Question

A point source of light $S$, placed at a distance $60\text{cm}$ in front of the centre of a plane mirror of width $50\text{cm}$, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance $1.2\text{m}$ from it $($see in the figure$)$. The distance between the extreme points where he can see the image of the light source in the mirror is ______$\text{cm}$.

Answer 1

The man will be able to see the image till he is present inside the field of view.
Here, $RPMQR$ is the field of view.
From similar triangles, $\Delta IMS$ and $\Delta IQT$,

$\frac{MS}{QT}=\frac{IS}{IT}$

$\Rightarrow \frac{25}{QT}=\frac{60}{60+120}$

$\Rightarrow QT=75\text{cm}$

Therefore, the distance between the extreme points where he can see the image of the light source in the mirror will be $QR=2\times 75=150\text{cm}$