• # question_answer One plano-convex and one plano-concave lens of same radius of curvature 'R' 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 : [JEE Main 10-4-2019 Morning] A) $\frac{R}{2-({{\mu }_{1}}-{{\mu }_{2}})}$                    B) $\frac{2R}{({{\mu }_{1}}-{{\mu }_{2}})}$C) $\frac{R}{2({{\mu }_{1}}-{{\mu }_{2}})}$                     D) $\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}$

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$\frac{1}{{{f}_{2}}}=\left( \frac{{{\mu }_{2}}-1}{1} \right)\left( \frac{1}{-R}-0 \right)=-\frac{{{\mu }_{2}}-1}{R}$ $\frac{1}{{{f}_{eq}}}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}$ $\frac{1}{{{f}_{eq}}}=\frac{R}{{{\mu }_{1}}-1}+\frac{R}{-({{\mu }_{2}}-1)}\Rightarrow \frac{1}{{{f}_{eq}}}=\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}$ Hence ${{f}_{eq}}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}$