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The radius of the convex surface of planoconvex lens is 20 cm and the refractive index of the material of the lens is 1.5. The focal length is
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40 cm
Using the relation for focal length of planoconvex lens
$=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
(Refractive index of material of lens $\mu=1.5$, $\left.R_1=20 \mathrm{~cm}, R_2=\infty\right)$
$\begin{aligned} \frac{1}{f} & =(1.5-1)\left(\frac{1}{20}-\frac{1}{\infty}\right) \\ \frac{1}{f} & =\frac{1}{40} \\ \text { or } \quad f & =40 \mathrm{~cm}\end{aligned}$
$=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
(Refractive index of material of lens $\mu=1.5$, $\left.R_1=20 \mathrm{~cm}, R_2=\infty\right)$
$\begin{aligned} \frac{1}{f} & =(1.5-1)\left(\frac{1}{20}-\frac{1}{\infty}\right) \\ \frac{1}{f} & =\frac{1}{40} \\ \text { or } \quad f & =40 \mathrm{~cm}\end{aligned}$
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