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A plane wave traveling in free space has an electric field phasor of $$ \textbf{E}(x,y,z) = E_{0}^{+}( \textbf{u}_{x} - \textbf{u}_{y} + \textbf{u}_{z})\cdot e^{-jk_{0}(x\sin \theta + z \cos \theta)} $$ and it hits a teflon-filled space with a $\theta = 45°$ angle. The power density is $1 \frac{KW}{m^{2}}$ and the frequency is $1 GHz$.

I have already found $E_{0}^{+}$ through the power density.

I have also found the reflected electric field expression splitting the given phasor into $TE$ and $TM$ parts and summing the results (since the wave travels on the $xz$ plane, $\textbf{u}_{x}$ and $\textbf{u}_{z}$ parts are in $TM$ mode, $\textbf{u}_{y}$ is in $TE$ mode. Is it correct?).

Now the last point is to find, if exists, the frequency at which the reflection coefficient in null. I have associated the reflection coefficient being null to the Brewster angle ($TM$ part), but I don't get why the frequency is asked since the Brewster angle has nothing to do with it. Or this reasoning is not correct and I have to think in terms of impedance matching?

ZeroTheHero
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wolph
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  • You are correct. Your setup suggests propagation in a lossless dielectric and then reflection coefficient depends only on incident angle and the impedances, which are real and independent of the frequencies... unless for one reason or another some parameters of your system like $\epsilon$ is given as frequency-dependent, which would then make the impedance of teflon frequency-dependent as well. Normally one chooses the incidence angle to extinguish the reflection with the Brewster angle. – ZeroTheHero Apr 21 '17 at 03:05
  • Thank you, I forgot to mention for teflon $\epsilon_{r} = 2$ – wolph Apr 21 '17 at 03:26
  • BTW... the direction of your $E$-field is not perpendicular to your direction of propagation so you're missing a small detail there... – ZeroTheHero Apr 21 '17 at 03:31
  • E-field propagates in $xz$ plane in $45°$ angle (given by the exponential part), but has components along all axis: $\textbf{u}{x}$ and $\textbf{u}{z}$ are part of the TM mode, $\textbf{u}_{y}$ the TE mode. – wolph Apr 21 '17 at 03:39
  • Your $\vec k= k_0(1,0,1)/\sqrt{2}$ is not orthogonal to $u_x+u_z$. – ZeroTheHero Apr 21 '17 at 03:45
  • Of course $\vec{k}$ belongs to $xz$ plane, but $\textbf{E}$ has components on all axis. – wolph Apr 21 '17 at 17:43
  • \begin{align} \vec E&=E_{0}^{+}( \textbf{u}{x} - \textbf{u}{y} + \textbf{u}{z})e^{-jk_0(x+z)/\sqrt{2}}, ,\ \vec k&=k_0( \textbf{u}{x} + \textbf{u}_{z})/\sqrt{2}, ,\ \vec E\cdot \vec k&=E_0^+k_0\sqrt{2}e^{-jk_0(x+z)/\sqrt{2}}\ne 0 \end{align} – ZeroTheHero Apr 21 '17 at 17:50
  • So what is wrong on what I have written in previous comments? Maybe I am missing something. – wolph Apr 21 '17 at 18:00
  • I'm not sure what's wrong... just sayin' your $\vec E$ is not $\perp$ to your $\vec k$, and they should be $\perp$ for a plane wave propagating in a lossless medium. – ZeroTheHero Apr 21 '17 at 18:17
  • Found this question explaining this and in my book. But some exercises with a phasor of this kind seemed to ignore the condition. So does it modify the solution in some way? – wolph Apr 21 '17 at 21:10

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