A lens behave as a converging lens in air and diverging lens in water then the refractive index of the lens is

Covering lens, f > 0,
Diverging lens f < 0
$$\begin{gathered} \begin{array}{l}\frac{1}{\mathrm{f}}=\left(\frac{\mathrm{\mu }\text{ material }}{\mathrm{\mu }\text{ medium }}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)\\ \frac{1}{\mathrm{f}}=\left(\frac{\mathrm{\mu }}{{\mathrm{\mu }}_{\text{air }}}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)>0\\ {\mathrm{\mu }}_{\text{air }}=1\\ \frac{1}{\mathrm{f}}=(\mathrm{\mu }-1)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)>0 .....\left(1\right)\\ {\mathrm{\mu }}_{\text{water }}=\frac{4}{3}=1.33\end{array} \\ \frac{1}{{\mathrm{f}}^{\mathrm{\prime }}}=\left(\frac{\mathrm{\mu }}{1.33}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)<0 ......\left(2\right) \end{gathered}$$
Comparing eq 1 & 2
$$\begin{gathered} \mathrm{\mu }-1>0\Rightarrow \mathrm{\mu }>1 \\ \begin{array}{l}\frac{\mathrm{\mu }}{1.33}-<0 \mathrm{\mu }<1.33\\ 1<\mathrm{\mu }<1.33\end{array} \end{gathered}$$