Question

The focal lengths of a lens are in the ratio $8:3$ when it is immersed in two different liquids of refractive indices $1.6$ and $1.2$ respectively. The refractive index of the material of the lens is

• A. $1.25$
• B. $1.5$
• C. $1.8$
• D. $2$

Answer 1

The correct option is D $2$

Given,
Refractive index of first liquid, ${\mu }_1=1.6$
Refractive index of second liquid, ${\mu }_2=1.2$

Let, refractive index of the lens be ${\mu }_l$,
and $f_1$ and $f_2$ be the focal lengths.
​​​
Using Lens maker's formula for a lens dipped in two different liquids,

$\frac{1}{f_1}=(\frac{{\mu }_l}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})....\left(1\right)$

$\frac{1}{f_2}=(\frac{{\mu }_l}{{\mu }_2}-1)(\frac{1}{R_1}-\frac{1}{R_2})....\left(2\right)$

Where,

$R_1$ and $R_2$ are radii of curvatures of lens.

From $\left(1\right)$ and $\left(2\right)$ ,

$\frac{f_1}{f_2}=\frac{\frac{{\mu }_l}{{\mu }_2}-1}{\frac{{\mu }_l}{{\mu }_1}-1}=\frac{{\mu }_1}{{\mu }_2}\left(\frac{{\mu }_l-{\mu }_2}{{\mu }_l-{\mu }_1}\right)$

From the data given in the question, $\frac{f_1}{f_2}=\frac{8}{3}$

$\frac{8}{3}=\frac{1.6}{1.2}\left(\frac{{\mu }_l-1.2}{{\mu }_l-1.6}\right)$

$\Rightarrow 2({\mu }_l-1.6)=({\mu }_l-1.2)$

$\Rightarrow {\mu }_l=3.2-1.2$

$\therefore {\mu }_l=2$

Hence, option (d) is the correct answer.