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

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Question

(B) For a plano-convex lens with refractive index $\mu_1$,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1)(\f

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a plano-convex lens with refractive index $\mu_1$,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 a plano-concave lens with refractive index $\mu_2$,the focal length $f_2$ is given by: $\frac{1}{f_2} = (\mu_2 - 1)(\frac{1}{-\infty} - \frac{1}{R}) = -\frac{\mu_2 - 1}{R}$.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,$F = \frac{R}{\mu_1 - \mu_2}$.

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