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Question

A point source of light S is placed at a distance L in front of the centre of plane mirror of length(d). 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


A
d2
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B
d
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C
3d
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D
2d
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Solution

The correct option is C 3d
The maximum distance over which man can see image of source S will correspond to the rays reflected from top most and bottom-most edge of mirror.


From the ray diagram shown,
ΔSAN and ΔSAC are similar.
SNSC=ANAC
L3L=(d/2)AC
AC=3d2
From symmetry of incident ray,
AC=CB or AB=2AC
AB=2×3d2=3d
Hence, option (c) is the correct answer.

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