1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook