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The refractive index of the material of a double convex lens is 1.5 and its focal length is $5 \mathrm{~cm}$. If the radii of curvature are equal, the value of the radius of curvature (in $\mathrm{cm}$ ) is
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$5.0$
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For double convex lens, $R_1=R, R_2=-R$
$\begin{aligned} & \therefore \quad \frac{1}{5}=(1.5-1)\left(\frac{1}{R}+\frac{1}{R}\right) \\ & \text { or } \quad \frac{1}{5}=0.5 \times \frac{2}{R} \\ & \text { or } \quad R=5 \mathrm{~cm} \\ & \end{aligned}$
For double convex lens, $R_1=R, R_2=-R$
$\begin{aligned} & \therefore \quad \frac{1}{5}=(1.5-1)\left(\frac{1}{R}+\frac{1}{R}\right) \\ & \text { or } \quad \frac{1}{5}=0.5 \times \frac{2}{R} \\ & \text { or } \quad R=5 \mathrm{~cm} \\ & \end{aligned}$
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