TM modes between 2 parallel plates with 2 dielectrics

[fluidistic]

Homework Statement

The space between 2 perfect conductor plates parallel to the x-z plane separated by a distance 2a are filled with 2 dielectric materials whose surface also lies in the x-z plane, at a distance equal to a from the plates. We're looking for TM modes propagating along the z direction.
1)Determine the TM modes and their cutoff frequencies.
2)Calculate the temporal mean of the power flux by mode and by unit of area of the surface.

Homework Equations

Helmholtz's equation for the transversal field $E_z$.

The Attempt at a Solution

I'm stuck on part 1), here is my attempt:
1)I am a bit confused on why they ask for several cutoff frequencies, because for TM modes I would have thought that there was a unique one.
So first I solve for the transversal field $E_z(x,y)$ which satisfies the Helmholtz's equation $(\nabla _\perp ^2 + \gamma ^2)E_z=0$ with the boundary conditions $E_z |_S=0$ and probably the condition $\varepsilon _1 E_{z1}=\varepsilon_2 E_{z2}$ at the surface between the 2 dielectric materials but I am not sure of it.
The solutions to the Helmholtz's equation is $E_z(x,y)=e^{ixm}\sin \left ( \frac{k\pi y}{2a} \right )$ where, if I'm not wrong, m=1, 2, 3, etc. and k=1, 2, 3, etc. This seem to satisfy the boundary condition.
This yields the eigenvalues $\gamma _{m,k}=\sqrt{m^2+\frac{k\pi^2}{2a}}$.
From this I am unsure how to get the cutoff frequency(ies). Is it just $\omega _c=\frac{\gamma}{\sqrt{\mu \varepsilon}}$ where $\gamma$ would be worth $\gamma _{m=1, k=1}$. And $\varepsilon$ would be worth $\varepsilon_1$ in region 1 and $\varepsilon_2$ in region 2, where both regions correspond to both dielectric materials.
Is this correct so far? I am unsure of my solution to the Helmholtz's equation, especially for the exponential part, maybe I should have taken either a sine or a cosine but I'm not sure, the exponential also seem to do the job and there's no particular boundary conditions that the function that depends on x must satisfy as far as I know; where I used separation of variables to solve it.

After this I can continue the problem with $\vec E_{\text{TM}}=[\frac{ih}{\gamma^2} \nabla _T E_z +E_z \hat z]e^{i(hz-\omega t)}$. Then I can calculate $\vec B_{\text{TM}}$ in a similar way and then finally get $\vec S$ the Poynting vector which I guess I must derivate with respect to time and divide by dxdy or something like that in order to answer to question 2).

 

[mfb]

Where do you take into account that the k to $\omega$ relation is different in the two dielectrics? Clearly the latter is fixed, so the first will be different in the two regions.

 

[fluidistic]

Where do you take into account that the k to $\omega$ relation is different in the two dielectrics? Clearly the latter is fixed, so the first will be different in the two regions.

I see. By k you mean the wavenumber right? Not the "k" I used as natural number I suppose. I think the wavenumber is what I denoted by h and so $h_i=\omega \sqrt{\mu_0 \varepsilon_i}$.
So in region 1 (where the 1st dielectric is), I have
$$\vec E_{\text{TM}}=[\frac{i\omega \sqrt{\mu_0 \varepsilon_1}}{\gamma^2} \nabla _\perp E_z +E_z \hat z]e^{i(\omega \sqrt{\mu_0 \varepsilon_1}z-\omega t)}$$.
In region 2, I'd have $$\vec E_{\text{TM}}=[\frac{i\omega \sqrt{\mu_0 \varepsilon_2}}{\gamma^2} \nabla _\perp E_z +E_z \hat z]e^{i(\omega \sqrt{\mu_0 \varepsilon_2}z-\omega t)}$$, does this look correct? I don't feel I used any equation that deals with what happens at the surface between the 2 dielectrics though...

 

[mfb]

By k you mean the wavenumber right? Not the "k" I used as natural number I suppose. I think the wavenumber is what I denoted by h

Ah okay. Then h, not k.

There is a boundary condition for the electric field at the surface between the dielectrics.

 

[fluidistic]

Ah okay. Then h, not k.

There is a boundary condition for the electric field at the surface between the dielectrics.

Thanks for still being with me on this problem!
Ok, maybe one of these is useful? : $\varepsilon_1 E_{1,\perp}=\varepsilon_2 E_{2,\perp}$ or $E_{1,\parallel}=E_{2,\parallel}$ where the perpendicular and parallel signs are with respect to the surface between the dielectrics. So the perpendicular part would be $E_{z}$.
That would make $E_{z,2}=\frac{\varepsilon_2}{\varepsilon_1}E_{z,1}$. Does this look correct?

 

[mfb]

I think the indices are wrong in the first equation. They are certainly inconsistent when compared to the last equation. Apart from that: yes.

Edit: fixed typo

 

[fluidistic]

I think the indices are wrong in the first equationm they are certainly inconsistent when compared to the last equation. Apart from that: yes.

Whoops, indeed. Thanks for all. The last one is the faulty one, I should get, I believe, $E_{z,2}=\frac{\varepsilon_1}{\varepsilon_2}E_{z,1}$.