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

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(D) For a plano-convex lens with refractive index $n_1$,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (n_1 - 1) \left(

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$\frac{R}{n_1-n_2}$

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(D) For a plano-convex lens with refractive index $n_1$,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (n_1 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{n_1 - 1}{R}$.For a plano-concave lens with refractive index $n_2$,the focal length $f_2$ is given by: $\frac{1}{f_2} = (n_2 - 1) \left( -\frac{1}{R} - \frac{1}{\infty} \right) = -\frac{n_2 - 1}{R}$.When these two lenses are combined,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Substituting the values: $\frac{1}{F} = \frac{n_1 - 1}{R} - \frac{n_2 - 1}{R} = \frac{n_1 - 1 - n_2 + 1}{R} = \frac{n_1 - n_2}{R}$.Therefore,the focal length of the combination is $F = \frac{R}{n_1 - n_2}$.

$A$ plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. The lenses are made of different materials with refractive indices $n_1$ and $n_2$,and $R$ is the radius of curvature of the curved surfaces. What is the focal length of the combination?

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