An equiconvex lens $(µ=\frac{3}{2})$ of focal length 20 cm in air is immersed in water of refractive index $(µ=\frac{4}{3})$. What is the percentage change in focal length of the lens ?

300%

Focal length in air is 

$ \frac{1}{f_a} = (\mu -1 ) (\frac{1}{R_1} - \frac{1}{R_2})= 0.5 \times (\frac{1}{R_1} - \frac{1}{R_2})$

Focal length in water is 

$ \frac{1}{f_w} = (\frac{\mu_g}{\mu_w} -1 ) (\frac{1}{R_1} - \frac{1}{R_2})= (\frac{9}{8} - 1)\times (\frac{1}{R_1} - \frac{1}{R_2})=\frac{1}{8}\times (\frac{1}{R_1} - \frac{1}{R_2}) $

$ f_w = 4 f_a$

$\Rightarrow \text{ increase in focal length is } 300\%$