POINT
【関連記事】
- 散乱問題と同じように，「定常的な状態を扱う方法」と「波を追跡する方法」がある．
- 前者では「定常的な解＋境界条件」をもとに解き，後者は「すべての反射波・透過波（この際，境界条件を考慮）を合成する」ことで解く．
【メモ】この記事の分量が多くなったため，他のパターン（斜め入射など）は，別記事で書く予定．
境界条件を課す方法
電磁波
$x$軸方向に直線的な偏りをもつ，角振動数$\omega_0$の平面波を$z$軸正方向に入射することを考える．境界面に対して接線成分だけがゼロでない値を持つから，境界条件は$\boldsymbol{E}(\boldsymbol{x},t), \boldsymbol{H}(\boldsymbol{x},t)$の接線成分の連続性を考える必要がある．
各誘電体中での電場を
\begin{aligned}
&E_x(z,t)\\
&=
\begin{cases}
\, E_\mathrm{i} e^{i(k_1 z-\omega_0 t)} + E_\mathrm{r} e^{-i(k_1 z+\omega_0 t)} & (z<0) \\
\, \tilde{E}_\mathrm{i} e^{i(k_2 z-\omega_0 t)} + \tilde{E}_\mathrm{r} e^{-i(k_2 z+\omega_0 t)} & (0 < z < d) \\
\, E_\mathrm{t} e^{i(k_3 z-\omega_0 t)} & (d < z)
\end{cases}
\end{aligned}
とすれば，関連記事[A]で見たように$\displaystyle \boldsymbol{B}=\frac{\boldsymbol{k}}{\omega_0} \times \boldsymbol{E}_0 e^{i (\boldsymbol{k}\cdot\boldsymbol{x}-\omega_0 t)}$が成り立つことから&E_x(z,t)\\
&=
\begin{cases}
\, E_\mathrm{i} e^{i(k_1 z-\omega_0 t)} + E_\mathrm{r} e^{-i(k_1 z+\omega_0 t)} & (z<0) \\
\, \tilde{E}_\mathrm{i} e^{i(k_2 z-\omega_0 t)} + \tilde{E}_\mathrm{r} e^{-i(k_2 z+\omega_0 t)} & (0 < z < d) \\
\, E_\mathrm{t} e^{i(k_3 z-\omega_0 t)} & (d < z)
\end{cases}
\end{aligned}
\begin{aligned}
&\boldsymbol{B}(\boldsymbol{x},t)
=B_y(z,t) \boldsymbol{e}_y, \\
&B_y(z,t) \\
&=
\begin{cases}
\, [E_\mathrm{i} e^{i(k_1 z-\omega_0 t)} - E_\mathrm{r} e^{-i(k_1 z+\omega_0 t)}]/v_1 & (z<0) \\
\, [\tilde{E}_\mathrm{i} e^{i(k_2 z-\omega_0 t)} - \tilde{E}_\mathrm{r} e^{-i(k_2 z+\omega_0 t)}]/v_2 & (0 < z < d) \\
\, [E_\mathrm{t} e^{i(k_3 z-\omega_0 t)}]/v_3 & (d < z)
\end{cases}
\end{aligned}
となる（$v_i(\omega)=1/\sqrt{\varepsilon_i(\omega)\mu_i(\omega)}=\omega/k_i$，$v_i=v_i(\omega_0)$）．&\boldsymbol{B}(\boldsymbol{x},t)
=B_y(z,t) \boldsymbol{e}_y, \\
&B_y(z,t) \\
&=
\begin{cases}
\, [E_\mathrm{i} e^{i(k_1 z-\omega_0 t)} - E_\mathrm{r} e^{-i(k_1 z+\omega_0 t)}]/v_1 & (z<0) \\
\, [\tilde{E}_\mathrm{i} e^{i(k_2 z-\omega_0 t)} - \tilde{E}_\mathrm{r} e^{-i(k_2 z+\omega_0 t)}]/v_2 & (0 < z < d) \\
\, [E_\mathrm{t} e^{i(k_3 z-\omega_0 t)}]/v_3 & (d < z)
\end{cases}
\end{aligned}
ここで，関連記事[A]で解説したように，分散性を持つ一様な誘電体中では（強誘電体や強磁性体を除き）$\boldsymbol{H}(\boldsymbol{x},t)=\boldsymbol{B}(\boldsymbol{x},t)/\mu(\omega_0)$となる．よって，
\begin{aligned}
&\boldsymbol{H}(\boldsymbol{x},t)
=H_y(z,t) \boldsymbol{e}_y, \\
&H_y(z,t) \\
&=
\begin{cases}
\, [E_\mathrm{i} e^{i(k_1 z-\omega_0 t)} - E_\mathrm{r} e^{-i(k_1 z+\omega_0 t)}]N_1 & (z<0) \\
\, [\tilde{E}_\mathrm{i} e^{i(k_2 z-\omega_0 t)} - \tilde{E}_\mathrm{r} e^{-i(k_2 z+\omega_0 t)}]N_2 & (0 < z < d) \\
