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 the figure. 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

The correct option is D $3d$

First of all lets draw the two extreme incident rays, which will give us two extreme reflected rays and we will be able to calculate the field of view. This field of view will give us the distance upto which the man will be able to see the image of the light source.

We see that the marked angles are equal
1.law of reflection and parallel lines and right angles.
2.Which makes the shaded triangles similar.
We get using the law of similarity;
$\frac{x}{2L}=\frac{d/2}{L}$
Which gives us,
x=d.
From the diagram, total field of view will be hence, d+d+d=3d.