Question

Two concave mirrors are placed $40\text{cm}$ apart and are facing each other. A point object lies between them at a distance $12\text{cm}$ from the mirror $M-1$. Find the location of the final image formed after two reflections, first at $M-1$ and second at $M-2$.

• A. $\frac{60}{7}\text{cm}$ left of $M-2$
• B. $\frac{60}{7}\text{cm}$ right of $M-2$
• C. $\frac{60}{7}\text{cm}$ right of $M-1$
• D. $\frac{60}{7}\text{cm}$ left of $M-1$

Answer 1

The correct option is A $\frac{60}{7}\text{cm}$ left of $M-2$

For first reflection at mirror $M-1$,
$u=-12\text{cm}$
$f=-10\text{cm}$
$[$as per sign convention$]$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{12}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-1}{60}$
$\Rightarrow v=-60\text{cm}$
This first image will act as the virtual object for the mirror $M-2$.
For second reflection at mirror $M-2$,
$u=+(60-40)=+20\text{cm}$
$f=-15\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{-1}{15}$
$\Rightarrow \frac{1}{v}=\frac{-35}{300}$
$\Rightarrow v=\frac{-60}{7}\text{cm}$
Thus, the final image is formed at $\frac{60}{7}\text{cm}$ left of $M-2$.