A plano-convex lens (focal length f_2, refrac | vedclass

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(B) For a plano-concave lens with refractive index $\mu_1$ and radius of curvature $R$,the focal length $f_1$ is given by the lens maker's formula: $\

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a plano-concave lens with refractive index $\mu_1$ and radius of curvature $R$,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = \frac{1 - \mu_1}{R}$.For a plano-convex lens with refractive index $\mu_2$ and radius of curvature $R$,the focal length $f_2$ is given by: $\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_2 - 1}{R}$.When the 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: $\frac{1}{f} = \frac{1 - \mu_1}{R} + \frac{\mu_2 - 1}{R} = \frac{1 - \mu_1 + \mu_2 - 1}{R} = \frac{\mu_2 - \mu_1}{R}$.Therefore,the focal length of the combination is $f = \frac{R}{\mu_2 - \mu_1}$.

Similar Questions

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