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EM wave at boundary

  1. May 8, 2009 #1
    I have a simple question about reflecting EM waves at dielectric boundaries. To best illustrate my question, consider normal incidence. The incident wave has the wavevector k positive, and the reflected has k negative. Since B = k x E , and k has changed sign, B must also change sign. This is my problem - why cant E change sign instead? This would satisfy the necessary equations. It also makes intuitive sense, at normal incidence E and B are both parallel to the plane, and flipping either would preserve the handedness. Why is B special?
     
  2. jcsd
  3. May 8, 2009 #2

    clem

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    It depends on the dielectric constant epsilon. If epsilon is greater than 1, which is the usual case, then the reflected E will not flip. For epsilon less than 1, E would flip.
    For instance, for a wave leaving a dielectric into air, the reflected E does flip.
     
  4. May 9, 2009 #3
    Thanks clem. Do you know how to show this from the boundary conditions?
     
  5. May 9, 2009 #4
    k is the propagation vector, it represents the direction of the wave. When the wave reflects, the direction is reversed, hence k changes sign.
    If you change the sign of E instead of that of k, you have a wave traveling in the same direction as initially.
    Clem, can you help me understand what role permittivity has to play in the direction of E? I don't know at all about it..
     
  6. May 9, 2009 #5
    ksac I think you have misunderstood my question. Given that k changes sign, E or B (but not both) could change sign to satisfy the B = k x E condition. A further condition is required to constrain which one will flip.
     
  7. May 9, 2009 #6

    Born2bwire

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    The Fresnel reflection coefficients have already done this for you.
     
  8. May 9, 2009 #7
    The Fresnel reflection coefficients are derived from the boundary conditions - they assume what I am trying to prove.
     
  9. May 9, 2009 #8

    clem

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    It is done in all textbooks.
    In a dielectric, the magnitudes of E and B are related by B=sqrt{epsilon mu} E.
    From Maxwell's equations, both E and B tangential are continuous at the interface.
    This means (with mu =1) that E_1+E_1'=sqrt{epsilon}E_2
    and E_1-E_1'=E_2.
    Solve for the reflected E_1'.
     
  10. May 9, 2009 #9
    This is exactly the problem! Why do you choose to put the negative sign on the second equation rather than the first? From what I can see you are choosing to flip the electric field in this case.
     
  11. May 10, 2009 #10

    Born2bwire

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    It doesn't matter, you get the same solution.
     
  12. May 10, 2009 #11
    Yeah thanks I figured it out. It seems to just be a convention. You write E as the one which doesn't change, and if this turns out to be false it goes negative, which in turn makes B go positive.
     
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