\, [E_\mathrm{t} e^{i(k_3 z-\omega_0 t)}]N_3 & (d < z)
\end{cases}
\end{aligned}
である（$\mu_i=\mu_i(\omega_0)$，$\varepsilon_i=\varepsilon_i(\omega_0)$，$N_i=\sqrt{\varepsilon_i/\mu_i}=1/(v_i \mu_i)$）．&\boldsymbol{H}(\boldsymbol{x},t)
=H_y(z,t) \boldsymbol{e}_y, \\
&H_y(z,t) \\
&=
\begin{cases}
\, [E_\mathrm{i} e^{i(k_1 z-\omega_0 t)} - E_\mathrm{r} e^{-i(k_1 z+\omega_0 t)}]N_1 & (z<0) \\
\, [\tilde{E}_\mathrm{i} e^{i(k_2 z-\omega_0 t)} - \tilde{E}_\mathrm{r} e^{-i(k_2 z+\omega_0 t)}]N_2 & (0 < z < d) \\
\, [E_\mathrm{t} e^{i(k_3 z-\omega_0 t)}]N_3 & (d < z)
\end{cases}
\end{aligned}
境界面における$\boldsymbol{E}(\boldsymbol{x},t), \boldsymbol{H}(\boldsymbol{x},t)$の接線成分の連続性から，
\begin{aligned}
&\underline{z=0:}\\
&\quad\text{(B1)}
\begin{cases}
\, E_\mathrm{i} + E_\mathrm{r} = \tilde{E}_\mathrm{i} + \tilde{E}_\mathrm{r} \\
\, N_1(E_\mathrm{i} - E_\mathrm{r}) = N_2(\tilde{E}_\mathrm{i} - \tilde{E}_\mathrm{r})
\end{cases}\\
&\underline{z=d:}\\
&\quad\text{(B2)}
\begin{cases}
\, \tilde{E}_\mathrm{i} e^{ik_2 d} + \tilde{E}_\mathrm{r} e^{-ik_2 d} = E_\mathrm{t} e^{i k_3 d} \\
\, N_2(\tilde{E}_\mathrm{i} e^{ik_2 d} - \tilde{E}_\mathrm{r} e^{-ik_2 d}) = N_3 E_\mathrm{t} e^{i k_3 d}
\end{cases}
\end{aligned}
となる．&\underline{z=0:}\\
&\quad\text{(B1)}
\begin{cases}
\, E_\mathrm{i} + E_\mathrm{r} = \tilde{E}_\mathrm{i} + \tilde{E}_\mathrm{r} \\
\, N_1(E_\mathrm{i} - E_\mathrm{r}) = N_2(\tilde{E}_\mathrm{i} - \tilde{E}_\mathrm{r})
\end{cases}\\
&\underline{z=d:}\\
&\quad\text{(B2)}
\begin{cases}
\, \tilde{E}_\mathrm{i} e^{ik_2 d} + \tilde{E}_\mathrm{r} e^{-ik_2 d} = E_\mathrm{t} e^{i k_3 d} \\
\, N_2(\tilde{E}_\mathrm{i} e^{ik_2 d} - \tilde{E}_\mathrm{r} e^{-ik_2 d}) = N_3 E_\mathrm{t} e^{i k_3 d}
\end{cases}
\end{aligned}
\begin{aligned}
\text{(B1)}
&\Leftrightarrow
\begin{cases}
\, \displaystyle
E_\mathrm{i}
= \frac{1}{2}\biggl[\biggl(1+\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{i}
+ \biggl(1-\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{r} \biggr] \\[8pt]
\, \displaystyle
E_\mathrm{r}
= \frac{1}{2}\biggl[\biggl(1-\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{i}
+ \biggl(1+\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{r} \biggr]
\end{cases}\\
\text{(B2)}
&\Leftrightarrow
\begin{cases}
\, \tilde{E}_\mathrm{i} e^{i(k_2-k_3) d} + \tilde{E}_\mathrm{r} e^{-i(k_2+k_3) d} = E_\mathrm{t} \\
\, \tilde{E}_\mathrm{i} e^{i(k_2-k_3) d} - \tilde{E}_\mathrm{r} e^{-i(k_2+k_3) d} = (N_3/N_2) E_\mathrm{t}
