Intensity of elliptically polarized light

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SUMMARY

The discussion centers on the intensity of elliptically polarized light, specifically the time-averaged norm of the Poynting vector. The correct formula for intensity, derived in SI units, is given by I(𝑥) = (cε₀/2)(E₀ₓ² + 2E₀ₓ·E₀ᵧcos(δ) + E₀ᵧ²). The user initially struggled to find a reliable source for this equation and mistakenly referenced an incorrect formula for linear polarization. The issue was resolved upon realizing that E₀ₓ·E₀ᵧ = 0, which clarified the calculations.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically Poynting vectors
  • Familiarity with elliptically polarized light and its mathematical representation
  • Knowledge of SI units in physics
  • Basic grasp of linear and elliptical polarization concepts
NEXT STEPS
  • Study the derivation of the Poynting vector in electromagnetic fields
  • Explore the mathematical treatment of elliptically polarized light
  • Investigate the implications of the intensity formula in practical applications
  • Review sources on electromagnetic wave propagation and polarization
USEFUL FOR

Physicists, optical engineers, and students studying electromagnetism or optics who are interested in the properties and calculations related to polarized light.

Wox
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The time averaged norm of the Poynting vector of this electromagnetic field (elliptically polarized light):
<br /> \begin{split}<br /> \bar{E}(t,\bar{x})=&amp;(\bar{E}_{0x}+\bar{E}_{0y}e^{i \delta})e^{\bar{k}\cdot\bar{x}-\omega t}\\<br /> \bar{B}(t,\bar{x})=&amp;\frac{1}{\omega}(\bar{k}\times\bar{E}(t,\bar{x}))<br /> \end{split}<br />
with \bar{E}\perp\bar{B}\perp\bar{k}, becomes (as I calculated in SI-units J/(m^{2}s))
<br /> I(\bar{x})=\left&lt;\left\|\bar{P}(t,\bar{x})\right\|\right&gt;=\frac{c\epsilon_{0}}{2}(\bar{E}_{0x}^{2}+2\bar{E}_{0x}\cdot\bar{E}_{0y}\cos\delta+\bar{E}_{0y}^{2})<br />
I have been trying to verify this, but I can't find a source that explicitly discusses this. For a linear polarized beam, \delta=0 so that I(\bar{x})=\frac{c\epsilon_{0}(\bar{E}_{0x}+\bar{E}_{0y})^{2}}{2}, which is correct. For general elliptical polarization I found this link which basically says that
<br /> I(\bar{x})=E_{x}E_{x}^{\ast}+E_{y}E_{y}^{\ast}= \bar{E}_{0x}^{2}+\bar{E}_{0y}^{2}<br />
which can't be right (as it doesn't work for linear polarized light). Does anyone know of a proper reference for this? Or even better, can someone verify my solution?
 
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Ok, this is embarassing. I didn't see that \bar{E}_{0x}\cdot\bar{E}_{0y}=0 which fixes the problem.
 

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