wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
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 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


d2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

d

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2d

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

3d

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

3d


First of all, let's 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 up to 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 are similar.
We get using the law of similarity;
x2L=d2L

Which gives us
x = d
From the diagram, total field of view will be hence, d + d + d = 3d


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Spherical Lens
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon