In the figure shown find the total magnification after two successive reflections first on $M_{1} &$ then on $M_{2}$ is

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Solution

For $M_{1}$ ,
We have

$u_{1} = - 30 c m, f_{1} = 20 c m$
$\frac{1}{v_{1}} + \frac{1}{u_{1}} = \frac{1}{f_{1}}$
$\Rightarrow \frac{1}{v_{1}} + \frac{1}{- 30} = \frac{1}{- 20}$

$\Rightarrow \frac{1}{v_{1}} = \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} = \frac{- 1}{60}$

$v_{1} = - 60 c m$

For $M_{2}$ ,
Image formed by $M_{1}$ will act as the object for the $M_{2}$
So, We have

$u_{2} = + (60 - (10 + 30)) = + 20 c m$
$f_{2} = + 10 c m$
$\frac{1}{v_{2}} + \frac{1}{20} = \frac{1}{10}$
$\Rightarrow v_{2} = + 20 c m$

Now for $M_{1}$ ,
$m_{1} = - \frac{v_{1}}{u_{1}}$
For $M_{2}$ ,
$m_{2} = - \frac{v_{2}}{u_{2}}$
We know,
Total $M_{T} = m_{1} \times m_{2} = \frac{v_{1}}{u_{1}} \times \frac{v_{2}}{u_{2}}$
Putting all the values
$M_{T}$ = $\frac{(- 60) (+ 20)}{(- 30) (+ 20)}$
$= + 2$
Hence, total magnification will be +2

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