Q 12-09-210JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
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 $\mu_1$, radius of curvature $R$). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:
Answer: (A) $\dfrac{R}{\mu_2 - \mu_1}$
Lens maker's formula for each lens (one plane face):
$$\frac{1}{f_2} = \frac{\mu_2 - 1}{R},\qquad \frac1{f_1} = -\frac{\mu_1 - 1}{R}$$
For lenses in contact:
$$\frac1F = \frac{\mu_2 - 1}{R} - \frac{\mu_1 - 1}{R} = \frac{\mu_2 - \mu_1}{R} \Rightarrow F = \frac{R}{\mu_2 - \mu_1}$$
Solution by Sreeraj P, M.Sc Physics