One plano-convex and one plano-concave lens of same radius of curvature R but of different materials are joined side by side as show in figure. If the refractive index of the material 1 is ${\mu }_1$ and that of 2 is ${\mu }_2,$ then the focal length of the combination is

Focal length of a lens is given as 

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

Focal length of plano-convex lens, i.e., lens 1,$R_1=\propto$Focal length of plano-convex lens, i.e., lens 1,and R = -R

$\Rightarrow \frac{1}{f_1}=\frac{{\mu }_1-1}{R}\text{ or }{f}_1=\frac{R}{\left({\mu }_1-1\right)}$----(i)

Similarly, focal length of plano-concave lens, i.e. lens 2     R₂=-R   Similarly, focal length of plano-concave lens, i.e. lens 2and $R_2=\propto$ and $\Rightarrow \frac{1}{f_2}=-\frac{\left({\mu }_2-1\right)}{R}\text{ or }{f}_2=\frac{-R}{\left({\mu }_2-1\right)}$-----(ii)

From Eqs.(i) and (ii) net focal length is $\frac{1}{f}=\frac{{\mu }_1-1}{R}-\frac{{\mu }_2-1}{R}=\frac{{\mu }_1-{\mu }_2}{R}$

From Eqs.(i) and (ii) net focal length is $\Rightarrow f=\frac{R}{{\mu }_1-{\mu }_2}$