Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is_______ cm

By lens maker formula, $\frac{1}{f}=\left({ }^g{\mu }_a-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

For Lens 1: 

$\frac{1}{f_1}=\left(1.5-1\right)\left(\frac{1}{R}-\frac{1}{-R}\right)\Rightarrow \frac{1}{f_1}=0.5\times \frac{2}{R}\Rightarrow f_1=R\Rightarrow R=15cm$

For Lens 2:

$\frac{1}{f_2}=\left(1.25-1\right)\left(\frac{1}{-R}-\frac{1}{+R}\right)=-0.25\times \frac{2}{R}=-\frac{0.5}{R}=\frac{-1}{2R}, \therefore f_2=-2R$

For Lens 3:

Similarly like lens $1,f_3=R$ So, $\frac{1}{f_{net}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{R}+\frac{1}{-2R}+\frac{1}{R}$

$=\frac{3}{2R}=\frac{2}{2\times 15}=\frac{1}{10}\therefore f_{net}=10cm$