An equiconvex lens of radius of curvature 14 cm is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of 40 cm , calculate the image distance.

From the diagram,
\(\frac{1}{f_1}=\left(\mu_1-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\mu_1-1\right)\left(\frac{1}{R}-\frac{1}{\infty}\right)\)
\(\frac{1}{f_1}=\frac{\mu_1-1}{R}\) and \(\frac{1}{f_2}=\left(\mu_2-1\right)\left(\frac{1}{\infty}-\frac{1}{-R}\right)\)
\(\frac{1}{f_2}=\frac{\mu_2-1}{R}\)
\(\therefore\) Combined focal length,
\(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{\mu_1-1}{R}+\frac{\mu_2-1}{R}\)
\(=\frac{\mu_1+\mu_2-2}{R}=\frac{1.5+1.2-2}{14}\)
\(\frac{1}{f}=\frac{0.7}{14}\)
\(\Rightarrow \quad f=20 \mathrm{~cm}\)
Again, \(v=-40 \mathrm{~cm}, v=\) ?
\(\therefore\) Using lens formula,
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\therefore \quad \frac{1}{20}=\frac{1}{v}-\frac{1}{(-40)} \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}\)
\(\Rightarrow \quad v=40 \mathrm{~cm}\)