Question

In the figure shown find the total magnification after two successive reflections first on $M_1$ and then on $M_2$.

• A. 1
• B. -2
• C. 2
• D. -1

Answer 1

The correct option is B -2

For reflection at $M_1$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}or\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-20}+\frac{1}{30}\therefore v=\frac{-20\times 30}{10}=-60cm$

which means image formed which will act as object on $M_2$ will be $20m$ left to $M_2$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}or\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{10}-\frac{1}{20}\therefore v=\frac{10\times 20}{10}=20cm$

$\therefore magnification=M_1\times M_2=-\left(\frac{-60}{-30}\right)\times -\left(\frac{20}{20}\right)=-2$