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(B) The light escapes from the surface only within a cone of semi-vertical angle $	heta_c$,where $	heta_c$ is the critical angle.Using Snell's law a

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$0.2$

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(B) The light escapes from the surface only within a cone of semi-vertical angle $	heta_c$,where $	heta_c$ is the critical angle.Using Snell's law at the critical angle: $\mu \sin 	heta_c = 1 \sin 90^{\circ} = 1$.Given $\mu = \frac{5}{4}$,we have $\frac{5}{4} \sin 	heta_c = 1$,which gives $\sin 	heta_c = \frac{4}{5}$.Thus,$\cos 	heta_c = \sqrt{1 - \sin^2 	heta_c} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The fraction of light energy escaping is given by the ratio of the solid angle of the cone to the total solid angle of a sphere $(4\pi)$:$	ext{Fraction} = \frac{2\pi(1 - \cos 	heta_c)}{4\pi} = \frac{1 - \cos 	heta_c}{2}$.Substituting the value of $\cos 	heta_c$: $	ext{Fraction} = \frac{1 - \frac{3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5} = 0.2$.

$A$ point source of light is placed at a depth $h = 0.5 \, m$ below the surface of a liquid $(\mu = \frac{5}{4})$. Then,the fraction of light energy that escapes directly from the liquid surface is:

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