question_answer

A point source of light, S is placed at a distance L in front of the centre 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 [JEE Main Online Paper Held On 12-Jan-2019 Morning]

A) 3d                                            

B) 2d

C) d                                 

D)   \[\frac{d}{2}\]

Correct Answer: A

Solution :

In the given figure \[\Delta AED\]and \[\Delta ABC\]are similar triangles. So,\[\frac{BC}{ED}=\frac{AC}{AD}\Rightarrow \frac{BC}{ED}=\frac{2L}{L}\]\[\Rightarrow BC=2ED\]                  ?(i) Also, \[\Delta AED\]and \[\Delta ASD\]are congruent triangles. So, ED = DS                                                                 ...(ii) Using (i) and (ii), BC = d So, the distance over which the man can see the image of the light source in the mirror is \[d+d+d=3d\]