A plano-convex thin lens is used to obtain th | vedclass

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(A) For a real image to be formed on a screen by a lens,the minimum distance $D$ between the object and the screen is $D = 4f$,where $f$ is the focal

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$10$

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(A) For a real image to be formed on a screen by a lens,the minimum distance $D$ between the object and the screen is $D = 4f$,where $f$ is the focal length of the lens.Given $D = 2.0 \ m = 200 \ cm$,we have $4f = 200 \ cm$,which gives $f = 50 \ cm$.Using the lens maker's formula for a plano-convex lens: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Here,$R_1 = R$ and $R_2 = \infty$,so $\frac{1}{f} = (\mu - 1) \frac{1}{R}$.Substituting the values: $\frac{1}{50} = (1.5 - 1) \frac{1}{R} \Rightarrow \frac{1}{50} = 0.5 	imes \frac{1}{R} \Rightarrow R = 25 \ cm$.The aperture radius $r$ of a plano-convex lens is related to its thickness $t$ and radius of curvature $R$ by the formula $t = R - \sqrt{R^2 - r^2}$. For a thin lens,this approximates to $t \approx \frac{r^2}{2R}$,so $r = \sqrt{2Rt}$.The aperture diameter $a = 2r = 2\sqrt{2Rt} = \sqrt{8Rt}$.Substituting the values: $a = \sqrt{8 	imes 25 	imes 0.5} = \sqrt{100} = 10 \ cm$.

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