An equiconvex lens is cut into two halves along (i) $XOX'$ and (ii) $YOY'$ as shown in the figure. Let ${f, f', f'' }$be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively.
Choose the correct statement from the following
Case I When lens is cut along $XOX'$, then each half is again equiconvex with $ R_1 = R, R_2 = - R $ $\therefore \frac{1}{f} = (\mu -1) \bigg[\frac{1}{R}-\frac{1}{(-R)}\bigg] $ $ = (\mu -1) \bigg[\frac{1}{R}+\frac{1}{R}\bigg] $ $ = (\mu -1) \frac{2}{R}+\frac{1}{f'} \Rightarrow \, \, f' = f $
Case II When lens is cut along $YOY'$, then each half becomes piano-convex with $ R_1 = R, R_2 = 8 $ $\therefore \frac{1}{f''} = (\mu -1) \bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg) $ $ = (\mu -1) \bigg(\frac{1}{R}+\frac{1}{8}\bigg) $ $ = \frac{(\mu -1)}{R}=\frac{1}{2f} $
Hence $ f'' = 2f$
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