Question

If 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 complete lens of each, 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)

$f^{\prime }=f,f^{\prime \prime }=2f$

The focal length of equiconvex lens
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$ .....(i)
For equiconvex lens, $R_1=R_2=R$
$\frac{1}{f}=(\mu -1)(\frac{1}{R}-\frac{1}{-R})=\frac{2(\mu -1)}{R}$
Case I When the lens is cut along $XO{X}^{\prime }$, then each half is given equiconvex with
$R_1=+R$, $R_2=-R$
$\therefore \frac{1}{f}=(\mu -1)[\frac{1}{R}-\left(\frac{1}{-R}\right)]\Rightarrow \frac{1}{f}=(\mu -1)(\frac{1}{R}+\frac{1}{R})$
$=(\mu -1)\frac{2}{R}=\frac{1}{f^{\prime }}\Rightarrow f^{\prime }=f$
Case II When lens is cut along $YO{Y}^{\prime }$, then each half becomes plano-convexed with
$R_1=+R$, $R_2=\infty$
$\therefore \frac{1}{f^{\prime \prime }}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
$=(\mu -1)(\frac{1}{R}-\frac{1}{\infty })=\left(\frac{\mu -1}{R}\right)=\frac{1}{2f}\Rightarrow f^{\prime \prime }=2f$