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A convex lens is placed somewhere in between an object and a screen. The distance between the object and screen is 48cm. If the numerical value of the magnification produced by the lens is 3, focal length of the lens is

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16 cm

4.5 cm

9 cm

12 cm

answer is C.

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Detailed Solution

$Given that u + v = 48 cm$

$Numerical value of m = 3 = \frac{v}{u} \Rightarrow v = 3 u$

$Substituting in u + v = 48 cm$

$We get u = 12 cm, v = 36 cm$

By sign convention

$u = - 12 cm, v = + 36 cm (real image formed)$

$We have \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{36} + \frac{1}{12} = \frac{1}{f}$

$∴ f = 9 cm$

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