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Question:
Curved surfaces of a plano-convex lens of refractive index $\mu_{1}$ and a plano-concave lens of refractive index $\mu_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.
Correct Option: , 2
Solution:
$\frac{1}{f_{1}}=\left(\mu_{1}-1\right)\left(\frac{1}{R}\right)$
$\frac{1}{f_{2}}=\left(\mu_{2}-1\right)\left(-\frac{1}{R}\right)$
$\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}=\frac{1}{\mathrm{f}_{\mathrm{cq}}}=\frac{\left(\mu_{1}-1\right)-\left(\mu_{2}-1\right)}{\mathrm{R}}$
$\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{\left(\mu_{1}-\mu_{2}\right)}{\mathrm{R}}$
$\frac{\mathrm{R}}{\mathrm{f}_{\mathrm{eq}}}=\left(\mu_{1}-\mu_{2}\right)$
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