z<0: \quad\left\{\begin{array}{l}\tilde{ E }_{I}(z, t)=\tilde{E}_{I} e^{i\left(k_{1} z-\omega t\right)} \hat{ x }, \quad \tilde{ B }_{I}(z, t)=\frac{1}{v_{1}} \tilde{E}_{I} e^{i\left(k_{1} z-\omega t\right)} \hat{ y } \\\tilde{ E }_{R}(z, t)=\tilde{E}_{R} e^{i\left(-k_{1} z-\omega t\right)} \hat{ x }, \tilde{ B }_{R}(z, t)=-\frac{1}{v_{1}} \tilde{E}_{R}e^{i\left(-k_{1} z-\omega t\right)} \hat{ y }\end{array}\right.
0<z<d: \begin{cases}\tilde{ E }_{r}(z, t)=\tilde{E}_{r} e^{i\left(k_{2} z-\omega t\right)} \hat{ x }, & \tilde{ B }_{r}(z, t)=\frac{1}{v_{2}} \tilde{E}_{r} e^{i\left(k_{2} z-\omega t\right)} \hat{ y } \\ \tilde{ E }_{l}(z, t)=\tilde{E}_{l} e^{i\left(-k_{2} z-\omega t\right)} \hat{ x }, & \tilde{ B }_{l}(z, t)=-\frac{1}{v_{2}} \tilde{E}_{l} e^{i\left(-k_{2} z-\omega t\right)} \hat{ y }\end{cases}
z>d: \quad\left\{\tilde{ E }_{T}(z, t)=\tilde{E}_{T} e^{i\left(k_{3} z-\omega t\right)} \hat{ x }, \quad \tilde{ B }_{T}(z, t)=\frac{1}{v_{3}} \tilde{E}_{T} e^{i\left(k_{3} z-\omega t\right)} \hat{ y }\right. .
Boundary conditions: E _{1}^{\|}= E _{2}^{\|}, B _{1}^{\|}= B _{2}^{\|} , at each boundary (assuming \mu_{1}=\mu_{2}=\mu_{3}=\mu_{0} ):
z=0:\left\{\begin{array}{l}\tilde{E}_{I}+\tilde{E}_{R}=\tilde{E}_{r}+\tilde{E}_{l} \\\frac{1}{v_{1}} \tilde{E}_{I}-\frac{1}{v_{1}} \tilde{E}_{R}=\frac{1}{v_{2}} \tilde{E}_{r}-\frac{1}{v_{2}} \tilde{E}_{l} \Rightarrow \tilde{E}_{I}-\tilde{E}_{R}=\beta\left(\tilde{E}_{r}-\tilde{E}_{l}\right), \text { where } \beta \equiv v_{1} / v_{2}\end{array}\right.
z=d:\left\{\begin{array}{l}\tilde{E}_{r} e^{i k_{2} d}+\tilde{E}_{l} e^{-i k_{2} d}=\tilde{E}_{T} e^{i k_{3} d} \\\frac{1}{v_{2}} \tilde{E}_{r} e^{i k_{2} d}-\frac{1}{v_{2}} \tilde{E}_{l} e^{-i k_{2} d}=\frac{1}{v_{3}} \tilde{E}_{T} e^{i k_{3} d} \Rightarrow\tilde{E}_{r} e^{i k_{2} d}-\tilde{E}_{l} e^{-i k_{2} d}=\alpha\tilde{E}_{T} e^{i k_{3} d}, \text { where } \alpha \equiv v_{2} / v_{3}\end{array}\right.
We have here four equations; the problem is to eliminate \tilde{E}_{R}, \tilde{E}_{r}, \text { and } \tilde{E}_{l} , to obtain a single equation for \tilde{E}_{T} \text { in terms of } \tilde{E}_{I}
Add the first two to eliminate \tilde{E}_{R}: \quad 2 \tilde{E}_{I}=(1+\beta) \tilde{E}_{r}+(1-\beta) \tilde{E}_{l} ;
Add the last two to eliminate \tilde{E}_{l}: \quad 2 \tilde{E}_{r} e^{i k_{2} d}=(1+\alpha) \tilde{E}_{T} e^{i k_{3} d} ;
Subtract the last two to eliminate \tilde{E}_{r}: 2 \tilde{E}_{l} e^{-i k_{2} d}=(1-\alpha) \tilde{E}^{T} e^{i k_{3} d}.
