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A double convex lens has focal length $25 \mathrm{~cm}$. The radius of curvature of one of the surfaces is doubled of the other. Find the radii, if the refractive index of the material of the lens is 1.5.
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$18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}$
\(\begin{aligned} & \frac{1}{\mathrm{f}}=(\mu-1)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & \frac{1}{25}=0.5\left[\frac{1}{\mathrm{R}}-\left(\frac{1}{2 R}\right)\right] \\ & \frac{1}{25}=\frac{0.5}{10}\left[\frac{3}{\mathrm{R}}\right] \\ & \mathrm{R}=\frac{75}{4}=18.75 \\ & \mathrm{R}_1=18.75 \mathrm{~cm}, \mathrm{R}_2=37.5 \mathrm{~cm}\end{aligned}\)
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