Curved surfaces of a plano-convex lens of ref | vedclass

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(B) For the 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_{

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$\mu_{1}-\mu_{2}$

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(B) For the plano-convex lens with refractive index $\mu_{1}$,the focal length $f_{1}$ is given by the lens maker's formula: $\frac{1}{f_{1}} = (\mu_{1}-1)(\frac{1}{R})$.For the plano-concave lens with refractive index $\mu_{2}$,the focal length $f_{2}$ is given by: $\frac{1}{f_{2}} = (\mu_{2}-1)(-\frac{1}{R})$.When these lenses are combined,the equivalent focal length $f_{eq}$ is given by: $\frac{1}{f_{eq}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.Substituting the expressions: $\frac{1}{f_{eq}} = (\mu_{1}-1)(\frac{1}{R}) + (\mu_{2}-1)(-\frac{1}{R}) = \frac{(\mu_{1}-1) - (\mu_{2}-1)}{R} = \frac{\mu_{1}-\mu_{2}}{R}$.Therefore,the ratio of the radius of curvature $R$ to the equivalent focal length $f_{eq}$ is: $\frac{R}{f_{eq}} = \mu_{1}-\mu_{2}$.

Curved surfaces of a plano-convex lens of refractive index $\mu_{1}$ and a plano-concave lens of refractive index $\mu_{2}$ have equal radius of curvature as shown in the figure. Find the ratio of the radius of curvature to the focal length of the combined lenses.

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