Question 9.39: According to Snell’s law, when light passes from an opticall...

$k_{T}\left(x \sin \theta_{T}+z \cos \theta_{T}\right)=x k_{T} \sin \theta_{T}+i z k_{T} \sqrt{\sin ^{2} \theta_{T}-1}=k x+i \kappa z$ , where

$k \equiv k_{T} \sin \theta_{T}=\left(\frac{\omega n_{2}}{c}\right) \frac{n_{1}}{n_{2}} \sin \theta_{I}=\frac{\omega n_{1}}{c} \sin \theta_{I}$,

$\kappa \equiv k_{T} \sqrt{\sin ^{2} \theta_{T}-1}=\frac{\omega n_{2}}{c} \sqrt{\left(n_{1} / n_{2}\right)^{2} \sin ^{2} \theta_{I}-1}=\frac{\omega}{c} \sqrt{n_{1}^{2} \sin ^{2} \theta_{I}-n_{2}^{2}}$ . So

$\tilde{ E }_{T}( r , t)=\tilde{ E }_{0_{T}} e^{-\kappa z} e^{i(k x-\omega t)}$ . qed

$\tilde{ E }_{T}( r , t)=\tilde{ E }_{0_{T}} e^{i\left( k _{T} \cdot r -\omega t\right)}, \quad \tilde{ B }_{T}( r , t)=\frac{1}{v_{2}}\left(\hat{ k }_{T} \times \tilde{ E }_{T}\right)$                         (9.91)

$\text { (b) } R=\left|\frac{\tilde{E}_{0_{R}}}{\tilde{E}_{0_{I}}}\right|^{2}=\left|\frac{\alpha-\beta}{\alpha+\beta}\right|$ . Here $\beta$ is real (Eq. 9.106) and $\alpha$ is purely imaginary (Eq. 9.108); write $\alpha=i a$ with a real: $R=\left(\frac{i a-\beta}{i a+\beta}\right)\left(\frac{-i a-\beta}{-i a+\beta}\right)=\frac{a^{2}+\beta^{2}}{a^{2}+\beta^{2}}=1$.

$\beta \equiv \frac{\mu_{1} v_{1}}{\mu_{2} v_{2}}=\frac{\mu_{1} n_{2}}{\mu_{2} n_{1}}$                        (9.106)

(c) From Prob. 9.17, $E_{0_{R}}=\left|\frac{1-\alpha \beta}{1+\alpha \beta}\right| E_{0_{I}}, \text { so } R=\left|\frac{1-\alpha \beta}{1+\alpha \beta}\right|^{2}=\left|\frac{1-i a \beta}{1+i a \beta}\right|^{2}=\frac{(1-i a \beta)(1+i a \beta)}{(1+i a \beta)(1-i a \beta)}=1$.

(d) From the solution to Prob. 9.17, the transmitted wave is

$\tilde{ E }( r , t)=\tilde{E}_{0_{T}} e^{i\left( k _{T} \cdot r -\omega t\right)} \hat{ y }, \quad \tilde{ B }( r , t)=\frac{1}{v_{2}} \tilde{E}_{0_{T}} e^{i\left( k _{T} \cdot r -\omega t\right)}\left(-\cos \theta_{T} \hat{ x }+\sin \theta_{T} \hat{ z }\right)$.

Using the results in (a): $k _{T} \cdot r =k x+i \kappa z, \sin \theta_{T}=\frac{c k}{\omega n_{2}}, \cos \theta_{T}=i \frac{c \kappa}{\omega n_{2}}$.

$\tilde{ E }( r , t)=\tilde{E}_{0_{T}} e^{-\kappa z} e^{i(k x-\omega t)} \hat{ y }, \quad \tilde{ B }( r , t)=\frac{1}{v_{2}} \tilde{E}_{0_{T}} e^{-\kappa z} e^{i(k x-\omega t)}\left(-i \frac{c \kappa}{\omega n_{2}} \hat{ x }+\frac{c k}{\omega n_{2}} \hat{ z }\right)$.

We may as well choose the phase constant so that $\tilde{E}_{0_{T}}$ is real. Then

$E ( r , t)=E_{0} e^{-\kappa z} \cos (k x-\omega t) \hat{ y }$;

$=\frac{1}{\omega} E_{0} e^{-\kappa z}[\kappa \sin (k x-\omega t) \hat{ x }+k \cos (k x-\omega t) \hat{ z }]$ . qed

$\text { (e) } \text { (i) } \nabla \cdot E =\frac{\partial}{\partial y}\left[E_{0} e^{-\kappa z} \cos (k x-\omega t)\right]=0$.

$=\frac{E_{0}}{\omega}\left[e^{-\kappa z} \kappa k \cos (k x-\omega t)-\kappa e^{-\kappa z} k \cos (k x-\omega t)\right]=0$.

$=\kappa E_{0} e^{-\kappa z} \cos (k x-\omega t) \hat{ x }-E_{0} e^{-\kappa z} k \sin (k x-\omega t) \hat{ z }$.

$=\kappa E_{0} e^{-\kappa z} \cos (k x-\omega t) \hat{ x }-k E_{0} e^{-\kappa z} \sin (k x-\omega t) \hat{ z }= \nabla \times E$.

$=\left[-\frac{E_{0}}{\omega} \kappa^{2} e^{-\kappa z} \sin (k x-\omega t)+\frac{E_{0}}{\omega} e^{-\kappa z} k^{2} \sin (k x-\omega t)\right] \hat{ y }=\left(k^{2}-\kappa^{2}\right) \frac{E_{0}}{\omega} e^{-\kappa z} \sin (k x-\omega t) \hat{ y }$.

$\text { Eq. } 9.202 \Rightarrow k^{2}-\kappa^{2}=\left(\frac{\omega}{c}\right)^{2}\left[n_{1}^{2} \sin ^{2} \theta_{I}-\left(n_{1} \sin \theta_{I}\right)^{2}+\left(n_{2}\right)^{2}\right]=\left(\frac{n_{2} \omega}{c}\right)^{2}=\omega^{2} \epsilon_{2} \mu_{2}$.

$=\epsilon_{2} \mu_{2} \omega E_{0} e^{-\kappa z} \sin (k x-\omega t) \hat{ y }$.

$\mu_{2} \epsilon_{2} \frac{\partial E }{\partial t}=\mu_{2} \epsilon_{2} E_{0} e^{-\kappa z} \omega \sin (k x-\omega t) \hat{ y }= \nabla \times B$.

(f)

$=\frac{E_{0}^{2}}{\mu_{2} \omega} e^{-2 \kappa z}\left[k \cos ^{2}(k x-\omega t) \hat{ x }-\kappa \sin (k x-\omega t) \cos (k x-\omega t) \hat{ z }\right]$.

Averaging over a complete cycle, using $\left\langle\cos ^{2}\right\rangle=1 / 2 \text { and }\langle\sin \cos \rangle=0,\langle S \rangle=\frac{E_{0}^{2} k}{2 \mu_{2} \omega} e^{-2 \kappa z} \hat{ x }$ . On average, then, no energy is transmitted in the z direction, only in the x direction (parallel to the interface). qed