Question

A biconvex glass lens ($R_1=R_2$) has a focal length $f$. Now if $R_1$ is doubled and $R_2$ is halved. Then its focal length becomes

• A. $1.25f$
• B. $0.8f$
• C. $2f$
• D. $2.2f$

Answer 1

The correct option is B $0.8f$

Using Lens makers formula,

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

Given, $R_1=R_2$

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{-R_2})=(\mu -1)(\frac{1}{R_1}+\frac{1}{R_1})=(\mu -1)(\frac{2}{R_1})$

$\frac{1}{f}=(\mu -1)(\frac{2}{R_1})$

Now, $R_1$ is doubled and $R_2$ is halved, we have

$\frac{1}{f_{new}}=(\mu -1)(\frac{1}{2R_1}-\frac{1}{(\frac{-R_2}{2})})$

$\frac{1}{f_{new}}=(\mu -1)(\frac{1}{2R_1}-\frac{2}{-R_2})=(\mu -1)(\frac{1}{2R_1}+\frac{2}{R_1})=(\mu -1)(\frac{5}{2R_1})$

$\frac{f_{new}}{f}=\frac{((\mu -1)(\frac{2}{R_1}))}{((\mu -1)(\frac{5}{2R_1}))}$

$\frac{f_{new}}{f}=\frac{(\frac{2}{R_1})}{(\frac{5}{2R_1})}$

$\frac{f_{new}}{f}=\frac{4}{5}$

$f_{new}=0.8f$
Hence, the correct answer is OPTION B.