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Boundary conditions - Fresnel equations

  1. Mar 3, 2012 #1
    whenever I come to the derivation of the Fresnel equations I get stuck on the boundary condition for the component of the E-Field that is parallel to the surface.

    I know for the parallel components Maxwell dictates that:

    [itex]E_{1t}[/itex] = [itex]E_{2t}[/itex].

    For the parallel incoming light field component [itex]E_{it}[/itex], the reflected component [itex]E_{rt}[/itex] and the refracted one [itex]E_{tt}[/itex] it holds that:

    [itex]E_{it}[/itex] + [itex]E_{rt}[/itex] = [itex]E_{tt}[/itex].

    I always think about time though. I have the sequence in my head: ray coming in and then we have the refracted and reflected beam. Does that not apply because we just assume, that everything is happening at once?

    Would be very nice if something could shed some light on this. Thank you very much in advance :) and have a good one,
  2. jcsd
  3. Mar 5, 2012 #2
    The way the Fresnel equations problem is set up, we are dealing with infinite plane waves that have always been propagating, always reflecting, and always transmitting (this makes the math easier). There is no moment of reflection. We are not dealing with pencil-thin beams of laser-pulsed waves. That problem is much harder. In the Fresnel equations approach, we are treating the waves as extended electromagnetic fields that must match up at boundaries, rather than balls bouncing around. The reason this approach is useful is because you can represent any beam as a sum of monochromatic plane waves. For instance, to deal with a pulse of light (what you seem to have in your head), you would decompose it into its frequency components, apply the Fresnel equations or whatever else to each component, then sum the results to get your final solution.
  4. Mar 16, 2012 #3
    Sorry for not replying earlier. Thank you very much for your reply :)! It is clear now.

    Have a good one,
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