Question 14 of 295

A point source of light, S is placed at a distance L in front of the center 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:

A$3d$B$2d$C$d$D$\dfrac{d}{2}$

Show answer and solution▾

Answer: (A) $3d$

By similar triangles

$\Delta AEC~\Delta IHA$

$\dfrac{y}{2L}=\dfrac{\dfrac{d}{2}}{L}$ (by properties of similar triangles)
$\Rightarrow y=d$
$∴$Total distance $\left(CD\right)=y+\dfrac{d}{2}+\dfrac{d}{2}+y=3d$