c
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 \(\frac{1}{{{f_2}}} = \left( {\frac{{{\mu _2} - 1}}{1}} \right)\) \(\left( {\frac{1}{{ - R}} - 0} \right) =  - \frac{{{\mu _2} - 1}}{R}\)

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

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