A point source of light S, placed at a
distance L in front of the centre of a mirror of width d,
hangs vertically on a wall. A man walks in front of the
mirror along a line parallel to th mirror at a distane
2L form it as shown.The greatest distance over which
he can see the image of the light source in the mirror is
(a) `d//2` (b) d (c) 2d (d) 3 d.

Explanation

Draw the rays BM and BN incident on the mirror edges, and draw the reflected rays (see Fig. ). From the figure, it is clear that the man can observe in the angular region `P^(')B^(')Q^(')`. The observer can see through the distance `P^(')Q^(')`, where reflected ray from B can meet the line.
From geometry,`CD=d=C^(')D^(')`
`BD=DP=d//2,CB=QC=d//2`
`P^(')Q^(')=C^(')D^(')+P^(')D^(')+Q^(')C^(')`
`rArr (ND)/(DP)=(ND^('))/(D^(')P^('))=(L)/(d//2)=(2L)/(D^(')P^('))rarr D^(')P^(')=d`
Similarly, `Q^(')C^(')`=d
Hence, distance through which the man can observe the image
=`Q^(')C^(')+C^(')D^(')+D^(')P^(')=d+d+d=3d`