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Byju's Answer

Standard XII

Physics

Refractive Index

One plano-con...

Question

One plano-convex and one plano-concave lens of same radius of curvature $^{'} R^{'}$ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of $1$ is $μ_{1}$ and that of $2$ is $μ_{2}$ is:

\frac{R}{2 - (μ_{1} - μ_{2})}

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\frac{2 R}{μ_{1} - μ_{2}}

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\frac{R}{2 (μ_{1} - μ_{2})}

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\frac{R}{μ_{1} - μ_{2}}

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Solution

The correct option is B $\frac{R}{μ_{1} - μ_{2}}$
For 1st lens $\frac{1}{f_{1}} = (\frac{μ_{1} - 1}{1}) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ_{1} - 1}{R}$

For 2nd lens $\frac{1}{f_{2}} = (\frac{μ_{2} - 1}{1}) (- \frac{1}{- R} - 0) = \frac{μ_{2} - 1}{R}$

$\frac{1}{f_{e q}}$ = $\frac{1}{f_{1}} + \frac{1}{f_{2}}$

$\frac{1}{f_{e} q} = \frac{R}{μ_{1} - 1} + \frac{R}{- (μ_{2} - 1)} \Rightarrow \frac{1}{f_{e q}} = \frac{R}{μ_{1} - μ_{2}}$

Hence $\frac{μ_{1} - μ_{2}}{R}$

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