One plano-convex and one plano-concave lens o | vedclass

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(C) For the $1^{	ext{st}}$ lens (plano-convex),using the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R}

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$\frac{R}{(\mu_1 - \mu_2)}$

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(C) For the $1^{	ext{st}}$ lens (plano-convex),using the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} ight) = \frac{\mu_1 - 1}{R}$.For the $2^{	ext{nd}}$ lens (plano-concave),using the lens maker's formula: $\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} ight) = -\frac{\mu_2 - 1}{R}$.The equivalent focal length $f_{eq}$ of the combination is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$.Substituting the values: $\frac{1}{f_{eq}} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.Therefore,$f_{eq} = \frac{R}{\mu_1 - \mu_2}$.

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 $\mu_1$ and that of $2$ is $\mu_2$,then the focal length of the combination is

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