Question

A thin lens of refractive index $1.5$ has a focal length of $15cm$ in air. When the lens is placed in a medium of refractive index $4/3$, its focal length will becomes _______ cm.

Answer 1

Case$I$;

$\frac{1}{f}=\frac{\mu -1}{1}(\frac{1}{R_1}-\frac{1}{R_2})\frac{1}{15}=\frac{0.5}{1}\frac{1}{2R}-\left(i\right)$

Case$II$;

$\frac{1}{f}=\frac{\frac{3}{2}-\frac{4}{3}}{\frac{4}{3}}\frac{1}{2R}-\left(ii\right)$

Evaluating $\left(i\right)$ and $\left(ii\right)$

$\frac{f}{15}=\frac{1}{2}\frac{6\times 4}{3}$

$f=15\times 4=60$cm