Attached are the drawings. Does the polarity(adsbygoogle = window.adsbygoogle || []).push({}); Eof depend on [itex] \Gamma =\frac {\eta_2-\eta_1}{\eta_2+\eta_1}[/itex]

So if [itex]\eta_2<\eta_1[/itex], the Γ is negative and the polarity of the reflected wave is opposite polarity?

I understand [itex]\theta_i=\theta_r[/itex] and all that. As shown in the drawing, if you look at the TEM wave travel from the left at z=-ve, in medium 1, hitting the boundary on xy plane at z=0, say it isperpendicular polarizationwhere [itex]\vec {E}= \hat {y} E(z)[/itex], which is parallel to the boundary. If [itex]\eta 1 > \eta 2\;\Rightarrow\; \Gamma=-ve[/itex], is [itex] \vec {E}_r[/itex] in opposite direction........[itex]\vec {E}_r=-\hat {y} E_r[/itex] as in the upper left drawing.

And if [itex]\eta 2 > \eta 1\;\Rightarrow\; \Gamma = +ve[/itex], then [itex] \vec {E}_r=\hat{y}E_r[/itex]. This is shown in the upper right drawing.

The upper left shows [itex]\eta 1 > \eta 2 [/itex] whereEi andEr are pointing inydirection. In drawing on top right where [itex]\eta 1 < \eta 2 [/itex],Er is in -veydirection.

I drew the case whereEi is on the xz plane as shown in lower left drawing, how do I even determine the direction of the reflection based on the intrinsic impedance of the two media? Please help. Please provide some article links if you can.

Thanks

Alan

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# Determine the polarity of the reflected EM wave.

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