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Homework Help: - Plane Wave Total Internal Reflection Problem

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A plane wave is incident at 70˚ to surface normal traveling within a medium with relative permittivity = 4, striking the boundary with air. Where is the nearest magnetic field max. to the boundary in the initial medium? Find the 1/e penetration depth of the evanescent wave.

    2. Relevant equations
    No equations are given, but i've been using:
    θC= sin-1(εr2/ εr1)^.5

    3. The attempt at a solution

    For the second part, i said that in the air, Et and Ht vary with the factor: exp(-α2z)exp(-jβ2xx), where
    α2 = β2(εr1/ εr2*sin2θi-1)^.5 = 1.59β2 the 1/e penetration distance is then just
    z = 1/(1.59β2) = .628β2

    The first part, however, is where I am having my main difficulty. I think I know how to do it were this to be a plane wave incident on a conductor, but I am not sure if I can use the same logic for the air interface given that I don't think I can assume that E = 0 at the boundary. ( for a conductor, i've been able to solve for H for a TE wave being H1=2*Ei0/Z1*cos(β1zcosθi)*exp(-jβ1xsinθi)
    from here I would just find where β1zcosθi = 0 and that would give the max. Does this still work for an air incidence though? And is there any max for a TM wave?

    Any help would be greatly appreciated! Thanks!
  2. jcsd
  3. Apr 30, 2010 #2
    The first part can be done by just considering that the field inside the medium can be represented as a sum of cosine/sine waves corresponding to the reflected and incident wave. The reflected ray will have experienced a phase delay upon reflection according to the Frensel equations. So the corresponding maximum (if we consider cosine waves and set the boundary at z = 0), will be the distance that the half of this phase delay corresponds to.

    I agree with your expression for [itex]\alpha_2[/itex] except for the [itex]\beta_2[/itex] term. Unless the material in which the wave is travelling is lossy, why would you expect there to be attenuation in the x direction?
  4. Apr 30, 2010 #3
    Sorry, missed the j term in the exponential with [itex]\beta_2[/itex]. But I still think it doesn't belong in the expression for [itex]\alpha_2[/itex]. It corresponds to the propagation constant in the x direction.
    Last edited: Apr 30, 2010
  5. Apr 30, 2010 #4
    Thanks for the help!

    Solving for the reflection coefficient, I get -.69-.73j which corresponds to an angle of -133.47˚. It then would make sense that the maximum would be at half of this away, ie 66.735˚. However, this should only correspond to .185λ. From adding up the equations for incident and reflected waves, however, I get a dependency on cos(β1*z*cos(θi)). Setting this = 1 gives me a result of z = 1.462, quite a different answer. Am I using the reflection coefficient correctly?

    Thanks again!
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