Question:
A point source of light, S is placed at 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
A point source of light, S is placed at 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
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The image is as far behind the mirror as S is in front; the man sees it only inside the cone from the image through the mirror edges.
Updated On: Nov 5, 2026
- \(d\)
- \(\frac{d}{2}\)
- \(3d\)
- \(2d\)
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The Correct Option is C
Solution and Explanation
The image S' is at distance L behind the mirror, on the axis through the mirror centre.
The man can see S' only if the line from S' to his eye passes through the mirror (width d). So the region is the cone from S' through the edges of the mirror.
The mirror is at distance L from S' and has half-width \(d/2\). The man's line is at distance \(L + 2L = 3L\) from S'.
By similar triangles, the half-width at that distance \(= \frac{d}{2}\cdot\frac{3L}{L} = \frac{3d}{2}\).
Total distance over which the image is visible \(= 2\times\frac{3d}{2} = 3d\).
Answer: (3) \(3d\)
The man can see S' only if the line from S' to his eye passes through the mirror (width d). So the region is the cone from S' through the edges of the mirror.
The mirror is at distance L from S' and has half-width \(d/2\). The man's line is at distance \(L + 2L = 3L\) from S'.
By similar triangles, the half-width at that distance \(= \frac{d}{2}\cdot\frac{3L}{L} = \frac{3d}{2}\).
Total distance over which the image is visible \(= 2\times\frac{3d}{2} = 3d\).
Answer: (3) \(3d\)
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