The focal length of a convex lens in air and in a liquid medium are 20 cm and 80 cm, respectively. If the absolute refractive index of lens is 1.5, then the refractive index of liquid is:

4/3

The correct answer is Option (2) → 4/3

Given:

Focal length in air: $f_{air}=20\ \text{cm}$

Focal length in liquid: $f_{liq}=80\ \text{cm}$

Refractive index of lens: $\mu_{lens}=1.5$

Lens maker formula (neglecting thickness):

$\frac{1}{f} = \left(\frac{\mu_{lens}}{\mu_{medium}} - 1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Let $K=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ (constant for same lens).

In air ($\mu_{medium}=1$):

$\frac{1}{f_{air}}=(\mu_{lens}-1)K$

$\frac{1}{20}=(1.5-1)K\ \Rightarrow\ \frac{1}{20}=0.5K$

$K=\frac{1}{10}$

In liquid:

$\frac{1}{f_{liq}}=\left(\frac{\mu_{lens}}{\mu_{liq}}-1\right)K$

$\frac{1}{80}=\left(\frac{1.5}{\mu_{liq}}-1\right)\frac{1}{10}$

$\frac{1}{80}=\frac{1.5-\mu_{liq}}{10\mu_{liq}}$

$\frac{1}{80}=\frac{1.5-\mu_{liq}}{10\mu_{liq}}$

Cross multiply:

$10\mu_{liq}\cdot\frac{1}{80}=1.5-\mu_{liq}$

$\frac{\mu_{liq}}{8}=1.5-\mu_{liq}$

$\frac{\mu_{liq}}{8}+\mu_{liq}=1.5$

$\frac{9\mu_{liq}}{8}=1.5$

$\mu_{liq}=\frac{1.5\times 8}{9}=\frac{12}{9}=\frac{4}{3}$

Answer: $\mu_{liq}=\frac{4}{3}=1.33$