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An object of height 1.2m is placed before a concave mirror of focal length 20cm so that a real image is formed at a distance of 60cm from it. Find the position of an object. What will be the height of the image formed?
Given: f = -20 cm; v = -60 cm
To find: Height of the image formed
Solution:
From mirror formula,
$\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$
$\frac{1}{u}=\frac{1}{-20}-\frac{1}{-60}$
$\frac{1}{u}=-\frac{1}{20}+\frac{1}{60}$
$\frac{1}{u}=-\frac{1}{30}$
So u = - 30 cm
Position of an object is 30 cm in front of the mirror
magnification, $m =\frac{h_{i}}{h_{o}} =\frac{-v }{u}$
$h_{i} =\frac{-(-60) }{-30} \times 1.2$
height of image = -2.4 m
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