There is an equiconvex glass lens with radius of each face as R and ${ }_a{\mu }_g=\frac{3}{2}$ and ${ }_a{\mu }_{\omega }=\frac{4}{3}$. If there is water in object space and air in image space, then the focal length is: 

Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium 

$$\begin{gathered} \frac{{\mu }_2-{\mu }_1}{R}=\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v_1} \\ \Rightarrow \frac{\left({\displaystyle \frac{3}{2}}\right)-\left({\displaystyle \frac{4}{3}}\right)}{R}=\frac{{\displaystyle \frac{4}{3}}}{\infty }+\frac{{\displaystyle \frac{3}{2}}}{v_1} \Rightarrow v_1=9R \end{gathered}$$

Now consider the refraction at the second surface of the lens i.e. refraction from denser medium to rarer medium

$\frac{1-{\displaystyle \frac{3}{2}}}{-R}=-\frac{{\displaystyle \frac{3}{2}}}{9R}+\frac{{\displaystyle 1}}{v_2} \Rightarrow v_2=\left(\frac{3}{2}\right)R$

The image will be formed at a distance do $\frac{3R}{2}$ . 

This is equal to the focal length of the lens.