An equiconvex lens is cut into two halves along $(i) {XOX}^{'} a n d (ii) YOY'$ , as shown in the figure. Let $f, f^{'}, f''$ be the focal lengths of complete lens, of each in case
$(i)$ and of each half in case $(ii)$ , respectively

Choose the correct statement from the following.

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The focal length of an equiconvex lens is

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ...(i)

For an equiconvex lens, $R_{1} = R_{2} = R$

$\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = \frac{2 (μ - 1)}{R}$

Case I: When the lens is cut along $XOX'$ , then each half becomes equiconvex with

$R_{1} = + R, R_{2} = - R$

$∴ \frac{1}{f} = (μ - 1) [\frac{1}{R} - (\frac{1}{- R})] \Rightarrow \frac{1}{f} (μ - 1) (\frac{1}{R} + \frac{1}{R})$

$= (μ - 1) \frac{2}{R} = \frac{1}{f^{'}} \Rightarrow f^{'} = f$

Case II: When the lens is cut along $YOY'$ , then each half becomes plano-convex with

$R_{1} = + R, R_{2} = \infty$

$∴ \frac{1}{f^{''}} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$= (μ - 1) (\frac{1}{R} - \frac{1}{\infty}) = (\frac{μ - 1}{R}) = \frac{1}{2 f} \Rightarrow f^{''} = 2 f$

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