question_answer

There is a small source of light at some depth below the surface of water (refractive index \[=\frac{4}{3}\]) in a tank of large cross-sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly) [Use the fact that surface area of a spherical cap of height h and radius of curvature r is\[2\pi rh\]] [JEE MAIN Held on 09-01-2020 Evening]

A) \[34%\]

B) \[21%\]

C) \[50%\]

D) \[17%\]

Correct Answer: D

Solution :

\[\sin \theta =\frac{3}{4}\] \[\cos \theta =\frac{\sqrt{7}}{4}\] \[\Omega =2\pi (1-\cos \theta )\] \[=2\pi \,\,(1-\sqrt{7}/4)\] Fraction of energy transmitted\[=\frac{1-\sqrt{7}/4}{2}\]                                                 \[=17%\]