A plano-convex lens fits exactly into a plano-concave lens. Their...

Focal length of combination

$\dfrac{1}{f} = \dfrac{1}{f1} + \dfrac{1}{f2} \ldots \left(i\right)$

Using lens maker's formula
$\dfrac{1}{f2} =\left({μ}_{2} -1\right)\left(\dfrac{1}{- R} - \dfrac{1}{\infty}\right) = - \left({μ}_{2} -1\right)R$

$\dfrac{1}{f1} = \left({μ}_{1} -1\right)\left(\dfrac{1}{\infty} - \dfrac{1}{- R}\right) = \dfrac{{μ}_{1} - 1}{R}$

or $\dfrac{1}{f1} = \left({μ}_{1} -1\right)R$

$\dfrac{1}{f2} = - \left({μ}_{2} -1\right)R$

Putting these values in Equation $( i )$, we get

$⇒ \dfrac{1}{f} = \left({μ}_{1} -1\right)R - \left({μ}_{2} -1\right)R$

$⇒ \dfrac{1}{f} = \left({μ}_{1} -1 - {μ}_{2} + 1\right)R$

$⇒ \dfrac{1}{f} = \dfrac{{μ}_{1} - {μ}_{2}}{R}$

$f = R \left({μ}_{1} - {μ}_{2}\right)</p>$