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Q. 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 $\ldots \ldots \ldots . cm$.
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JEE MainJEE Main 2021Ray Optics and Optical Instruments

Solution:

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$\tan \theta=\frac{25}{60}=\frac{ x }{180}$
$x =75 cm$
so distance between extreme point $=2 x =2 \times$
$75=150 cm$