How to Reduce Glare off Water by Factor of 1.33: Polarization Physics Problem

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SUMMARY

The discussion focuses on reducing glare off water by a factor of 1.33, specifically when the sun is at 45 and 60 degrees above the horizon, utilizing the refractive index of water, which is 1.33. The participants derived Fresnel reflection coefficients for both perpendicular and parallel electric field (E-field) orientations. It was established that glare is primarily caused by the horizontally polarized E-field due to Rayleigh scattering, necessitating the consideration of both E polarizations at the water surface for accurate calculations.

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  • Understanding of Fresnel reflection coefficients
  • Knowledge of polarization physics
  • Familiarity with Rayleigh scattering principles
  • Basic grasp of refractive indices, specifically for water
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"by what factor may the glare off water be reduced when the sun is at

a) 45 degrees above the horizon
b)60 degrees above the horizon

n.b refractive index of water is 1.33"

I have derived the fresnel reflection coefficients for the cases when E is perpendicular to the plane of incidence, and when E is parallel to the plane of incidence. I am also assuming that "glare" is caused by the E-field. But I'm not sure which reflection coeffient to use i.e do we assume that E is parallel to plane of incidence of perp.?

thanks
 
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The E field from the blue sky is predominantly horizontally polarized.
This is because the Rayleigh scattering polarizes the scattered radiation.
The horizontal polarization comes out to be
P=sin^2\theta/(1+cos\theta)^2.
You will have to use both E polarizations at the water surface.
The above factor gives the polarized horizontal E^2. Then you have to include both E polarizations with E^2 = one half of the unpolarized component.
 

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