A plano-convex lens fits exactly into a plano | vedclass

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(C) The combination of two lenses $1$ and $2$ is as shown in the figure.For a combination of thin lenses in contact,the equivalent focal length $f$ is

Vedclass · Accepted Answer

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

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(C) The combination of two lenses $1$ and $2$ is as shown in the figure.For a combination of thin lenses in contact,the equivalent focal length $f$ is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$.According to the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For the plano-convex lens (lens $1$): $R_1 = \infty$,$R_2 = -R$. Thus,$\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R}$.For the plano-concave lens (lens $2$): $R_1 = -R$,$R_2 = \infty$. Thus,$\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R}$.Substituting these into the combination formula:$\frac{1}{f} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.Therefore,$f = \frac{R}{\mu_1 - \mu_2}$.

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