Percentage of Energy: Total Internal Reflection

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SUMMARY

The fraction of light energy, f, that escapes from a point source of light submerged at depth h in water is given by the formula f=0.5(1-\sqrt{1-1/n^2}), where n represents the index of refraction of water. This fraction is independent of the depth h and assumes negligible absorption and reflection at the surface. The critical angle is determined using Snell's Law, specifically n1sin(theta1) = n2sin(90), leading to theta1 = arcsin(1/n). Understanding solid angles is crucial for calculating the fraction of light escaping, as it involves the geometry of light propagation.

PREREQUISITES
  • Understanding of Snell's Law and critical angles
  • Familiarity with the concept of solid angles in three-dimensional geometry
  • Knowledge of the index of refraction and its implications in optics
  • Basic principles of light behavior at interfaces, including refraction and reflection
NEXT STEPS
  • Study the derivation of Snell's Law and its applications in optics
  • Learn about solid angles and their calculation in three-dimensional space
  • Explore the concept of total internal reflection and its conditions
  • Investigate the effects of absorption in different media on light propagation
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding light behavior in different media, particularly in aquatic environments.

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Homework Statement



A point source of light is at depth h below the surface of a large and deep lake. Show that the fraction, f, of the light energy that escapes directly form the water surface is independent of h and is given by

f=0.5(1-\sqrt{1-1/n^2})

where n is the index of refraction of the water. Absorption within the water and reflection at the surface (except where it is total) have been neglected.

Homework Equations



n1sin(theta1) = n2sin(theta2)
E=hv?

The Attempt at a Solution



First step would be to find the critical angle:
nsin(theta1)=sin(90) (n2=1 for air)
theta1 = arcsin(1/n)

So anything with an incident angle greater than theta1, will be totally reflected. The problem seems to assume that all of the light with angle less than theta1 will be refracted (even though in reality some will be reflected, too).

So here's where I am going wrong: I tried to assume that all of the light would be equally split up between all angles of radiation. So out of 180 degrees, arcsin(1/n) would escape.
So my fraction and answer would be arcsin(1/n)/180, but that's obviously wrong.

I am having trouble relating energy to this. Thanks in advance.
 
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You have to think in 3-dimensions. The light escaping is inside a cone of half angle equal to the critical angle. Therefore the fraction escaping is the solid angle subtended by the cone divided by the total solid angle which in this case is 2π.
 
Wow, thank you very much. I figured out the problem with the information you gave. I had never even heard of a solid angle before so I had some learning to do.
 

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