Two identical glass $\left(\mu_g=\frac{3}{2}\right)$ equiconvex lenses of focal length f are kept in contact. The space between the two lenses in filled with water $\left(\mu_w=\frac{4}{3}\right)$. The focal length of combination is:

$\frac{3 f}{4}$

Let R be the radius of curvature of each surface.

$\frac{1}{f}=(1.5-1)\left(\frac{1}{R}+\frac{1}{R}\right)$    ∴ R = f

For the water lens, $\frac{1}{f'}=\left(\frac{4}{3}-1\right)\left(-\frac{1}{R}-\frac{1}{R}\right)=\frac{1}{3}\left(-\frac{2}{f}\right)$

Or $\frac{1}{f'}=-\frac{2}{3 f}$

Using, $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f'_3}$

$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}+\frac{1}{f'}=\frac{2}{f}-\frac{2}{3 f}=\frac{4}{3 f}$

∴ $F=\frac{3 f}{4}$