Question

A thin convex lens made from crown glass $\left(\mu=\frac{3}{2}\right)$ 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 lengths is

• A. $f_1 - f_2
• B. $f_2 >\, f$ and $f_2$ becomes negative
• C. $f_2 >\, f$ and $f_1$ becomes negative
• D. $f_1$ and $f_2$ both become negative

Answer 1

Answer: (B)

$f_2 >\, f$ and $f_2$ becomes negative

Answer 2

By Lens maker's formula
$\frac{1}{f_{1}}=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$\frac{1}{f_{2}}=\left(\frac{3 / 2}{5 / 3}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$\frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$\Rightarrow f_{1}=4 f$& $f_{2}=-5 f$