\end{cases}\\
&\Leftrightarrow
\begin{cases}
\, \displaystyle
\tilde{E}_\mathrm{i}
= \frac{1}{2}\biggl(1+\frac{N_3}{N_2} \biggr) E_\mathrm{t} e^{i(-k_2+k_3) d} \\[8pt]
\, \displaystyle
\tilde{E}_\mathrm{r}
= \frac{1}{2}\biggl(1-\frac{N_3}{N_2} \biggr) E_\mathrm{t} e^{i(k_2+k_3) d}
\end{cases}
\end{aligned}
であるから，\text{(B1)}
&\Leftrightarrow
\begin{cases}
\, \displaystyle
E_\mathrm{i}
= \frac{1}{2}\biggl[\biggl(1+\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{i}
+ \biggl(1-\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{r} \biggr] \\[8pt]
\, \displaystyle
E_\mathrm{r}
= \frac{1}{2}\biggl[\biggl(1-\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{i}
+ \biggl(1+\frac{N_2}{N_1} \biggr)\tilde{E}_\mathrm{r} \biggr]
\end{cases}\\
\text{(B2)}
&\Leftrightarrow
\begin{cases}
\, \tilde{E}_\mathrm{i} e^{i(k_2-k_3) d} + \tilde{E}_\mathrm{r} e^{-i(k_2+k_3) d} = E_\mathrm{t} \\
\, \tilde{E}_\mathrm{i} e^{i(k_2-k_3) d} - \tilde{E}_\mathrm{r} e^{-i(k_2+k_3) d} = (N_3/N_2) E_\mathrm{t}
\end{cases}\\
&\Leftrightarrow
\begin{cases}
\, \displaystyle
\tilde{E}_\mathrm{i}
= \frac{1}{2}\biggl(1+\frac{N_3}{N_2} \biggr) E_\mathrm{t} e^{i(-k_2+k_3) d} \\[8pt]
\, \displaystyle
\tilde{E}_\mathrm{r}
= \frac{1}{2}\biggl(1-\frac{N_3}{N_2} \biggr) E_\mathrm{t} e^{i(k_2+k_3) d}
\end{cases}
\end{aligned}
\begin{aligned}
E_\mathrm{i}
&=\frac{1}{4}\biggl[\biggl(1+\frac{N_2}{N_1} \biggr) \biggl(1+\frac{N_3}{N_2} \biggr)e^{-ik_2d}
+\biggl(1-\frac{N_2}{N_1} \biggr) \biggl(1-\frac{N_3}{N_2} \biggr)e^{ik_2d} \biggr] e^{ik_3 d}E_\mathrm{t} \\
&=\underbrace{
\frac{1}{2}\biggl[\biggl(1+\frac{N_3}{N_1} \biggr) \cos(k_2 d)
-i\biggl(\frac{N_2}{N_1}+\frac{N_3}{N_2} \biggr)\sin(k_2 d) \biggr] e^{ik_3 d}}_{=A} E_\mathrm{t} \\
E_\mathrm{r}
&=\frac{1}{4}\biggl[\biggl(1-\frac{N_2}{N_1} \biggr) \biggl(1+\frac{N_3}{N_2} \biggr)e^{-ik_2d}
+\biggl(1+\frac{N_2}{N_1} \biggr) \biggl(1-\frac{N_3}{N_2} \biggr)e^{ik_2d} \biggr] e^{ik_3 d}E_\mathrm{t} \\
&=\underbrace{
\frac{1}{2}\biggl[\biggl(1-\frac{N_3}{N_1} \biggr) \cos(k_2 d)
+i\biggl(\frac{N_2}{N_1}-\frac{N_3}{N_2} \biggr)\sin(k_2 d) \biggr] e^{ik_3 d}}_{=B} E_\mathrm{t}
\end{aligned}
となる．E_\mathrm{i}
&=\frac{1}{4}\biggl[\biggl(1+\frac{N_2}{N_1} \biggr) \biggl(1+\frac{N_3}{N_2} \biggr)e^{-ik_2d}
+\biggl(1-\frac{N_2}{N_1} \biggr) \biggl(1-\frac{N_3}{N_2} \biggr)e^{ik_2d} \biggr] e^{ik_3 d}E_\mathrm{t} \\
&=\underbrace{
\frac{1}{2}\biggl[\biggl(1+\frac{N_3}{N_1} \biggr) \cos(k_2 d)
