A person stands straight infront of a convex mirror at a distance of 30 m away from it. He sees his erect image whose height is $\frac{1}{6}$ th of his real height. Find the focal length of the convex mirror.

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Solution

So the image is $1 / 6$ of the original height.
u is distance of object from mirror, u=-30 cm
Magnification $m = \frac{1}{6} = - \frac{v}{- 30}$
$v = 5$ vm
Now $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{f} = \frac{1}{5} + \frac{1}{- 30}$
$f = 6 c m$

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