Question

A equiconvex lens is cut into two halves along (i) XOX' and (ii) YOY' as shown in the figure.Let f,f′,f"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. $f^{\prime }=f,f"=2f$
• B. $f^{\prime }=2f,f"=f$
• C. $f^{\prime }=f,f"=f$
• D. $f^{\prime }=2f,f"=2f$

Answer 1

The correct option is D $f^{\prime }=2f,f"=2f$

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

$R_1=+R$, $R_2=\infty$

$\therefore \frac{1}{f^{\prime \prime }}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

$=(\mu -1)(\frac{1}{R}-\frac{1}{\infty })=\left(\frac{\mu -1}{R}\right)=\frac{1}{2f}\Rightarrow f^{\prime \prime }=2f$