-i\biggl(\frac{N_2}{N_1}+\frac{N_3}{N_2} \biggr)\sin(k_2 d) \biggr] e^{ik_3 d}}_{=A} E_\mathrm{t} \\
E_\mathrm{r}
&=\frac{1}{4}\biggl[\biggl(1-\frac{N_2}{N_1} \biggr) \biggl(1+\frac{N_3}{N_2} \biggr)e^{-ik_2d}
+\biggl(1+\frac{N_2}{N_1} \biggr) \biggl(1-\frac{N_3}{N_2} \biggr)e^{ik_2d} \biggr] e^{ik_3 d}E_\mathrm{t} \\
&=\underbrace{
\frac{1}{2}\biggl[\biggl(1-\frac{N_3}{N_1} \biggr) \cos(k_2 d)
+i\biggl(\frac{N_2}{N_1}-\frac{N_3}{N_2} \biggr)\sin(k_2 d) \biggr] e^{ik_3 d}}_{=B} E_\mathrm{t}
\end{aligned}
以上より，反射波・透過波の振幅を「入射波の振幅」を用いて表すと
振幅
\begin{aligned}
E_\mathrm{r}
&=\frac{B}{A} E_\mathrm{i} \\
&=\frac{\bigl(1-\frac{N_2}{N_1} \bigr) \bigl(1+\frac{N_3}{N_2} \bigr)e^{-ik_2d}
+\bigl(1+\frac{N_2}{N_1} \bigr) \bigl(1-\frac{N_3}{N_2} \bigr)e^{ik_2d}}
{\bigl(1+\frac{N_2}{N_1} \bigr) \bigl(1+\frac{N_3}{N_2} \bigr)e^{-ik_2d}
+\bigl(1-\frac{N_2}{N_1} \bigr) \bigl(1-\frac{N_3}{N_2} \bigr)e^{ik_2d}}
E_\mathrm{i} \\
E_\mathrm{t}
&=\frac{1}{A} E_\mathrm{i} \\
&=\frac{4 e^{-ik_3 d}}
{\bigl(1+\frac{N_2}{N_1} \bigr) \bigl(1+\frac{N_3}{N_2} \bigr)e^{-ik_2d}
+\bigl(1-\frac{N_2}{N_1} \bigr) \bigl(1-\frac{N_3}{N_2} \bigr)e^{ik_2d}} E_\mathrm{i}
\end{aligned}
E_\mathrm{r}
&=\frac{B}{A} E_\mathrm{i} \\
&=\frac{\bigl(1-\frac{N_2}{N_1} \bigr) \bigl(1+\frac{N_3}{N_2} \bigr)e^{-ik_2d}
+\bigl(1+\frac{N_2}{N_1} \bigr) \bigl(1-\frac{N_3}{N_2} \bigr)e^{ik_2d}}
{\bigl(1+\frac{N_2}{N_1} \bigr) \bigl(1+\frac{N_3}{N_2} \bigr)e^{-ik_2d}
+\bigl(1-\frac{N_2}{N_1} \bigr) \bigl(1-\frac{N_3}{N_2} \bigr)e^{ik_2d}}
E_\mathrm{i} \\
E_\mathrm{t}
&=\frac{1}{A} E_\mathrm{i} \\
&=\frac{4 e^{-ik_3 d}}
{\bigl(1+\frac{N_2}{N_1} \bigr) \bigl(1+\frac{N_3}{N_2} \bigr)e^{-ik_2d}
+\bigl(1-\frac{N_2}{N_1} \bigr) \bigl(1-\frac{N_3}{N_2} \bigr)e^{ik_2d}} E_\mathrm{i}
\end{aligned}
\begin{aligned}
E_\mathrm{r}
&=\frac{\bigl(1 - \frac{N_{3}}{N_{1}}\bigr) \cos(k_{2}d) + i \bigl(\frac{N_2}{N_1} - \frac{N_3}{N_2} \bigr) \sin(k_{2}d)}
{\bigl(1 + \frac{N_{3}}{N_{1}}\bigr) \cos(k_{2}d) - i \bigl(\frac{N_2}{N_1} + \frac{N_3}{N_2} \bigr) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{(N_{1}N_{2} - N_{2}N_{3}) \cos(k_{2}d) + i(N_{2}^{2} - N_{1}N_{3}) \sin(k_{2}d)}
{(N_{1}N_{2} + N_{2}N_{3}) \cos(k_{2}d) - i(N_{2}^{2} + N_{1}N_{3}) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{(N_{2}^{-1}N_{3}^{-1} - N_{1}^{-1}N_{2}^{-1}) \cos(k_{2}d) + i(N_{1}^{-1} N_{3}^{-1} - N_{2}^{-2}) \sin(k_{2}d)}
{(N_{2}^{-1}N_{3}^{-1} + N_{1}^{-1}N_{2}^{-1}) \cos(k_{2}d) - i(N_{1}^{-1} N_{3}^{-1} + N_{2}^{-2}) \sin(k_{2}d)}
E_{\mathrm{i}}
\end{aligned}
