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(C) Let $h$ be the depth of the source,$h = 4 \ m$. Let $n$ be the refractive index of the liquid,$n = 5/3$. To cut off all light,the disc must cover

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$6$

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(C) Let $h$ be the depth of the source,$h = 4 \ m$. Let $n$ be the refractive index of the liquid,$n = 5/3$. To cut off all light,the disc must cover the area corresponding to the critical angle $C$. Using Snell's law at the critical angle: $n \sin C = 1 	imes \sin 90^{\circ}$. $\sin C = 1/n = 1 / (5/3) = 3/5$. From the geometry of the problem,$	an C = r / h$,where $r$ is the radius of the disc. Since $\sin C = 3/5$,we have $\cos C = \sqrt{1 - (3/5)^2} = 4/5$. Thus,$	an C = \sin C / \cos C = (3/5) / (4/5) = 3/4$. Therefore,$r / h = 3/4 \Rightarrow r = (3/4) 	imes 4 = 3 \ m$. The diameter of the disc is $D = 2r = 2 	imes 3 = 6 \ m$.

$A$ small source of light is $4 \ m$ below the surface of a liquid of refractive index $5/3$. In order to cut off all the light coming out of the liquid surface,the minimum diameter of the disc placed on the surface of the liquid is .......... $m$.

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