Question

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

• A. ${f' = f, f" = f}$
• B. ${f' = 2f, f" = 2f}$
• C. ${f' = f, f" = 2f}$
• D. ${f' =2f, f" = f}$

Answer 1

Answer: (C)

The focal length of equiconvex lens
$ \frac{1}{f} = (\mu -1) \bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)$...(i)
$ \frac{1}{f} = (\mu -1) \bigg(\frac{1}{R}-\frac{1}{-R}\bigg) = \frac{2 (\mu =- 1)}{R}$
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 = \infty $
$\therefore \frac{1}{f''} = (\mu -1) \bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg) $
$ = (\mu -1) \bigg(\frac{1}{R}+\frac{1}{\infty}\bigg) $
$ = \frac{(\mu -1)}{R}=\frac{1}{2f} $
Hence $ f'' = 2f$