E_\mathrm{r}
&=\frac{\bigl(1 - \frac{N_{3}}{N_{1}}\bigr) \cos(k_{2}d) + i \bigl(\frac{N_2}{N_1} - \frac{N_3}{N_2} \bigr) \sin(k_{2}d)}
{\bigl(1 + \frac{N_{3}}{N_{1}}\bigr) \cos(k_{2}d) - i \bigl(\frac{N_2}{N_1} + \frac{N_3}{N_2} \bigr) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{(N_{1}N_{2} - N_{2}N_{3}) \cos(k_{2}d) + i(N_{2}^{2} - N_{1}N_{3}) \sin(k_{2}d)}
{(N_{1}N_{2} + N_{2}N_{3}) \cos(k_{2}d) - i(N_{2}^{2} + N_{1}N_{3}) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{(N_{2}^{-1}N_{3}^{-1} - N_{1}^{-1}N_{2}^{-1}) \cos(k_{2}d) + i(N_{1}^{-1} N_{3}^{-1} - N_{2}^{-2}) \sin(k_{2}d)}
{(N_{2}^{-1}N_{3}^{-1} + N_{1}^{-1}N_{2}^{-1}) \cos(k_{2}d) - i(N_{1}^{-1} N_{3}^{-1} + N_{2}^{-2}) \sin(k_{2}d)}
E_{\mathrm{i}}
\end{aligned}
\begin{aligned}
E_\mathrm{t}
&=\frac{2 e^{-ik_3 d}}
{\bigl(1 + \frac{N_{3}}{N_{1}}\bigr) \cos(k_{2}d) - i \bigl(\frac{N_2}{N_1} + \frac{N_3}{N_2} \bigr) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{2 N_{1}N_{2} e^{-ik_3 d}}
{(N_{1}N_{2} + N_{2}N_{3}) \cos(k_{2}d) - i(N_{2}^{2} + N_{1}N_{3}) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{2 N_{2}^{-1}N_{3}^{-1} e^{-ik_3 d}}
{(N_{2}^{-1}N_{3}^{-1} + N_{1}^{-1}N_{2}^{-1}) \cos(k_{2}d) - i(N_{1}^{-1} N_{3}^{-1} + N_{2}^{-2}) \sin(k_{2}d)}
E_{\mathrm{i}}
\end{aligned}
とも表せる．E_\mathrm{t}
&=\frac{2 e^{-ik_3 d}}
{\bigl(1 + \frac{N_{3}}{N_{1}}\bigr) \cos(k_{2}d) - i \bigl(\frac{N_2}{N_1} + \frac{N_3}{N_2} \bigr) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{2 N_{1}N_{2} e^{-ik_3 d}}
{(N_{1}N_{2} + N_{2}N_{3}) \cos(k_{2}d) - i(N_{2}^{2} + N_{1}N_{3}) \sin(k_{2}d)}
E_{\mathrm{i}} \\
&=\frac{2 N_{2}^{-1}N_{3}^{-1} e^{-ik_3 d}}
{(N_{2}^{-1}N_{3}^{-1} + N_{1}^{-1}N_{2}^{-1}) \cos(k_{2}d) - i(N_{1}^{-1} N_{3}^{-1} + N_{2}^{-2}) \sin(k_{2}d)}
E_{\mathrm{i}}
\end{aligned}
エネルギーの反射率$r$と透過率$t$（定義や表式は関連記事[A]参照）は，それぞれ
反射率・透過率
\begin{aligned}
r&=\biggl| \frac{E_\mathrm{r}}{E_\mathrm{i}}\biggr|^2
=\biggl| \frac{B}{A}\biggr|^2 \\
&=\frac{\displaystyle -2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)}
{\displaystyle 2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)} \\
t&=\frac{N_3}{N_1} \biggl| \frac{E_\mathrm{t}}{E_\mathrm{i}}\biggr|^2
=\frac{N_3}{N_1} \biggl| \frac{1}{A}\biggr|^2\\
&=\frac{4}{\displaystyle 2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)}
\end{aligned}
r&=\biggl| \frac{E_\mathrm{r}}{E_\mathrm{i}}\biggr|^2
=\biggl| \frac{B}{A}\biggr|^2 \\
&=\frac{\displaystyle -2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)}
