Question

A thin glass lens with relative refractive index ${\mu }_a^g=1.5$ , has an optical power of $-5\text{D}$ in air. Its optical power in a liquid medium with relative refractive index ${\mu }_a^l=1.6$ will be

[Assume, lens is fully immersed in the liquid]

• A. $-\frac{5}{8}\text{D}$
• B. $+\frac{5}{8}\text{D}$
• C. $-\frac{8}{5}\text{D}$
• D. $+\frac{8}{5}\text{D}$

Answer 1

The correct option is B $+\frac{5}{8}\text{D}$

From lens makers' formula,

$\frac{1}{f}=(\frac{{\mu }_1}{{\mu }_2}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

When the lens is in air medium,

$\frac{1}{f_a}=(\frac{1.5}{1}-1)(\frac{1}{R_1}-\frac{1}{R_2})........\left(1\right)$

When the lens is in liquid medium,

$\frac{1}{f_l}=(\frac{1.5}{1.6}-1)(\frac{1}{R_1}-\frac{1}{R_2})........\left(2\right)$

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

$\frac{f_l}{f_a}=\frac{P_a}{P_l}=\left(\frac{1.5-1}{\frac{1.5}{1.6}-1}\right)=-8$

Given, $P_a=-5\text{D}$

$\Rightarrow P_l=\frac{-P_a}{8}=+\frac{5}{8}$

Hence, option (b) is the correct answer.