JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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 \(\ldots \ldots \ldots\, cm.\)

- A \(75\)
- B \(135\)
- C \(150\)
- D \(210\)
Answer & Solution
Correct Answer
(C) \(150\)
Step-by-step Solution
Detailed explanation
\(\tan \theta=\frac{25}{60}=\frac{ x }{180}\) \(x =75 \,cm\) so distance between extreme point \(=2 x =2 \times\) \(75=150\, cm\)
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