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A convex lens of focal length f is placed somewhere in between the object and a screen. The distance between object and screen is x . If magnification produced is m, the focal length of the lens is

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$\frac{m x}{(m + 1)^{2}}$

$\frac{m x}{(m - 1)^{2}}$

$\frac{(m + 1)^{2}}{m} x$

$\frac{(m - 1)^{2}}{m} x$

answer is A.

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Detailed Solution

$u + v = x \to (1); \frac{v}{u} = m \to (2);$

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \to (3)$

$Solving (1), (2), (3), f = \frac{m x}{(m + 1)^{2}}$

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