A concave lens of glass, refractive index $1.5$, has both surfaces of the same radius of curvature $R$. On immersion in a medium of refractive index $1.75$ it will behave as a

The focal length of a lens of refractive index surrounded by a medium of refractive index ${\mu }_1$ is given
by

$\frac{1}{f}=\left(\frac{{\mu }_2-{\mu }_1}{{\mu }_1}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
 

Hence, we have
$$\begin{gathered} -\frac{1}{f}=\left(\frac{1.5-1.75}{1.75}\right)\left(\frac{1}{R}+\frac{1}{R}\right) \\ =\left(\frac{-0.25}{1.75}\right)\left(\frac{2}{R}\right) \end{gathered}$$

which gives $f = 3.5 R$. Since the focal length is positive, the lens acts like a convergent lens.