If a convex lens having focal length f and refractive index 1.5 is immersed in water then new focal length will be ($\mu_w = 4/3$)

4f

$\text{Focal length of lens in air is }\frac{1}{f_a} = (\mu_g - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

$\text{Focal length of lens in liquid is }\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

$\frac{f_l}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_l} - 1} = \frac{1.5 - 1}{\frac{3/2}{4/3} - 1} = \frac{0.5}{1/8} = 4$

$\Rightarrow f_l = 4f$