Q. A point source of light $B$ is placed at a distance $L$ in front of the centre of a mirror width $d$ hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance $2L$ from it as shown. The greater distance over which he can see the image of the light source in the mirror is
A) $d/2$
B) $d$
C) $2d$
D) $3d$
Explanation
(d) $HI = AB = d$
and $DS = CD = \frac{d}{2}$
$\because AH = 2AD$
$\Rightarrow GH = 2CD = \frac{2d}{2} = d$
Similarly $IJ = d$ so $GJ = GH + HI + IJ = d + d + d = 3d$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.