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

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(D) The equivalent focal length $f_{eq}$ of two thin lenses in contact is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$.For the plano-co

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

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(D) The equivalent focal length $f_{eq}$ of two thin lenses in contact is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$.For the plano-convex lens (refractive index $\mu_1$): The first surface is plane $(R_1 = \infty)$ and the second surface is convex $(R_2 = -R)$. Using the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R}$.For the plano-concave lens (refractive index $\mu_2$): The first surface is concave $(R_1 = -R)$ and the second surface is plane $(R_2 = \infty)$. Using the lens maker's formula: $\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R}$.Adding the two focal powers:$\frac{1}{f_{eq}} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.Therefore,$f_{eq} = \frac{R}{\mu_1 - \mu_2}$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

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