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A spherical convex surface of power 5 dioptre separates object and image space of refractive indices $1.0$ and $\frac{4}{3}$ respectively. The radius of curvature of the surface is
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$5 \mathrm{~cm}$

$P=5 D$
$u=\infty$ then
$\frac{1}{\infty}-\frac{4 / 3}{f}=\frac{1-4 / 3}{R}$
$-\frac{4}{3 f}=-\frac{1}{3 R}$
$\frac{f}{4}=R \quad \quad f=4 R$
$\therefore \frac{100}{5}=4 R \Rightarrow R=\frac{25}{5}=5 \mathrm{~cm}$
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