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(-α_{2}z)exp(-jβ_{2x}x), where α_{2} = β_{2}(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 H_{1}=2*E_{i0}/Z_{1}*cos(β_{1}zcosθ_{i})*exp(-jβ_{1}xsinθ_{i}) from here I would just find where β_{1}zcosθ_{i} = 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!
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?
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.
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(β_{1}*z*cos(θ_{i})). Setting this = 1 gives me a result of z = 1.462, quite a different answer. Am I using the reflection coefficient correctly? Thanks again!