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Update 2022-03-07 20:40

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Jean-Sébastien
2022-03-07 20:40:36 +01:00
parent 21bf9fdba5
commit 4808df71e6
194 changed files with 1487 additions and 5980 deletions
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build/emdm_emwm_refl_ni.html
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-02 Wed 15:45 -->
<!-- 2022-03-07 Mon 20:38 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1098,14 +1098,6 @@ Table of contents
<li>
<a href="./emdm_emwm_refl_oi.html#emdm_emwm_refl_oi">Oblique Incidence</a><span class="headline-id">emdm.emwm.refl.oi</span>
</li>
<li>
<a href="./emdm_emwm_refl_Fe.html#emdm_emwm_refl_Fe">Fresnel's Equations</a><span class="headline-id">emdm.emwm.refl.Fe</span>
</li>
<li>
<a href="./emdm_emwm_refl_Ba.html#emdm_emwm_refl_Ba">Brewster's Angle</a><span class="headline-id">emdm.emwm.refl.Ba</span>
</li>
</ul>
@@ -1629,7 +1621,7 @@ in \(x\) direction approaches surface:
{\boldsymbol B}_I (z,t) = \frac{1}{v_1} E_{0_I} e^{i (k_1 z - \omega t)} \hat{\boldsymbol y}
\label{Gr(9.75)}
\]
(in which we have used \(B_0 = \frac{1}{v} E_0\), \ref{Gr(9.47)}).
(in which we have used \(B_0 = \frac{1}{v} E_0\), <a href="./emd_emw_mpw.html#EBmpw">EBmpw</a>.
</p>
<p>
@@ -1644,19 +1636,19 @@ Reflected wave:
<p>
For the transmitted wave, we put
\[
{\boldsymbol E}_T (z,t) = E_{0_T} e^{i (k_2 z - \omega t)} \hat{\boldsymbol x}, \hspace{1cm}
{\boldsymbol B}_T (z,t) = \frac{1}{v_2} E_{0_T} e^{i (k_2 z - \omega t)} \hat{\boldsymbol y}.
\label{Gr(9.75)}
\]
</p>
\begin{align*}
{\boldsymbol E}_T (z,t) &amp;= E_{0_T} e^{i (k_2 z - \omega t)} \hat{\boldsymbol x}, \\
{\boldsymbol B}_T (z,t) &amp;= \frac{1}{v_2} E_{0_T} e^{i (k_2 z - \omega t)} \hat{\boldsymbol y}.
\end{align*}
<p>
Our boundary is by choice of coordinate system at \(z = 0\). Our setup calls for solving the boundary conditions <a href="./emdm_Me_bc.html#disc_nfc">disc_nfc</a> with \({\boldsymbol E}_I + {\boldsymbol E}_R\) and \({\boldsymbol B}_I + {\boldsymbol B}_R\) on one side, and \({\boldsymbol E}_T\) and \({\boldsymbol B}_T\) on the other.
</p>
<p>
Our boundary is by choice of coordinate system at \(z = 0\). Our setup calls for solving the boundary conditions \ref{eq:EMBdryCondAtMediumInterface} with \({\boldsymbol E}_I + {\boldsymbol E}_R\) and \({\boldsymbol B}_I + {\boldsymbol B}_R\) on one side, and \({\boldsymbol E}_T\) and \({\boldsymbol B}_T\) on the other.
</p>
<p>
At normal incidence, there are no perpendicular components of the fields relative to the surface, so \ref{eq:EMBdryCondAtMediumInterface} (i) and (ii) are obeyed. (iii) means that
At normal incidence, there are no perpendicular components of the fields relative to the surface, so <a href="./emdm_Me_bc.html#disc_nfc">disc_nfc</a> (i) and (ii) are obeyed. (iii) means that
\[
E_{0_I} + E_{0_R} = E_{0_T}
\label{Gr(9.78)}
@@ -1677,10 +1669,27 @@ E_{0_I} - E_{0_R} = \beta E_{0_T}, \hspace{1cm}
<p>
Solving these coupled equations, we can write
outgoing amplitudes in terms of incident ones:
</p>
<div class="eqlabel" id="org653a14a">
<p>
<a id="ERT"></a><a href="./emdm_emwm_refl_ni.html#ERT"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org5690572">
<ul class="org-ul">
<li>Gr (9.82)</li>
</ul>
</div>
</div>
<p>
\[
E_{0_R} = \frac{1 - \beta}{1 + \beta} E_{0_I}, \hspace{10mm}
E_{0_T} = \frac{2}{1 + \beta} E_{0_I}.
\label{Gr(9.82)}
\tag{ERT}\label{ERT}
\]
</p>
@@ -1701,12 +1710,14 @@ R \equiv \frac{I_R}{I_I} = \frac{E^2_{0_R}}{E^2_{0_I}} = \left( \frac{1 - \beta}
\label{Gr(9.86)}
\]
while the transmitted intensity is
\[
</p>
\begin{equation}
T \equiv \frac{I_T}{I_I} = \frac{v_2 \varepsilon_2}{v_1 \varepsilon_1} \frac{E^2_{0_T}}{E^2_{0_I}}
= \sqrt{\frac{\mu_1 \varepsilon_2}{\mu_2 \varepsilon_1}} \frac{E^2_{0_T}}{E^2_{0_I}} = \frac{4 \beta}{(1 + \beta)^2}.
\label{Gr(9.86)}
\]
We thus have that the {\bf reflection} and {\bf transmission coefficients} satisfy
\end{equation}
<p>
We thus have that the <b>reflection</b> and <b>transmission coefficients</b> satisfy
\[
R + T = 1.
\label{Gr(9.88)}
@@ -1732,7 +1743,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
</div>
<div id="postamble" class="status">
<p class="author">Author: Jean-Sébastien Caux</p>
<p class="date">Created: 2022-03-02 Wed 15:45</p>
<p class="date">Created: 2022-03-07 Mon 20:38</p>
<p class="validation"></p>
</div>