Question

Question 2
A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl 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

Answer 1

Given: Height of the object, $h_0$ = 10 mm
Height of the image, $h_i$ = 5 mm
Image distance, v = -30 cm
$m=\frac{h_i}{h_0}=-\frac{5}{10}=-\frac{v}{u}$
$-\frac{1}{2}=-\frac{(-30)}{u}$
$u=-60cm$
Now applying the mirrors formula,
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\frac{1}{f}=\frac{1}{-60}+\frac{1}{-30}$
$\frac{1}{f}=\frac{1}{-60}+\frac{2}{-60}$
$\frac{1}{f}=\frac{3}{-60}$
$\Rightarrow$ Focal length, f = -20 cm