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Focal length of convex lens is 'f'. Find the position of real object for which virtual image is 4 times the size of object.

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$- f$

$- 0.25 f$

$-0.50 f$

$- 0.75 f$

answer is D.

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Detailed Solution

$\begin{array}{l} n = \frac{v}{u} \\ \Rightarrow \frac{1}{nu} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{u} (\frac{1 - n}{n}) = \frac{1}{f} \\ \Rightarrow u = \frac{f (1 - n)}{n} = \frac{f (1 - 4)}{4} = - \frac{3 f}{4} = - 0.75 \end{array}$

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