Question

An equiconvex lens is cut into two halves along (i) $XO{X}^{\prime }$ and (ii) $YO{Y}^{\prime }$ as shown in the figure. Let $f,f^{\prime },f^{\prime \prime }$ 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. $f^{\prime }=f,f^{\prime \prime }=f$
• B. $f^{\prime }=2f,f^{\prime \prime }=2f$
• C. $f^{\prime }=f,f^{\prime \prime }=2f$
• D. $f^{\prime }=2f,f^{\prime \prime }=f$

Answer 1

Answer: (C)

Initially, the focal length of equiconvex lens is
$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots$ (i)
$\frac{1}{f}=(\mu -1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(\mu -1)}{R}$
Case I: When lens is cut along $XO{X}^{\prime }$
then each half is again equiconvex with
$R_1=+R,R_2=-R$
Thus, $\frac{1}{f}=(\mu -1)\left[\frac{1}{R}-\frac{1}{(-R)}\right]$
$=(\mu -1)\left[\frac{1}{R}+\frac{1}{R}\right]$
$=(\mu -1)\frac{2}{R}=\frac{1}{f^{\prime }}$
$\Rightarrow f^{\prime }=f$
Case II: When lens is cut along $YO{Y}^{\prime }$,
then each half becomes plano-convex with
$R_1=+R,R_2=\infty$
Thus, $\frac{1}{f^{\prime \prime }}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$=(\mu -1)\left(\frac{1}{R}-\frac{1}{\infty }\right)$
$=\frac{(\mu -1)}{R}=\frac{1}{2f}$
Hence, $f^{\prime }=f,f^{\prime \prime }=2f$