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(C) Given, radius $R=0.05 \ m = 5 	imes 10^{-2} \ m$. Refractive index $n=1.5$. The critical angle $c$ is given by $\sin c = \frac{1}{n} = \frac{1}{1

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$(3 \sqrt{5}-5) 	imes 10^{-2} \ m$

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(C) Given, radius $R=0.05 \ m = 5 	imes 10^{-2} \ m$. Refractive index $n=1.5$. The critical angle $c$ is given by $\sin c = \frac{1}{n} = \frac{1}{1.5} = \frac{2}{3}$. From the geometry of the problem, the ray that strikes the curved surface at the critical angle $c$ will emerge tangent to the surface and hit the table at a distance $x$ from the edge of the cylinder. In the right-angled triangle formed by the radius $R$ and the distance $(R+x)$, the angle at the center is $c$. Thus, $\cos c = \frac{R}{R+x}$. Since $\sin c = \frac{2}{3}$, we have $\cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. Equating the two expressions for $\cos c$: $\frac{\sqrt{5}}{3} = \frac{R}{R+x}$. $\sqrt{5}(R+x) = 3R$. $\sqrt{5}x = 3R - \sqrt{5}R = R(3 - \sqrt{5})$. $x = R \frac{3 - \sqrt{5}}{\sqrt{5}} = R \left( \frac{3}{\sqrt{5}} - 1 ight) = R \left( \frac{3\sqrt{5}}{5} - 1 ight)$. Wait, re-evaluating the geometry: The distance from the center to the point where the ray hits the table is $d = \frac{R}{\cos c}$. So, $R+x = \frac{R}{\cos c} = \frac{R}{\sqrt{5}/3} = \frac{3R}{\sqrt{5}} = \frac{3\sqrt{5}R}{5}$. $x = \frac{3\sqrt{5}R}{5} - R = R \left( \frac{3\sqrt{5}-5}{5} ight)$. Given $R = 5 	imes 10^{-2} \ m$, we get $x = 5 	imes 10^{-2} \left( \frac{3\sqrt{5}-5}{5} ight) = (3\sqrt{5}-5) 	imes 10^{-2} \ m$.

$A$ parallel beam of light is incident on a glass prism in the shape of a quarter cylinder of radius $R=0.05 \ m$ and refractive index $n=1.5$, placed on a horizontal table as shown in the figure. Beyond the cylinder, a patch of light is found whose nearest distance $x$ from the cylinder is

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