question

For the arrangement shown in the figure, find the final magnification of the extended object $AB$ after two successive reflections. Take the first reflection from mirror $M_1$.

• A. $+1$
• B. $+2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is B $+2$

For mirror $M_1$,
$u=-30\text{cm}$
$f=-20\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{30}=\frac{-1}{20}$
$\Rightarrow \frac{1}{v}=\frac{-1}{60}$
$\Rightarrow v=-60\text{cm}$
So, $m_1=\frac{-v}{u}=\frac{-(-60)}{-30}=-2$
For mirror $M_2$,
The image formed by mirror $M_1$ will act as a virtual object for mirror $M_2$,
$u=+(60-40)=+20\text{cm}$
$f=+10\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{1}{10}$
$\Rightarrow \frac{1}{v}=\frac{1}{20}$
$\Rightarrow v=20\text{cm}$
So, $m_2=\frac{-v}{u}=\frac{-20}{20}=-1$
Now, final magnification of the extended object,
$m=m_1{m}_2=-2\times -1=+2$

Why this question ? Tip: Image formed by one mirror act as object for other mirror.