A 2 cm high object is placed on the principal axis of a concave mirror at a distance of 12 cm from the pole. If the image is inverted, real and 5 cm high. The location of the image and the focal length of the mirror is

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8.6 cm, 30 cm

30 cm, 10 cm

30 cm, 8.6 cm

10 cm, 30 cm

answer is A.

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Detailed Solution

U = - 12 cm

$m = \frac{height of image}{height of object} = \frac{- 5}{2}$

$m = - 2.5 cm$

$m = \frac{- v}{u}$

$v = 2.5 \times (- 12)$

$v = - 30 cm$

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$f = \frac{- 12 \times 30}{- (12 + 30)} = \frac{12 \times 30}{42} = \frac{60}{7} = 8.6 cm$

Image distance = 30 cm

Focal length = 8.6 cm

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