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Reflected wave at complex refrective index

  1. Sep 8, 2011 #1
    A plane wave can be described by the real part of the exponential wave equation:

    [tex]\mathbf{E}=E_{0}e^{i(kz-wt)}[/tex]

    Adding the subscript i or r for incident or reflected waves, the ratio of the amplitude of reflected to incident wave is given by:

    [tex]\frac{E_{r0}}{E_{i0}} = \frac{n_1-n_2}{n_1+n_2}[/tex]

    But if n2 is complex, then this leads to a complex Er0. What does this mean for the physical wave, the real part of E?

    [tex]\mathbf{E}=E_{r0}e^{i(kz-wt)}=(Re\{E_{r0}\}+iIm\{E_{r0}\})e^{i(kz-wt)}=Re\{E_{r0}\}e^{i(kz-wt)}+Im\{E_{r0}\}e^{i(kz-wt+\pi/2)}[/tex]

    To me it looks like 2 out of phase waves are reflected. If this is right can you point me somewhere I can read up more about it? Or have I abused some notation somewhere?

    Thanks for your help understanding whats going on.
     
  2. jcsd
  3. Sep 8, 2011 #2

    DrDu

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    Science Advisor

    No, as you said, you have to take the real part of E, only. A complex E_0 can be writen as [itex] |E_0|\exp(i\phi)[/itex]. Hence the reflected wave will be out of phase with the incident one.
    For non-perpendicular incidence this leads to the reflected wave becoming partly circularly polarized and is used in elipsometry.
     
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