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Homework Help: Fresnel Coefficient for Unpolarised Light

  1. Sep 4, 2012 #1
    The problem statement, all variables and given/known data

    A homework problem asks me to find the Fresnel coefficient for a linearly polarised plane wave, which is incident under an angle theta, whose electric field vector can be given as:


    [itex]\hat{s}[/itex] and [itex]\hat{p}[/itex] are the unit vectors for s-polarised and p-polarised waves.

    The attempt at a solution

    The previous question on my homework sheet asked me to derive the Fresnel equations for s and p-polarised waves, which I did starting with the bondary conditions. Here, I beleive I have an unpolarised wave. My initial assumption was to take the average of both s and p-polarised light that I worked out previously.

    For example,




    However, I'm concerned this is an oversimplification and a more robust approach is in order. I'd love a nudge in the right direction or a thumbs up if my initial approach looks fine.

  2. jcsd
  3. Sep 4, 2012 #2
    For your purposes (this question) unpolarized fields do not exist. You may combine the polarizations; such as what peice of film or a camera does. But there is no unpolarized light.
  4. Sep 4, 2012 #3
    A poor choice of word by me, but I did account for the polarisation components due to the wave's orientation by averaging the Fresnel coefficients in my attempt above.

    I note an error in my opening post as the wave should be E=Ecos[itex]\alpha[/itex][itex]\hat{s}[/itex]+Esin[itex]\alpha[/itex][itex]\hat{p}[/itex].

    The wave is alinged by [itex]\alpha[/itex] with the two terms representing the relevant polarisation component projections. Therefore, the s and p-polarised components could be expressed by their own Fresnel equations - I just need to combine them. I'm just not sure how (I should have taken a Maths minor...) I'd appretiate any help.
    Last edited: Sep 4, 2012
  5. Sep 4, 2012 #4
    Averaging fields in not ideal; you should combine them by summing squares (and perhaps taking the root of the sum) because the two polarizations cannot interfere but their separate powers would add.

    This is what an "unpolarized detector" would measure as the reflection.
  6. Sep 7, 2012 #5
    Thanks for the tip. That looks like the way to go.
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