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Question: Answered & Verified by Expert
A plano-convex lens of refractive index ( $\mu_1$ ' fits exactly into a plano-concave lens of refractive index $\mu_2$. Their plane surfaces are parallel to each other. ' $R$ ' is the radius of curvature of the curved surface of the lenses. The focal length of the combination is
PhysicsRay OpticsMHT CETMHT CET 2021 (24 Sep Shift 1)
Options:
  • A $\frac{\mathrm{R}}{\mu_1-\mu_2}$
  • B $\frac{R}{2\left(\mu_1+\mu_2\right)}$
  • C $\frac{2 R}{\mu_1-\mu_2}$
  • D $\frac{R}{2\left(\mu_1-\mu_2\right)}$
Solution:
1854 Upvotes Verified Answer
The correct answer is: $\frac{\mathrm{R}}{\mu_1-\mu_2}$
$\begin{aligned} & \frac{1}{\mathrm{f}_1}=\left(\mu_1-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=\frac{\left(\mu_1-1\right)}{\mathrm{R}} \\ & \frac{1}{\mathrm{f}_2}=\left(\mu_2-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=\frac{\left(\mu_2-1\right)}{-\mathrm{R}} \\ & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}=\frac{\mu_1-1}{\mathrm{R}}-\frac{\mu_2-1}{\mathrm{R}}=\frac{1}{\mathrm{R}}\left[\mu_1-1-\mu_2+1\right]=\frac{\mu_1-\mu_2}{\mathrm{R}} \\ & \mathrm{f}=\frac{\mathrm{R}}{\mu_1-\mu_2}\end{aligned}$

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