Q112 / 125 · BITSAT 2015 · Physics
A planoconvex lens fits exactly into a planoconcave lens. Their plane surface are parallel to each other. If the lenses are made of different materials of refractive indices $\mu_{1} \& \mu_{2}$ and $R$ is the radius of curvature of the curved surface of the lenses, then focal length of combination is
- Correct answerA$\frac{R}{\mu_{1}-\mu_{2}}$
- B$\frac{2 R}{\mu_{1}-\mu_{2}}$
- C$\frac{R}{2\left(\mu_{1}-\mu_{2}\right)}$
- D$\frac{R}{2-\left(\mu_{1}+\mu_{2}\right)}$
Solution
If F be the equivalent focal length, then
$$
\begin{aligned}
\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \quad \frac{1}{F}=\left(\mu_{1}-1\right)\left(\frac{1}{\infty}+\frac{1}{R}\right) \\
+\left(\mu_{2}-1\right)\left(\frac{1}{-R}-\frac{1}{\infty}\right)=\frac{\mu_{1}-\mu_{2}}{R} \\
F=\frac{R}{\mu_{1}-\mu_{2}}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.