A thin lens is constructed from a material with a refractive index...

Using the lens maker's formula:

$\dfrac{1}{f} = \left( \dfrac{\mu_L}{\mu_M} -1 \right)\left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)<br><br>For a biconvex lens,$R_1$is positive and$R_2$is negative, so the term \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right) is positive.<br><br>Given the refractive index of the lens$\mu_L = 1.5$and the refractive index of the surrounding medium (water)$\mu_M = 1.33$.<br><br>Since$\mu_L > \mu_M$, the relative refractive index \dfrac{\mu_L}{\mu_M}$> 1$, making the term$\left( \dfrac{\mu_L}{\mu_M} -1 \right) positive.

Thus, the focal length $f$ remains positive.

A lens with a positive focal length acts as a convergent lens. Therefore, the lens continues to act as a convergent lens when immersed in water.