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Intensity of elliptically polarized light

  1. Jul 13, 2012 #1

    Wox

    User Avatar

    The time averaged norm of the Poynting vector of this electromagnetic field (elliptically polarized light):
    [tex]
    \begin{split}
    \bar{E}(t,\bar{x})=&(\bar{E}_{0x}+\bar{E}_{0y}e^{i \delta})e^{\bar{k}\cdot\bar{x}-\omega t}\\
    \bar{B}(t,\bar{x})=&\frac{1}{\omega}(\bar{k}\times\bar{E}(t,\bar{x}))
    \end{split}
    [/tex]
    with [itex]\bar{E}\perp\bar{B}\perp\bar{k}[/itex], becomes (as I calculated in SI-units [itex]J/(m^{2}s)[/itex])
    [tex]
    I(\bar{x})=\left<\left\|\bar{P}(t,\bar{x})\right\|\right>=\frac{c\epsilon_{0}}{2}(\bar{E}_{0x}^{2}+2\bar{E}_{0x}\cdot\bar{E}_{0y}\cos\delta+\bar{E}_{0y}^{2})
    [/tex]
    I have been trying to verify this, but I can't find a source that explicitly discusses this. For a linear polarized beam, [itex]\delta=0[/itex] so that [itex]I(\bar{x})=\frac{c\epsilon_{0}(\bar{E}_{0x}+\bar{E}_{0y})^{2}}{2}[/itex], which is correct. For general elliptical polarization I found this link which basically says that
    [tex]
    I(\bar{x})=E_{x}E_{x}^{\ast}+E_{y}E_{y}^{\ast}= \bar{E}_{0x}^{2}+\bar{E}_{0y}^{2}
    [/tex]
    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?
     
  2. jcsd
  3. Jul 13, 2012 #2

    Wox

    User Avatar

    Ok, this is embarassing. I didn't see that [itex]\bar{E}_{0x}\cdot\bar{E}_{0y}=0[/itex] which fixes the problem.
     
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