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A point source of light S, placed at a distance 60 cm infront of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks infront of the mirror along a line parallel to the mirror at a distance 1.2 m 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
Correct Answer: 150
Distance of virtual image $$I$$ behind the mirror: $$v = u = 60\ \text{cm}$$
Total distance from virtual image to the observer's line: $$x = v + D = 60 + 120 = 180\ \text{cm}$$
Using similar triangles properties: $$\frac{L}{d} = \frac{x}{v}$$
$$\implies \frac{L}{50} = \frac{180}{60}$$
$$\implies L = 50 \times 3 = 150\ \text{cm}$$
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