Question 9.36: Light of (angular) frequency ω passes from medium 1, through...

Light of (angular) frequency ω passes from medium 1, through a slab (thickness d) of medium 2, and into medium 3 (for instance, from water through glass into air, as in Fig. 9.27). Show that the transmission coefficient for normal incidence is given by

T^{-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(\frac{n_{2} \omega d}{c}\right)\right].                       (9.199)

[Hint: To the left, there is an incident wave and a reflected wave; to the right, there is a transmitted wave; inside the slab, there is a wave going to the right and a wave going to the left. Express each of these in terms of its complex amplitude, and relate the amplitudes by imposing suitable boundary conditions at the two interfaces. All three media are linear and homogeneous; assume \mu_{1}=\mu_{2}=\mu_{3}=\mu_{0}

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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] .

9.36

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