1. Tardigrade
  2. Question
  3. Physics
  4. There is a small source of light at some depth below the surface of water (refractive index =(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 2π rh]

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 MainJEE Main 2020Ray Optics and Optical Instruments

Solution:

$\frac{4}{3}sin\,\theta = 1\,sin\,90^{\circ}$
$sin\,\theta = \frac{3}{4}$
Area of sphere in which light spread $= 4\pi R^{2}$
$\Omega = 2\pi\left(1-cos\,\theta \right)$
$\Omega = 2\pi \left(1-\frac{\sqrt{7}}{4}\right)$
$P \to 4\pi$ steradians
$P' \to \frac{P}{4\pi} \left(1-cos\,\theta \right)$
Ratio $= \frac{P'}{P} = \frac{ 2\pi \left(1-cos\,\theta \right)}{4\pi} = \frac{ \left(1-cos\,\theta \right)}{2} = \frac{1-\frac{\sqrt{7}}{4}}{2}$
$= \frac{0.33}{2} = 0.17$

Solution Image