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

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{1}{\mathrm{f}_1}=\left(\mu_1-1ight)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight)=\frac{\left(\mu_1-1ight)}{\mathrm{R}} \ & \frac{1}{\mathrm{f}_2}=\left(\mu_2-1ight)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight)=\frac{\left(\mu_2-1ight)}{-\mathrm{R}} \ & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}=\frac{\mu_1-1}{\mathrm{R}}-\frac{\mu_2-1}{\mathrm{R}}=\frac{1}{\mathrm{R}}\left[\mu_1-1-\mu_2+1ight]=\frac{\mu_1-\mu_2}{\mathrm{R}} \ & \mathrm{f}=\frac{\mathrm{R}}{\mu_1-\mu_2}\end{aligned}$

A plano-convex lens of refractive index ( $\mu_1$ ' fits exactly into a plano-concave lens of refractive…

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