{\displaystyle 2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)} \\
t&=\frac{N_3}{N_1} \biggl| \frac{E_\mathrm{t}}{E_\mathrm{i}}\biggr|^2
=\frac{N_3}{N_1} \biggl| \frac{1}{A}\biggr|^2\\
&=\frac{4}{\displaystyle 2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)}
\end{aligned}
\begin{aligned}
&\Biggl|\biggl(1\pm\frac{N_3}{N_1} \biggr) \cos(k_2 d)
-i\biggl(\pm\frac{N_2}{N_1}+\frac{N_3}{N_2} \biggr)\sin(k_2 d) \Biggr|^2 \\
&=\Biggl[1\pm2\frac{N_3}{N_1} + \biggl(\frac{N_3}{N_1}\biggr)^2\Biggr] \cos^2(k_2 d)
+\Biggl[\biggl(\frac{N_2}{N_1}\biggr)^2\pm2\frac{N_3}{N_1}+\biggl(\frac{N_3}{N_2}\biggr)^2\Biggr] \sin^2(k_2 d) \\
&=\frac{N_3}{N_1}
\Biggl[\pm 2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)\Biggr]
\end{aligned}
を用いた．&\Biggl|\biggl(1\pm\frac{N_3}{N_1} \biggr) \cos(k_2 d)
-i\biggl(\pm\frac{N_2}{N_1}+\frac{N_3}{N_2} \biggr)\sin(k_2 d) \Biggr|^2 \\
&=\Biggl[1\pm2\frac{N_3}{N_1} + \biggl(\frac{N_3}{N_1}\biggr)^2\Biggr] \cos^2(k_2 d)
+\Biggl[\biggl(\frac{N_2}{N_1}\biggr)^2\pm2\frac{N_3}{N_1}+\biggl(\frac{N_3}{N_2}\biggr)^2\Biggr] \sin^2(k_2 d) \\
&=\frac{N_3}{N_1}
\Biggl[\pm 2+\biggl(\frac{N_3}{N_1}+\frac{N_1}{N_3}\biggr)\cos^2(k_2 d)
+\biggl(\frac{N_2^2}{N_1 N_3}+\frac{N_1 N_3}{N_2^2}\biggr)\sin^2(k_2 d)\Biggr]
\end{aligned}
音波
関連記事[A]で触れたように，電磁波の計算で$E\to P, N_i\to 1 / Z_i$と置き換えれば良い．波の反射・透過を追跡して合成する方法
関連記事[A]で扱っている2層の反射・透過の問題を繰り返し適用し，全ての波を合成しても上と同じ結果が得られます（定常的な状態は，無限に長い時間が経過した後に達すると考えられるわけですが，無限回の反射も無限に長い時間を考えているわけです）．以下を繰り返すことで計算できる．
- 各境界面を原点$z=0$として2層の問題を適用する（2層・垂直入射の反射と透過（電磁波・音波・量子力学） - Notes_JP）．
- 次の境界まで波が進んだときの$e^{-i\omega t}$の係数を求める（獲得する位相の計算）．これが次の境界面を$z = 0$としたときの$e^{\pm i(kz \mp \omega t)}$の係数となる．
層$i$から層$j$へ平面波が入射するときの「振幅」反射率$R_{ij}$と「振幅」透過率$T_{ij}$は，
\begin{aligned}
R_{ij}
&= \frac{N_{i} - N_{j}}{N_{i} + N_{j}} \\
T_{ij}
&= \frac{2N_{i}}{N_{i} + N_{j}}
\end{aligned}
となる（2層・垂直入射の反射と透過（電磁波・音波・量子力学） - Notes_JP）．R_{ij}
&= \frac{N_{i} - N_{j}}{N_{i} + N_{j}} \\
T_{ij}
&= \frac{2N_{i}}{N_{i} + N_{j}}
\end{aligned}
これより，
\begin{aligned}
\frac{2}{T_{ij}}
&= 1 + \frac{N_{j}}{N_{i}} \\
\frac{2R_{ij}}{T_{ij}}
&= 1 - \frac{N_{j}}{N_{i}}
\end{aligned}
が成り立つ．この関係式は後で使う．\frac{2}{T_{ij}}
&= 1 + \frac{N_{j}}{N_{i}} \\
\frac{2R_{ij}}{T_{ij}}
&= 1 - \frac{N_{j}}{N_{i}}
\end{aligned}
上図より，透過振幅は
\begin{aligned}
E_\mathrm{t}
&=E_\mathrm{i} \Bigl[T_{12} e^{ik_{2}d} T_{23}
+ T_{12} e^{ik_{2}d} \overbrace{ R_{23} e^{-ik_{2}(-d)} R_{21} e^{ik_{2}d}}^{\mathrlap{=\xi=-R_{12}R_{23} e^{i2 k_{2}d}}} T_{23} \\
& \qquad
