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Measuring intensity of superposed waves using complex amplitudes

  1. May 9, 2012 #1
    Hi,

    Suppose we have the x and y components of the electric field being described as [itex](E_{x}e^{i(kz-\omega t)}, E_{y}e^{i(kz-\omega t +\phi)})[/itex], what is the intensity?

    I think the correct answer is [itex]E_{x}^{2} +E_{y}^{2} + 2E_{x}E_{y}\cos\phi[/itex]. However, I am not sure how to deal with this using the Jones formalism. In that, the intensity is given by [itex]E^{\dagger}E[/itex] which would give

    [tex]\begin{pmatrix} E_{x}e^{-i(kz-\omega t)} & E_{y}e^{-i(kz-\omega t +\phi)} \end{pmatrix} \begin{pmatrix} E_{x}e^{i(kz-\omega t)}\\ E_{y}e^{i(kz-\omega t +\phi)}\end{pmatrix} =E_{x}^{2} +E_{y}^{2} [/tex]

    Clearly, the above answer is independent of the relative phase and I think it cannot be right because of that. So what is the correct way to calculate the intensity using Jones formalism? Thank you
     
  2. jcsd
  3. May 10, 2012 #2

    mfb

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    How do you get your first result?
    If you flip the sign of Ey, you modify your calculated intensity, which is wrong.
     
  4. May 10, 2012 #3
    The first result was using

    [itex]I = (E_{x}e^{-i(kz-\omega t)}+ E_{y}e^{-i(kz-\omega t +\phi)})(E_{x}e^{i(kz-\omega t)}+ E_{y}e^{i(kz-\omega t +\phi)})[/itex]

    Now that I think of it, this doesn't seem correct either. What is the correct expression for intensity?
     
  5. May 11, 2012 #4

    mfb

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    E_x^2 + E_y^2 should be correct.
     
  6. May 11, 2012 #5
    Are you sure about that? Is the phase completely irrelevant in the intensity calculation?
     
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