Question

A biconvex lens has a focal length of $10$ cm. It is cut in half and two pieces are placed as shown. The focal length of the final combination in cm is

• A. $10$
• B. $20$
• C. $40$
• D. Not a lens

Answer 1

The correct option is A $10$

Before breaking the focal length of the lens can be found out as shown below
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})=(\mu -1)[\frac{1}{R_1}-(-\frac{1}{R_2})]$
$\frac{1}{f}=(\mu -1)\frac{2}{R}\Rightarrow f=\frac{R}{2(\mu -1)}$
After breaking, let the focal length of the leftmost lens be $f_1$
$\frac{1}{f_1}=(\mu -1)(\frac{1}{\infty }-(-\frac{1}{R}))$
$\frac{1}{f_1}=(\mu -1)\left(\frac{1}{R}\right)$
$f_1=\frac{R}{(\mu -1)}$
$f_1=2f=20\text{cm}$
let the focal length of the rightmost lens be $f_2$
$\frac{1}{f_2}=(\mu -1)(\frac{1}{R}-\frac{1}{\infty })$
$\frac{1}{f_2}=(\mu -1)\left(\frac{1}{R}\right)$
$f_2=\frac{R}{(\mu -1)}$
$f_2=2f=20\text{cm}$
$\frac{1}{f_{\text{net}}}=\frac{1}{f_1}+\frac{1}{f_2}$
$\frac{1}{20}+\frac{1}{20}=\frac{1}{10}$
$f_{\text{net}}=10\text{cm}$