Normal incidence of a plane polarized wave through multiple mediums

[DannyJ108]

Homework Statement

Find the amplitudes of the wave in each medium

Relevant Equations

$E_i = E_i e^{-jK_i z} +E^'_i e^{jK_i z}$

Hello everybody,

I have to find the amplitudes of a wave that goes through 4 different mediums in terms of $E_0$, suffering reflection in the first three but not the last one. I calculated the corresponding reflection indexes of the three mediums (all of them real).
Following calculations, I find that the first and second amplitudes are real, but I get that the ones in the third and last medium are complex. I developed Eulers equation $e^{jx}$ and got a real and imaginary component for these amplitudes. Is this correct? Or how can I interpret this?

Thank you.

 

[berkeman]

Can you write the reflection equation that you are using? And what are the dielectric constants of each medium? Can you show your calculations in detail? (preferably using LaTeX) Thank you.

 

[DannyJ108]

Assuming the wave travels in direction $z$ and $E$ in direction $x$, and $H$ in direction $y$, putting the system reference $z=0$ at the first boundary between mediums 0 and 1, and being:

$Medium 1 = 2cm$
$Medium 2 =3 cm$
$Medium 3 = Indefinite$

For mediums 0-3 (being $n_0=1$ (air), $n_1=3, n_2=2, n_3=1$), just substitute $i$ for each medium:

$E_{ix} = E_i e^{-jK_iz} + E^{'}_i e^{jK_iz}$
$H_{iy} = \frac {E_i} {\eta_i}e^{-jK_iz} + \frac {E^{'}_i} {\eta_i}e^{jK_iz}$

For medium 4 (no reflection):

$E_{3x} = E_3 e^{-jK_3z}$
$H_{3y} = \frac {E_3} {\eta_3}e^{-jK_3z}$

Using the following and taking d as 0,01 m:

$\rho_{01}= \left. \frac {E^{'}_0 e^{jK_0z}} {E_0 e^{-jK_0z}} \right|_{z=0} \Rightarrow E^{'}_0 = E_0 \rho_{01}$
$\rho_{12}= \left. \frac {E^{'}_1 e^{jK_1z}} {E_1 e^{-jK_1z}} \right|_{z=2d} \Rightarrow E^{'}_1 = E_1 \rho_{12} e^{-4jK_1d}$
$\rho_{23}= \left. \frac {E^{'}_2 e^{jK_2z}} {E_2 e^{-jK_2z}} \right|_{z=5d} \Rightarrow E^{'}_2 = E_2 \rho_{23} e^{-10jK_1d}$

Aplying continuity conditions $E_i(z)=E_{i+1}(z)$ at $z=0, z=2d, z=5d$ and solving the equation systems, I get that the reflection coefficients are:

$\rho_{01}= - \frac 1 2$
$\rho_{12}= \frac 1 2$
$\rho_{23}= \frac 1 3$

Taking into account that $k_i = \frac \omega {v_i}$, operating using relations between $\eta, \epsilon, c$ and $n$, I can figure out each $k$.
So, substituting in my $E^{'}_0, E^{'}_1, E^{'}_2$ formulas and then in the field equations I get:

$E_{0x} = E_0 e^{-jK_0z} - \frac {E_0} 2 e^{jK_0z}$
$H_{0y} = \frac {E_0} {\eta_0}e^{-jK_0z} + \frac {E_0} {2 \eta_i}e^{jK_0z}$

$E_{1x} = E_1 e^{-jK_1z} + \frac {E_1} 2 e^{jK_1z}$
$H_{1y} = \frac {E_1} {\eta_1}e^{-jK_1z} - \frac {E_1} {2 \eta_i}e^{jK_1z}$

$E_{2x} = E_2 e^{-jK_2z} + \frac {E_2} 3 \left(- \frac 1 2 -j \frac {\sqrt3} 2 \right)e^{jK_2z}$
$H_{1y} = \frac {E_2} {\eta_2}e^{-jK_2z} -\frac {E_2} {3 \eta_2} \left(- \frac 1 2 -j \frac {\sqrt3} 2 \right)e^{jK_2z}$

$E_{3x} = E_3 e^{-jK_3z}$
$H_{3y} = \frac {E_3} {\eta_3}e^{-jK_3z}$

So applying again continuityconditions, but with these equations at the layer boundaries, I get that the amplitudes in terms of $E_0$ are:

$E_1= \frac {E_0} 3$
$E_2= - \frac 3 {16} E_0 (1+j\sqrt3)$
$E_3 = \frac {E_0} 4 (-1 +j\sqrt3)$

I don't know if this is correct or not.
I'm so frustrated with this exercise, I can't figure out if this is correct and don't know how to interpret it or if it's plain wrong. Please help. I put a lot of effort into writing all this in LaTex, didn't have much idea how to first. Thank you in advance kind sir!

 

[DannyJ108]

No worries. I found the solution on my own