Question

For the given system of mirrors, the image formed after two reflections is on the source $S$ itself. Then, the distance between the two mirrors is (Take first reflection from mirror $M_2$)

• A. $70\text{cm}$
• B. $60\text{cm}$
• C. $40\text{cm}$
• D. $50\text{cm}$

Answer 1

The correct option is D $50\text{cm}$

Let us suppose, distance of source $S$ from $M_2$ is $d$.
For first reflection from $M_2$,
$u=-d;f=+20\text{cm}$

From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}-\frac{1}{d}=\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{1}{d}+\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{20+d}{20d}$

$\Rightarrow v=\frac{20d}{20+d}$


Now for mirror $M_1$, image $I$ will act as a virtual object.
So, $u=-(30+d+\frac{20d}{20+d}).....\left(1\right)$
Given, image formed after second reflection is at $S$ itself.
$\Rightarrow v=-30\text{cm}$
& $f=-20\text{cm}$

From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{-30}+\frac{1}{u}=\frac{1}{-20}\Rightarrow \frac{1}{u}=\frac{20-30}{600}$

$\Rightarrow u=-60\text{cm}.......\left(2\right)$

From (1) and (2),

$30+d+\frac{20d}{20+d}=60$

$\Rightarrow d^2+10d-600=0$

$\Rightarrow d^2-20d+30d-600=0$

$\Rightarrow (d-20)(d+30)=0$

$\Rightarrow d=20\text{cm}\text{or}d=-30\text{cm}$

Distance cannot be negative.
So, $d=20\text{cm}$

Hence, distance between the mirrors is $30+d=30+20=50\text{cm}$

So, option $\left(d\right)$ is correct.