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An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be
Explanation
Image moves from 24 cm to 60 cm in front of the mirror.
Detailed Solution
$\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u}$
$-\frac{1}{15} = \frac{1}{v_1} - \frac{1}{40}$
$\frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} \Rightarrow v_1 = -24$ cm
When the object is displaced by 20 cm towards the mirror, $u_2 = -20$ cm
$\frac{1}{f} = \frac{1}{v_2} + \frac{1}{u_2} \Rightarrow \frac{1}{-15} = \frac{1}{v_2} - \frac{1}{20}$
$\frac{1}{v_2} = \frac{1}{20} - \frac{1}{15} \Rightarrow v_2 = -60$ cm
So, the image shifts away from the mirror by $60 - 24 = 36$ cm.