A plano convex lens fits exactly in to a plano concave lens. Their plane surfaces are parallel to each other. Lenses are made up of different materials of refractive indices ' $n_1$ ' $n_2$ ' and ' $R$ ' is the radius of curvature of the curved surface of lenses. Focal length of the combination is

Focal length analysis of the lens combination

The focal length of each lens is determined using the Lens Maker's Formula: $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, assuming light travels left to right.

For the plano-convex lens, the first surface is flat ($R_1 = \infty$) and the second surface is convex with $R_2 = -R$.

$\frac{1}{f_1} = (n_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = \frac{n_1 - 1}{R}$

The plano-concave lens interfaces with the convex surface, giving $R_1 = -R$ for its concave surface, while its second surface is flat ($R_2 = \infty$).

$\frac{1}{f_2} = (n_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = -\frac{n_2 - 1}{R}$

Combining the thin lenses in contact:

$\frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{n_1 - 1}{R} - \frac{n_2 - 1}{R} = \frac{n_1 - n_2}{R}$

$F_{eq} = \frac{R}{n_1 - n_2}$

This result corresponds to option D.