Q. A plano-convex lens (focal length $f_2$, refractive index $\mu_2$, radius of curvature $R$) fits exactly into a plano-concave lens (focal length $f_1$, refractive index $\mu_2$, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be :
Solution:
$\frac{1}{F} \, = \, \frac{1}{f_1} + \frac{1}{f_2} = \frac{1-\mu_1}{R} + \frac{\mu_2 - 1}{R}$
