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A point source of light S, placed at a distance 60cm in front of the centre of a plane mirror of width 50cm, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror a distance 1.2m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is ____________cm.

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answer is 150.

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Detailed Solution

$Using properties of similar triangles,$

$\begin{array}{l} \frac{25}{60} = \tan θ = \frac{y}{1.2} \\ \Rightarrow y = 0.5 m = 50 cm \\ Thus, length of field of view = y + 50 + y = 150 cm \end{array}$

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