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When an object is placed at a distance of 25 cm from a concave mirror, the magnification is $m_{1}$ . The object is moved 15 cm farther away with respect to the earlier position, and the magnification becomes $m_{2}$ . If $m_{1} {/m}_{2} = 4$ the focal length of the mirror is (Assume that image is real and $m_{1} {,m}_{2}$ are numerical values)

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30 cm

20 cm

10 cm

15 cm

answer is D.

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Detailed Solution

m = \frac{- v}{u} = - (\frac{f}{u - f})

$Now m_{1} = - (\frac{f}{25 - f}) \to (1)$

$m_{2} = - (\frac{f}{40 - f}) \to (2)$

$\frac{m_{1}}{m_{2}} = \frac{40 - f}{25 - f}$

$\frac{4}{1} = \frac{40 - f}{25 - f}$

$\Rightarrow 100 - 4 f = 40 - f$

$60 = 3 f$

$f = \frac{60}{3} = 20 cm$

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