JEE Main & Advanced
JEE Main Paper (Held On 12-Jan-2019 Morning)
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 \[d+d+d=3d\]