An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance x in front of the second wall. The required focal length of the lens will be :

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more than $\frac{x}{4}$ but less than $\frac{x}{2}$

$\frac{x}{2}$

less than $\frac{x}{4}$

$\frac{x}{4}$

answer is C.

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Detailed Solution

When of an equal size image is formed the
magnifi cation is unity.
i.e., v = u
image distance = object distance
Here, v = x
but according to lens formula,
$\begin{array}{l} \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\ \frac{2}{v} = \frac{1}{f} \\ f = \frac{v}{2} = \frac{x}{2} = focal length of the lens is x / 2 . \end{array}$