Question 8.18: The uniform plane wave with parallel polarization, E^i = 4 a......
The uniform plane wave with parallel polarization, \pmb{E}^{i} = 4 \pmb{a}_{E} e ^{- j (2x + 3z)}, propagates in free space( z < 0 ), and impinges obliquely on a lossless dielectric of \varepsilon _{r} = 2.25 occupying the region z ≥ 0 . Find
(a) expression for \pmb{a}_{E} in Cartesian coordinates,
(b) \pmb{E}^{r} and
(c) \pmb{E}^{t} .
(a) From the wavevector \pmb{k}_{i} = 2 \pmb{a}_{x} + 3 \pmb{a}_{z}, the angle of incidence is
\theta _{i} = \tan ^{-1} \frac{2}{3} \cong 33.7^{\circ }The direction of the parallel polarization is determined from the two conditions; that is, \pmb{a}_{E} \pmb{\cdot } \pmb{k}_{i} = 0 and (\pmb{a}_{E})_{x} \parallel (\pmb{k}_{i})_x.
From the first condition : \pmb{a}_{E} \parallel \pm (3 \pmb{a}_{x} – 2 \pmb{a}_{z} )
From the second condition : \pmb{a}_{E} \parallel (3 \pmb{a}_{x} – 2 \pmb{a}_{z} )
Thus,
\pmb{a}_{E} = \frac{1}{\sqrt{13} } (3 \pmb{a}_{x} – 2 \pmb{a}_{z} )The electric field phasor of the incident wave is
\pmb{E}^{i} = \frac{4}{\sqrt{13} } (3 \pmb{a}_{x} – 2 \pmb{a}_{z} ) e^{-j (2x + 3z)} (8-153a)
(b) Substituting n_{1} = 1 and n_{2} = \sqrt{2.25} = 1.5 into Snell’s law of refraction, n_{1} \sin \theta _{i} = n_{2} \sin \theta _{t}, we obtain
\theta _{t} = \sin^{-1} \left\lgroup \frac{1}{1.5} \sin 33.7^{\circ }\right\rgroup \cong 21.7^{\circ }From Eq. (8-151a), by use of \eta = \eta_{o}/n, we obtain
\boxed{\Gamma_{\parallel} = \frac{E^{r}_{o}}{E^{i}_{o}} = \frac{\eta_{2} \cos \theta_{t} – \eta _{1} \cos \theta _{i}}{ \eta _{2} \cos \theta_{t} + \eta _{1} \cos \theta _{i}}} (8-151a)
\Gamma_{\parallel} = \frac{n_{1} \cos \theta_{t} – n_{2} \cos \theta _{i}}{n_{1} \cos \theta_{t} + n_{2} \cos \theta _{i}} = \frac{\cos 21.7^{\circ } – 1.5 \times \cos 33.7^{\circ }}{\cos 21.7^{\circ } + 1.5 \times \cos 33.7^{\circ }} = -0.15Substituting E^{i}_{o} = 4, \Gamma _{\parallel } = – 0.15, k_{r} = k_{i} = \sqrt{13} and \theta_{r} =\theta _{i} = 33.7^{\circ } into Eq. (8-148b), we obtain
\pmb{E}^{r} = (\pmb{a}_{x} \cos \theta _{r} + \pmb{a}_{z} \sin \theta _{r}) E^{r}_{o} e^{ – j[(k_{r} \sin \theta _{r})x -(k_{r} \cos \theta _{r})z]} (8-148b)
\pmb{E}^{r} = (\pmb{a}_{x} \cos \theta _{r} + \pmb{a}_{z} \sin \theta _{r}) \Gamma _{\parallel } E^{i}_{o} e^{ – j[(k_{r} \sin \theta _{r})x -(k_{r} \cos \theta _{r})z]} \\ \quad = (0.83 \pmb{a}_{x} + 0.55 \pmb{a}_{z})( – 0.15 \times 4) e^{- j [2x – 3z]} \\ \quad = – (0.50 \pmb{a}_{x} + 0.33 \ \pmb{a}_{z}) e^{-j [2x – 3z]} (8-153b)
(c) From Eq. (8-151b) we obtain
\boxed{\tau_{\parallel} = \frac{E^{t}_{o}}{E^{i}_{o}} = \frac{2 \eta_{2} \cos \theta_{i}}{ \eta _{2} \cos \theta_{t} + \eta _{1} \cos \theta _{i}}} (8-151b)
\tau _{\parallel} = \frac{2 n_{1} \cos \theta_{i} }{n_{1} \cos \theta_{t} + n_{2} \cos \theta _{i}} = \frac{2 \cos 33.7^{\circ } }{0.929 + 1.5 \times \cos 33.7^{\circ }} = 0.76Substituting k_{t} = k_{i}\sqrt{\varepsilon _{r}} = 1.5 \sqrt{13} and \theta_{t} =21.7^{\circ } into Eq. (8-148c), we obtain
\pmb{E}^{t} = (\pmb{a}_{x} \cos \theta _{t} – \pmb{a}_{z} \sin \theta _{t}) E^{t}_{o} e^{ – j[(k_{t} \sin \theta _{t})x + (k_{t} \cos \theta _{t})z]} (8-148c)
\pmb{E}^{t} = (\pmb{a}_{x} \cos \theta _{t} – \pmb{a}_{z} \sin \theta _{t}) \tau _{\parallel } E^{i}_{o} e^{ – j[(k_{t} \sin \theta _{t})x +(k_{t} \cos \theta _{t})z]} \\ \quad = (0.93 \pmb{a}_{x} – 0.37 \pmb{a}_{z}) 0.76 \times 4 e^{- j [2.00x + 5.03z]} \\ \quad = (2.84 \pmb{a}_{x} – 1.13 \pmb{a}_{z}) e^{-j [2.00x + 5.03z]} (8-153c)