Question

Find the position of the final image of the object $O$ after three successive reflections taking first reflection on $m_1$.

• A. $12.5\text{cm}$ in front of mirror $m_1$.
• B. $10\text{cm}$ in front of mirror $m_1$.
• C. $12.5\text{cm}$ in front of mirror $m_2$.
• D. $10\text{cm}$ in front of mirror $m_2$.

Answer 1

The correct option is A $12.5\text{cm}$ in front of mirror $m_1$.

For first reflection at mirror $m_1$,
$u=-15\text{cm}$
$f=\frac{-20}{2}=-10\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{15}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-1}{30}$
$\Rightarrow v=-30\text{cm}$
This first image will act as the object for the plane mirror $m_2$.
For second reflection at plane mirror $m_2$,
$u=-(40-30)=-10\text{cm}$
So, $v=+10\text{cm}$
This second image will act as the virtual object for the mirror $m_1$.
For third reflection at mirror $m_1$,
$u=-(40+10)=-50\text{cm}$
$f=\frac{-20}{2}=-10\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{50}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-2}{25}$
$\Rightarrow v=-12.5\text{cm}$
Thus, the final image is formed at $12.5\text{cm}$ in front of mirror $m_1$.