Plug the last two of these into the first:
2 \tilde{E}_{I}=(1+\beta) \frac{1}{2} e^{-i k_{2} d}(1+\alpha) \tilde{E}_{T} e^{i k_{3} d}+(1-\beta) \frac{1}{2} e^{i k_{2} d}(1-\alpha) \tilde{E}_{T} e^{i k_{3} d}
4 \tilde{E}_{I}=\left[(1+\alpha)(1+\beta) e^{-i k_{2} d}+(1-\alpha)(1-\beta) e^{i k_{2} d}\right] \tilde{E}_{T} e^{i k_{3} d}
=\left[(1+\alpha \beta)\left(e^{-i k_{2} d}+e^{i k_{2} d}\right)+(\alpha+\beta)\left(e^{-i k_{2} d}-e^{i k_{2} d}\right)\right] \tilde{E}_{T} e^{i k_{3} d}
=2\left[(1+\alpha \beta) \cos \left(k_{2} d\right)-i(\alpha+\beta) \sin \left(k_{2} d\right)\right] \tilde{E}_{T} e^{i k_{3} d}
Now the transmission coeffcient is T=\frac{v_{3} \epsilon_{3} E_{T_{0}}^{2}}{v_{1} \epsilon_{1} E_{I_{0}}^{2}}=\frac{v_{3}}{v_{1}}\left(\frac{\mu_{0} \epsilon_{3}}{\mu_{0} \epsilon_{1}}\right) \frac{\left|\tilde{E}_{T}\right|^{2}}{\left|\tilde{E}_{I}\right|^{2}}=\frac{v_{1}}{v_{3}} \frac{\left|\tilde{E}_{T}\right|^{2}}{\left|\tilde{E}_{I}\right|^{2}}=\alpha \beta \frac{\left|\tilde{E}_{T}\right|^{2}}{\left|\tilde{E}_{I}\right|^{2}} , so
T^{-1}=\frac{1}{\alpha \beta} \frac{\left|\tilde{E}_{I}\right|^{2}}{\left|\tilde{E}_{T}\right|^{2}}=\frac{1}{\alpha \beta}\left|\frac{1}{2}\left[(1+\alpha \beta) \cos \left(k_{2} d\right)-i(\alpha+\beta) \sin \left(k_{2} d\right)\right] e^{i k_{3} d}\right|^{2}
=\frac{1}{4 \alpha \beta}\left[(1+\alpha \beta)^{2} \cos ^{2}\left(k_{2} d\right)+(\alpha+\beta)^{2} \sin ^{2}\left(k_{2} d\right)\right] . \quad \text { But } \cos ^{2}\left(k_{2} d\right)=1-\sin ^{2}\left(k_{2} d\right).
=\frac{1}{4 \alpha \beta}\left[(1+\alpha \beta)^{2}+\left(\alpha^{2}+2 \alpha \beta+\beta^{2}-1-2 \alpha \beta-\alpha^{2} \beta^{2}\right) \sin ^{2}\left(k_{2} d\right)\right]
=\frac{1}{4 \alpha \beta}\left[(1+\alpha \beta)^{2}-\left(1-\alpha^{2}\right)\left(1-\beta^{2}\right) \sin ^{2}\left(k_{2} d\right)\right].
\text { But } n_{1}=\frac{c}{v_{1}}, n_{2}=\frac{c}{v_{2}}, n_{3}=\frac{c}{v_{3}}, \text { so } \alpha=\frac{n_{3}}{n_{2}}, \beta=\frac{n_{2}}{n_{1}} .
=\frac{1}{4 n_{1} n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}} \sin ^{2}\left(k_{2} d\right)\right] .