+ T_{12} e^{ik_{2}d} \xi^2 T_{23}
+T_{12} e^{ik_{2}d} \xi^3 T_{23}
+\cdots \Bigr] \\
&=T_{12} e^{ik_{2}d}
\biggl( \underbrace{\sum_{n=0}^{\infty}\xi^n}_{=\frac{1}{1-\xi}} \biggr)
T_{23} E_\mathrm{i} \\
&=T_{12} e^{ik_{2}d}
\frac{1}{1 +R_{12} R_{23} e^{i2 k_{2}d}}
T_{23} E_\mathrm{i} \\
&=\frac{4}
{\frac{2}{T_{12}} \frac{2}{T_{23}} e^{-ik_2 d}
+\frac{2 R_{12}}{T_{12}} \frac{2R_{23}}{T_{23}} e^{ik_2 d}}
E_\mathrm{i} \\
&=\frac{4}
{\bigl(1 + \frac{N_2}{N_1}\bigr) \bigl(1 + \frac{N_3}{N_2}\bigr) e^{-ik_2 d}
+\bigl(1 - \frac{N_2}{N_1}\bigr) \bigl(1 - \frac{N_3}{N_2}\bigr) e^{ik_2 d} }
E_\mathrm{i}
\end{aligned}
となる（$R_{21}=-R_{12}$）．これは$z = d$を原点としたときの表式だから，「境界条件を課す方法」のように$z = 0$を原点とした表式にするとE_\mathrm{t}
&=E_\mathrm{i} \Bigl[T_{12} e^{ik_{2}d} T_{23}
+ T_{12} e^{ik_{2}d} \overbrace{ R_{23} e^{-ik_{2}(-d)} R_{21} e^{ik_{2}d}}^{\mathrlap{=\xi=-R_{12}R_{23} e^{i2 k_{2}d}}} T_{23} \\
& \qquad
+ T_{12} e^{ik_{2}d} \xi^2 T_{23}
+T_{12} e^{ik_{2}d} \xi^3 T_{23}
+\cdots \Bigr] \\
&=T_{12} e^{ik_{2}d}
\biggl( \underbrace{\sum_{n=0}^{\infty}\xi^n}_{=\frac{1}{1-\xi}} \biggr)
T_{23} E_\mathrm{i} \\
&=T_{12} e^{ik_{2}d}
\frac{1}{1 +R_{12} R_{23} e^{i2 k_{2}d}}
T_{23} E_\mathrm{i} \\
&=\frac{4}
{\frac{2}{T_{12}} \frac{2}{T_{23}} e^{-ik_2 d}
+\frac{2 R_{12}}{T_{12}} \frac{2R_{23}}{T_{23}} e^{ik_2 d}}
E_\mathrm{i} \\
&=\frac{4}
{\bigl(1 + \frac{N_2}{N_1}\bigr) \bigl(1 + \frac{N_3}{N_2}\bigr) e^{-ik_2 d}
+\bigl(1 - \frac{N_2}{N_1}\bigr) \bigl(1 - \frac{N_3}{N_2}\bigr) e^{ik_2 d} }
E_\mathrm{i}
\end{aligned}
\begin{aligned}
E_\mathrm{t} e^{i[k_{3}(z - d) - \omega t]}
&= E_\mathrm{t} e^{-ik_{3}d} \cdot e^{i(k_{3}z - \omega t)}
\end{aligned}
となり，$ e^{-ik_{3}d}$がつく．E_\mathrm{t} e^{i[k_{3}(z - d) - \omega t]}
&= E_\mathrm{t} e^{-ik_{3}d} \cdot e^{i(k_{3}z - \omega t)}
\end{aligned}
また，反射振幅は
\begin{aligned}
E_\mathrm{r}
&=E_\mathrm{i} \Bigl[R_{12}
+ \overbrace{T_{12} e^{ik_{2}d} R_{23} e^{-ik_{2}(-d)}}^{=\eta} T_{21} \\
&\qquad\qquad
+\eta \overbrace{R_{21} e^{ik_{2}d} R_{23} e^{-ik_{2}(-d)}}^{=\xi} T_{21} \\
&\qquad\qquad
+\eta \xi^2 T_{21} + \eta \xi^3 T_{21}
+\cdots \Bigr] \\
&=\biggl[R_{12}
+\eta \biggl( \underbrace{\sum_{n=0}^{\infty}\xi^n}_{=\frac{1}{1-\xi}} \biggr) T_{21} \biggr] E_\mathrm{i} \\
&=\frac{R_{12} (1+R_{12} R_{23} e^{i 2k_{2} d}) + T_{12} R_{23}T_{21} e^{i2k_{2} d}}
{1+R_{12} R_{23} e^{i2k_{2} d}} E_\mathrm{i} \\
&=\frac{R_{12} + (R_{12}^2 + T_{12}T_{21})R_{23} e^{i2k_{2} d}}
{1+R_{12} R_{23} e^{i 2k_{2} d}} E_\mathrm{i} \\
&=\frac{\frac{2 R_{12}}{T_{12}} \frac{2}{T_{23}} e^{-ik_2 d}
