A point source of light $S$, placed at a distance $60$ cm in front of the centre of a plane mirror of width $50$ cm, hangs vertically on a wall. A man walks in front 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
Numerical value type. Enter your answer.
Answer: 150
The image $S'$ is $60$ cm behind the centre of the mirror. The man sees it only along lines from $S'$ that pass through the mirror.
Lines from $S'$ through the two edges of the mirror (half-width $25$ cm at distance $60$ cm) reach the man's line, which is $60 + 120 = 180$ cm from $S'$. By similar triangles the half-width there is
$$25\times\frac{180}{60} = 75\ \text{cm}$$
Required distance $= 2\times75 = 150$ cm.
Solution by Sreeraj P, M.Sc Physics