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. $f^{\prime }=f,f"=f$
• B. $f^{\prime }=2f,f"=2f$
• C. $f^{\prime }=f,f"=2f$
• D. $f^{\prime }=2f,f"=f$

Answer 1

Answer: (C)

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