A plano-convex lens of refractive index mu_1 | vedclass

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Question

(A) For the plano-convex lens,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1)(\frac{1}{R} - \frac{1}{\infty}

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

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(A) For the plano-convex lens,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{\mu_1 - 1}{R}$.For the plano-concave lens,the focal length $f_2$ is given by: $\frac{1}{f_2} = (\mu_2 - 1)(\frac{1}{-\infty} - \frac{1}{-R}) = \frac{\mu_2 - 1}{-R} = -\frac{\mu_2 - 1}{R}$.The focal length $f$ of the combination is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$.Substituting the values: $\frac{1}{f} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.Therefore,$f = \frac{R}{\mu_1 - \mu_2}$.

$A$ 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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