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

f 1.5 f $\frac{4}{3}f$ 4 f

4 f

$\frac{1}{f}=(μ_g-1)(\frac{1}{R_1}-\frac{1}{R_1})$ . . . (1) $\frac{1}{f_w}=(\frac{μ_g-μ_w}{μ_w})(\frac{1}{R_1}-\frac{1}{R_1})$  . . . (2) $⇒f_w=4f$