Question

Figure shows an over-head view of a corridor with a plane mirror $M$ mounted at one end. A burglar $B$ sneaks along the corridor directly toward the centre of the mirror. If $d=3.0\text{m}$, how far from the mirror will he be, when the security guard $S$ first observes him in the mirror ?

• A. $2\text{m}$
• B. $1.5\text{m}$
• C. $2\text{m}$
• D. $6\text{m}$

Answer 1

The correct option is B $1.5\text{m}$


The security person will see the burglar when the rays coming from the burglar gets reflected at the corner $C$ of the mirror.
Let at this point the burglar is at $B^{\prime }$
$\angle AC{B}^{\prime }=\angle DCS$ (Law of reflaction)
From properties of triangle, $\angle CSD={45}^{\circ }$
Also, $\angle A{B}^{\prime }C={45}^{\circ }$

In $\Delta A{B}^{\prime }C$ and $\Delta DSC$:
In $\angle A{B}^{\prime }C=\angle DSC$
$\angle CA{B}^{\prime }=\angle CDS$ (both ${90}^{\circ }$)
$\Rightarrow \Delta CA{B}^{{}^{\prime }}\sim \Delta CDS$ ($AAA$ similarity)
$\Rightarrow \frac{CA}{CD}=\frac{A{B}^{{}^{\prime }}}{DS}$
$\Rightarrow \frac{d}{2d}=\frac{A{B}^{{}^{\prime }}}{d}$
$\Rightarrow A{B}^{{}^{\prime }}=\frac{d}{2}=\frac{3}{2}=1.5\text{m}$
So, the security guard will see the burglar when burglar is $1.5\text{m}$ away from the mirror.

$\begin{array}{c}\text{Why this question?}\\ \text{Hint: Earliest view of the burglar will be seen when the}\\ \text{light gets reflected from the rightmost portion of the mirror.}\end{array}$