Q. A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If the 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 focal length of the combination is
A) $\frac{R}{2{(\mu}_{1} + \mu_{2})}$
B) $\frac{R}{2{(\mu}_{1} - \mu_{2})}$
C) $\frac{R}{{(\mu}_{1} - \mu_{2})}$
D) $\frac{2R}{{(\mu}_{2} - \mu_{1})}$
Explanation
(c) The combination of two lenses is
As $\frac{4}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\therefore\ \frac{1}{F} = \left( u_{1} - 1 \right)\left( \frac{1}{\infty} + \frac{1}{R} \right) + (u_{2} - 1)\left( \frac{1}{- R} - \frac{1}{\infty} \right)$
$= \frac{u_{1} - 1}{R} + \frac{u_{2} - 1}{R}$
$\frac{1}{F} = \frac{u_{1} - u_{2}}{R}$
Or $F = \frac{R}{u_{1} - u_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.