Question

An thin convex lens made from crown glass $(\mu =\frac{3}{2})$ has focal length $f$. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$. It has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal length is:

• A. $f_1=f_2<f$
• B. $f_2>f$ and $f_1$ becomes negative
• C. $f_1$ and $f_2$ both become negative
• D. $f_1>f$ and $f_2$ become negative

Answer 1

The correct option is D $f_1>f$ and $f_2$ become negative

By lens maker's formula:
$\frac{1}{f}=(\frac{3}{2}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Let, $(\frac{1}{x}=\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=\frac{1}{2x}\Rightarrow f=2x...\left(1\right)$

Given:
${\mu }_o=\frac{3}{2},{\mu }_1=\frac{4}{3},{\mu }_2=\frac{5}{2}$

$\frac{1}{f_1}=(\frac{\frac{3}{2}}{\frac{4}{3}}-1)\frac{1}{x}$

$\frac{1}{f_1}=\frac{1}{8x}$

On comparing with equation (1),
$\frac{1}{f_1}=\frac{1}{4\left(2x\right)}=\frac{1}{4f}$
$\Rightarrow f_1=4f$

$\frac{1}{f_2}=(\frac{\frac{3}{2}}{\frac{5}{3}}-1)\left(\frac{1}{x}\right)$
$\Rightarrow f_2$ is negative


Analytically, If a lens is inserted in a denser surrounding the sign of focal length changes and if lens is inserted in a rarer surrounding, the sign of focal length remain same.
If lense is inserted in rarer medium the focal length increases.

Hence, option $B$ is correct.