Question

A thin lens made of glass $(\text{refractive index}=1.5)$ of focal length $f=16\text{cm}$ is immersed in a liquid of refractive index $1.42$. If its focal length in liquid is $f_1$, then the ratio $f_1/f$ is closest to the integer:

• A. $1$
• B. $9$
• C. $5$
• D. $17$

Answer 1

The correct option is B $9$

Using the lens maker's formula

$\frac{1}{f}=(\frac{{\mu }_g}{{\mu }_a}-1)[\frac{1}{R_1}-\frac{1}{R_2}]$
Here, ${\mu }_g$ and ${\mu }_a$ are the refractive index of glass and air respectively

$\Rightarrow \frac{1}{f}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\Rightarrow \frac{1}{f}=\left(0.5\right)(\frac{1}{R_1}-\frac{1}{R_2})---\left(1\right)$

When immersed in liquid,

$\frac{1}{f_1}=(\frac{{\mu }_g}{{\mu }_l}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

[Here, ${\mu }_l\to$refractive index of liquid]

$\Rightarrow \frac{1}{f_1}=(\frac{1.5}{1.42}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\Rightarrow \frac{1}{f_1}=\left(\frac{0.08}{1.42}\right)(\frac{1}{R_1}-\frac{1}{R_2})---\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$

$\Rightarrow \frac{f_1}{f}=\frac{0.5\times 1.42}{0.08}\approx 9$

Hence, option $\left(B\right)$ is the correct answer.