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

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(C) \(\frac{R}{{\left( {{\mu _1} - {\mu _2}} \right)}}\)

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For \(1^{\text {st }}\) 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 \(2^{n d}\) lens \(\frac{1}{{{f_2}}} = \left( {\frac{{{\mu _2} - 1}}{1}} \right)\) \(\left( {\frac{1}{{ - R}} - 0} \right) = - \frac{{{\mu _2} - 1}}{R}\)

\(\frac{1}{f_{e q}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

\(\frac{1}{{{f_{eq}}}} = \frac{{{\mu _1} - 1}}{R} + \frac{{ - \left( {{\mu _2} - 1} \right)}}{R}\) \( \Rightarrow {f_{eq}} = \frac{R}{{{\mu _1} - {\mu _2}}}\)

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