Plane Wave Total Internal Reflection Problem

[marksambob]

Homework Statement

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.

Homework Equations

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

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(-jβ_2xx), 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*E_i0/Z₁*cos(β₁zcosθ_i)*exp(-jβ₁xsinθ_i)
from here I would just find where β₁zcosθ_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!

 

[BerryBoy]

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 \alpha_2 except for the \beta_2 term. Unless the material in which the wave is traveling is lossy, why would you expect there to be attenuation in the x direction?

 

[BerryBoy]

Sorry, missed the j term in the exponential with \beta_2. But I still think it doesn't belong in the expression for \alpha_2. It corresponds to the propagation constant in the x direction.

 

[marksambob]

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(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!