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(D) The solid angle subtended by a cone of semi-vertical angle $	heta$ is given by $\Omega = 2\pi(1 - \cos 	heta)$.The total solid angle around a po

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$17$

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(D) The solid angle subtended by a cone of semi-vertical angle $	heta$ is given by $\Omega = 2\pi(1 - \cos 	heta)$.The total solid angle around a point source is $4\pi$ steradians.The fraction of light energy that escapes is the ratio of the solid angle of the cone defined by the critical angle $C$ to the total solid angle $4\pi$.Fraction of energy $= \frac{2\pi(1 - \cos C)}{4\pi} = \frac{1}{2}(1 - \cos C)$.From Snell's law at the critical angle $C$,$\mu_w \sin C = \mu_a \sin 90^{\circ}$.Given $\mu_w = 4/3$ and $\mu_a = 1$,we have $\frac{4}{3} \sin C = 1$,so $\sin C = 3/4$.Then,$\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - (3/4)^2} = \sqrt{1 - 9/16} = \sqrt{7/16} = \frac{\sqrt{7}}{4}$.Substituting this into the fraction formula:Fraction $= \frac{1}{2}(1 - \frac{\sqrt{7}}{4}) = \frac{1}{2}(1 - \frac{2.646}{4}) = \frac{1}{2}(1 - 0.6615) = \frac{0.3385}{2} \approx 0.16925$.Converting to percentage,we get approximately $17\%$.

$A$ point source of light is placed at a distance $h$ below the surface of a large deep lake. What is the percentage of light energy that escapes directly from the water surface if the refractive index $\mu$ of the water is $4/3$ (in $\%$)? (Neglect partial reflection)

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