Question

A point source of light B is placed at a distance L in front of the center of a mirror of 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 in fig. The greatest distance over which he can see the image of the light source in the mirror is:

• A. $\frac{d}{2}$
• B. $d$
• C. $2d$
• D. $3d$

Answer 1

Answer: (D)

$3d$

As the person moves up at some he receives the light ray which is reflected from the mirror this is the starting point at which he receives and as when moves on at some point he just receives the light ray that is the endpoint and this area is field of view for the person
from the figure triangle $ABG$ and $ADF$ are similar

$\frac{AG}{AF}=\frac{BG}{DF}$

$\frac{L}{3L}=\frac{\left(d/2\right)}{DF}$

$DF=\frac{3d}{2}$

So $\frac{3d}{2}$ above and below the point D
Hence a total distance of $\frac{3d}{2}+\frac{3d}{2}=3d$