Question

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

• A. $\frac{R}{2-\left({\mu }_1-{\mu }_2\right)}$
• B. $\frac{2R}{{\mu }_1-{\mu }_2}$
• C. $\frac{R}{2\left({\mu }_1-{\mu }_2\right)}$
• D. $\frac{R}{{\mu }_1-{\mu }_2}$

Answer 1

Answer: (D)

For 1st lens $\frac{1}{f_1}=\left(\frac{{\mu }_1-1}{1}\right)\left(\frac{1}{\infty }-\frac{1}{-R}\right)=\frac{{\mu }_1-1}{R}$
for 2nd lens $\frac{1}{f_2}=\left(\frac{{\mu }_2-1}{1}\right)\left(\frac{1}{-R}-0\right)=-\frac{{\mu }_2-1}{R}$
$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$
$\frac{1}{f_{eq}}=\frac{R}{{\mu }_1-1}+\frac{R}{-\left({\mu }_2-1\right)}\Rightarrow \frac{1}{f_{eq}}=\frac{R}{{\mu }_1-{\mu }_2}$
Hence $f_{eq}=\frac{{\mu }_1-{\mu }_2}{R}$