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 $2\pi rh$ ]

• A

  $21\%$
• B

  $34\%$
• Correct answer
  C

  $17\%$
• D

  $50\%$

Solution

$sin\beta =\frac{3}{4},\cos \beta =\frac{\sqrt{7}}{4}$
Solid angle $d\mathrm{\Omega }=2\pi R^2\left(1-\cos \beta \right)$
Percentage of light $=\frac{2\pi R^2\left(1-\cos \beta \right)}{4\pi R^2}\times 100$
$=\frac{1-cos\beta }{2}\times 100=\left(\frac{4-\sqrt{7}}{8}\right)\times 100\approx 17\%$

Hence, the correct answer is option: 'C'.