+ \Bigl(R_{12}\frac{2 R_{12}}{T_{12}} + 2 T_{21} \Bigr)
\frac{2R_{23}}{T_{23}} e^{ik_2 d}}
{\frac{2}{T_{12}} \frac{2}{T_{23}} e^{-ik_2 d}
+\frac{2 R_{12}}{T_{12}} \frac{2R_{23}}{T_{23}} e^{ik_2 d}}
E_\mathrm{i} \\
&=\frac{\bigl(1 - \frac{N_2}{N_1}\bigr) \bigl(1 + \frac{N_3}{N_2}\bigr) e^{-ik_2 d}
+ \bigl(1+\frac{N_2}{N_1}\bigr) \bigl(1 - \frac{N_3}{N_2}\bigr)e^{ik_2 d}}
{\bigl(1 + \frac{N_2}{N_1}\bigr) \bigl(1 + \frac{N_3}{N_2}\bigr) e^{-ik_2 d}
+\bigl(1 - \frac{N_2}{N_1}\bigr) \bigl(1 - \frac{N_3}{N_2}\bigr) e^{ik_2 d} }
E_\mathrm{i}
\end{aligned}
となる．（ここでE_\mathrm{r}
&=E_\mathrm{i} \Bigl[R_{12}
+ \overbrace{T_{12} e^{ik_{2}d} R_{23} e^{-ik_{2}(-d)}}^{=\eta} T_{21} \\
&\qquad\qquad
+\eta \overbrace{R_{21} e^{ik_{2}d} R_{23} e^{-ik_{2}(-d)}}^{=\xi} T_{21} \\
&\qquad\qquad
+\eta \xi^2 T_{21} + \eta \xi^3 T_{21}
+\cdots \Bigr] \\
&=\biggl[R_{12}
+\eta \biggl( \underbrace{\sum_{n=0}^{\infty}\xi^n}_{=\frac{1}{1-\xi}} \biggr) T_{21} \biggr] E_\mathrm{i} \\
&=\frac{R_{12} (1+R_{12} R_{23} e^{i 2k_{2} d}) + T_{12} R_{23}T_{21} e^{i2k_{2} d}}
{1+R_{12} R_{23} e^{i2k_{2} d}} E_\mathrm{i} \\
&=\frac{R_{12} + (R_{12}^2 + T_{12}T_{21})R_{23} e^{i2k_{2} d}}
{1+R_{12} R_{23} e^{i 2k_{2} d}} E_\mathrm{i} \\
&=\frac{\frac{2 R_{12}}{T_{12}} \frac{2}{T_{23}} e^{-ik_2 d}
+ \Bigl(R_{12}\frac{2 R_{12}}{T_{12}} + 2 T_{21} \Bigr)
\frac{2R_{23}}{T_{23}} e^{ik_2 d}}
{\frac{2}{T_{12}} \frac{2}{T_{23}} e^{-ik_2 d}
+\frac{2 R_{12}}{T_{12}} \frac{2R_{23}}{T_{23}} e^{ik_2 d}}
E_\mathrm{i} \\
&=\frac{\bigl(1 - \frac{N_2}{N_1}\bigr) \bigl(1 + \frac{N_3}{N_2}\bigr) e^{-ik_2 d}
+ \bigl(1+\frac{N_2}{N_1}\bigr) \bigl(1 - \frac{N_3}{N_2}\bigr)e^{ik_2 d}}
{\bigl(1 + \frac{N_2}{N_1}\bigr) \bigl(1 + \frac{N_3}{N_2}\bigr) e^{-ik_2 d}
+\bigl(1 - \frac{N_2}{N_1}\bigr) \bigl(1 - \frac{N_3}{N_2}\bigr) e^{ik_2 d} }
E_\mathrm{i}
\end{aligned}
\begin{aligned}
&R_{12}\frac{2 R_{12}}{T_{12}} + 2 T_{21} \\
&=\frac{N_1 - N_2}{N_1 + N_2} \frac{N_1 - N_2}{N_1}
+\frac{4N_2}{N_1 + N_2} \\
&=\frac{(N_1 + N_2)^2}{N_1(N_1 + N_2)}\\
&=1+\frac{N_2}{N_1}
\end{aligned}
を用いた）．反射波については，「境界条件を課す方法」のように$z = 0$を原点とした表式になっている．&R_{12}\frac{2 R_{12}}{T_{12}} + 2 T_{21} \\
&=\frac{N_1 - N_2}{N_1 + N_2} \frac{N_1 - N_2}{N_1}
+\frac{4N_2}{N_1 + N_2} \\
&=\frac{(N_1 + N_2)^2}{N_1(N_1 + N_2)}\\
&=1+\frac{N_2}{N_1}
\end{aligned}
以上で求めた振幅の表式は，上で「境界条件を課す方法」で導いたものと一致している．
参考文献
[1]詳解電磁気学演習：第10章$\text{\sect} 1.$ 問題{1.3}（3層・垂直入射）．※「理論電磁気学 第8章 問題(9)」と同じもの．[2]理論電磁気学：第8章$\text{\sect} 4$「電磁波の反射と屈折」
