A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices $\mu_1$ and $\mu_2$ and $R$ is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

Let the focal length of the plano-convex lens be $f_1$ and its refractive index be $\mu_1$.

Applying the lens maker's formula for the plano-convex lens, assuming light is incident on the plane surface:
$\dfrac{1}{f_1} = (\mu_1 - 1) \left(\dfrac{1}{\infty} - \dfrac{1}{-R}\right) = \dfrac{\mu_1 - 1}{R}$

Let the focal length of the plano-concave lens be $f_2$ and its refractive index be $\mu_2$. Since the plano-convex lens fits exactly into it, the concave surface of the second lens has the same radius of curvature $R$ and matches the convex surface of the first lens.

Applying the lens maker's formula for the plano-concave lens:
$\dfrac{1}{f_2} = (\mu_2 - 1) \left(\dfrac{1}{-R} - \dfrac{1}{\infty}\right) = -\dfrac{(\mu_2 - 1)}{R}$

The equivalent focal length $F$ of the combination of thin lenses in contact is given by:
$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$

Substituting the values of $\dfrac{1}{f_1}$ and $\dfrac{1}{f_2}$:
$\dfrac{1}{F} = \dfrac{\mu_1 - 1}{R} - \dfrac{\mu_2 - 1}{R}$

$\dfrac{1}{F} = \dfrac{\mu_1 - 1 - \mu_2 + 1}{R}$

$\dfrac{1}{F} = \dfrac{\mu_1 - \mu_2}{R}$

$F = \dfrac{R}{\mu_1 - \mu_2}$.

Answer: $\dfrac{R}{(\mu_1 - \mu_2)}$