A point source of light $S$ is placed at a distance $L$ in front of the centre of plane mirror of length$\left(d\right)$. A man walks in front of the mirror along a line parallel to the mirror, at a distance $2L$ as shown below. The distance over which the man can see the image of the light source in the mirror is

The correct option is C $3d$
The maximum distance over which man can see image of source $S$ will correspond to the rays reflected from top most and bottom-most edge of mirror.


From the ray diagram shown,
$\Delta S^{\prime }{A}^{\prime }N$ and $\Delta S^{\prime }AC$ are similar.
$\Rightarrow \frac{S^{\prime }N}{S^{\prime }C}=\frac{A^{\prime }N}{AC}$
$\Rightarrow \frac{L}{3L}=\frac{\left(d/2\right)}{AC}$
$\Rightarrow AC=\frac{3d}{2}$
From symmetry of incident ray,
$\Rightarrow AC=CB$ or $AB=2AC$
$\therefore AB=2\times \frac{3d}{2}=3d$
Hence, option $\left(c\right)$ is the correct answer.