Question

If a lens of focal length, $f$, made of glass of refractive index $\frac{5}{3}$ is placed in a liquid of refractive index, $4\frac{5}{3}$, then the focal length of the lens becomes

• A. $2f$
• B. $-4f$
• C. $4f$
• D. $f$

Answer 1

Answer: (C)

In Air

$\frac{1}{f}=({\mu }_g-1)[\frac{1}{R_1}+\frac{1}{R_2}]$

In water,

$\frac{1}{f^{\prime }}=(\frac{{\mu }_g}{{\mu }_w}-1)[\frac{1}{R_1}+\frac{1}{R_2}]$

$\frac{f^{\prime }}{f}=\left(\frac{{\mu }_g-1}{{\mu }_g-{\mu }_w}\right)\times {\mu }_w$

$f^{\prime }=\left(\frac{\frac{5}{3}-1}{\frac{5}{3}-4\frac{5}{3}}\right)\times 4\frac{5}{3}f=-\frac{-17}{18}f$