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(A) The light source emits light in all directions,covering a total solid angle of $4\pi$ steradians.Light emerges from the water surface only if the

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) The light source emits light in all directions,covering a total solid angle of $4\pi$ steradians.Light emerges from the water surface only if the angle of incidence is less than or equal to the critical angle $	heta_c$.Using Snell's law at the critical angle: $\mu \sin	heta_c = 1 \sin 90^{\circ}$,where $\mu = 4/3$.$\sin	heta_c = 1 / (4/3) = 3/4$.Then,$\cos	heta_c = \sqrt{1 - \sin^2	heta_c} = \sqrt{1 - 9/16} = \sqrt{7}/4$.The solid angle $\Omega$ subtended by the cone of light that emerges is $\Omega = 2\pi(1 - \cos	heta_c)$.Substituting the value of $\cos	heta_c$: $\Omega = 2\pi(1 - \sqrt{7}/4) = 2\pi(1 - 2.646/4) = 2\pi(1 - 0.6615) = 2\pi(0.3385) = 0.677\pi$.The fraction of light emerging is $\frac{\Omega}{4\pi} = \frac{0.677\pi}{4\pi} \approx 0.169$.Converting to percentage: $0.169 	imes 100 \approx 17\%$.

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