JEE Main 12 Jan 2019 Morning Question 6
Question: 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]
Options:
A) 3d
B) 2d
C) d
D) $ \frac{d}{2} $
Show Answer
Answer:
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 $
