A thin lens is made with a material having refractive index $\mu =1.5.$. Both the sides are convex. It is dipped in water ($\mu =$ 1.33). It will behave like

consider a lens of focal length l with radius of curvature of both the surfaces R₁ and R₂ $\frac{1}{f}= (1.5-1)(\frac{1}{R_1}+\frac{1}{R_2}) = 0.5(\frac{1}{R_1}+\frac{1}{R_2})$ 

if the lens is placed in water then the focal length f' is given by 

$\frac{1}{f\text{'}}=(\frac{1.5}{1.33}-1)(\frac{1}{R_1}+\frac{1}{R_2}) = \frac{0.17}{1.33}\left(\frac{2}{f}\right)$

since f is positive f' is also positive hence the lens is converging lens