Q 12-09-208JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
A point source of light, S is placed at a distance $L$ in front of the centre of plane mirror of width $d$ which is hanging vertically on a wall. 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:
Answer: (A) $3d$
The image S′ is at distance $L$ behind the mirror. The man sees it when the line from S′ to his eye passes through the mirror.
Rays from S′ through the two edges of the mirror (separated by $d$, at distance $L$ from S′) reach the man's line at distance $L + 2L = 3L$ from S′. By similar triangles the width there is
$$d\times\frac{3L}{L} = 3d$$
Solution by Sreeraj P, M.Sc Physics