Question:
A 10 mm long owl pin is placed vertically in front of a concave mirror. A 5 mm long image of the owl pin is formed at 30 cm in front of the mirror. The
focal length of this mirror is
(a) – 30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm
Solution:
(b).
Explanation : $\mathrm{m}=\frac{h^{\prime}}{h}=\frac{-v}{u}$ or
$u=\frac{-0 \times b}{h^{\prime}}=\frac{-(-30 \mathrm{~cm}) \times(10 \mathrm{~mm})}{-5 \mathrm{~mm}}$
$=-60 \mathrm{~cm}$
$\therefore \frac{1}{f}=\frac{1}{u}+\frac{1}{v}=-\frac{1}{60}-\frac{1}{30}=-\frac{1}{20}$ or $f=-20 \mathrm{~cm}$
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