Distance of an object from the first focus of an equiconvex lens is 10 cm and the distance of its real image from second focus is 40 cm. the focal length of the lens is

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25cm

10cm

20cm

40cm

answer is C.

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Detailed Solution

In the figure

$\begin{array}{l} u = (- 10 + f) \\ v = (40 + f) \\ and f = + f \\ using \frac{1}{v} - \frac{1}{u} = \frac{1}{f} we have \\ \frac{1}{40 + f} + \frac{1}{10 + f} = \frac{1}{f} \\ solving this we get f = 20 cm \end{array}$

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