A plano-convex lens,when silvered on the plan | vedclass

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(D) The equivalent focal length $F$ of a silvered lens is given by $\frac{1}{F} = \frac{2}{f} + \frac{1}{f_m}$,where $f$ is the focal length of the le

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$1.5$

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(D) The equivalent focal length $F$ of a silvered lens is given by $\frac{1}{F} = \frac{2}{f} + \frac{1}{f_m}$,where $f$ is the focal length of the lens and $f_m$ is the focal length of the mirror formed by silvering.Case $1$: Silvered on the plane side. Here,the mirror is plane,so $f_m = \infty$. Given $F = -30 \ cm$ (concave mirror).$\frac{1}{-30} = \frac{2}{f} + \frac{1}{\infty} \Rightarrow \frac{1}{-30} = \frac{2}{f} \Rightarrow f = -60 \ cm$.Case $2$: Silvered on the convex side. Here,the mirror is convex with radius $R$,so $f_m = \frac{R}{2}$. Given $F = -10 \ cm$.Using the lens maker's formula for a plano-convex lens,$\frac{1}{f} = (\mu - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{\mu - 1}{R}$. Thus,$R = f(\mu - 1)$.Substituting into the mirror formula: $\frac{1}{F} = \frac{2}{f} + \frac{2}{R} \Rightarrow \frac{1}{-10} = \frac{2}{-60} + \frac{2}{R} \Rightarrow \frac{1}{R} = \frac{1}{2}(\frac{1}{-10} + \frac{1}{30}) = \frac{1}{2}(\frac{-3+1}{30}) = -\frac{1}{30}$.Since $R$ is a radius of curvature,we take magnitude $R = 30 \ cm$.Using $\frac{1}{f} = \frac{\mu - 1}{R}$,we have $\frac{1}{-60} = \frac{\mu - 1}{-30} \Rightarrow \frac{1}{2} = \mu - 1 \Rightarrow \mu = 1.5$.

$A$ plano-convex lens,when silvered on the plane side,behaves like a concave mirror of focal length $30 \ cm$. However,when silvered on the convex side,it behaves like a concave mirror of focal length $10 \ cm$. The refractive index of its material is:

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