Question 8.17: A uniform plane wave, E^i = 10 ay cos(ωt - √3 x -z), propaga......
A uniform plane wave, \mathscr{E}^{i} = 10 \pmb{a}_{y} \cos (\omega t – \sqrt{3} x – z), propagates in free space(z < 0) and impinges on a lossless dielectric(z ≥ 0) of \varepsilon _{r} = 4 . Find
(a) \pmb{k}_{i} and \theta _{i},
(b)\mathscr{E}^{r}, and
(c) total electric field intensity in the region z < 0 .
(a) Phasor form of \mathscr{E}^{i} , is
\pmb{E}^{i} = 10 \pmb{a}_{y} e^{-j(\sqrt{3} x + z )}The wavevector of \pmb{E}^{i} is
\pmb{k}_{i} = \sqrt{3} \pmb{a}_{x} + \pmb{a}_{z}The angle of incidence is the smaller angle between \pmb{k}_{i} and \pmb{a}_{z}
\theta_{i} = \tan ^{-1} \frac{\sqrt{3} }{1} = 60^{\circ }(b) Inserting \theta_{i} = 60^{\circ }, n_{1} = 1, and n_{2}= 2 , into Snell’s law of refraction (n_{1} \sin \theta_{i} = n_{2}\sin \theta _{t} ), we have
\cos \theta_{t} = \sqrt{1 -\left\lgroup \frac{n_{1}}{n_{2}}\sin \theta _{i}\right\rgroup^{2} } = 0.901Rewriting Eq. (8-146a) by use of \eta = \eta _{o}/ n , we have
\boxed{\Gamma_{\bot } = \frac{E^{r}_{o}}{E^{i}_{o}} = \frac{\eta _{2} \cos \theta _{i} – \eta _{1} \cos \theta _{t}}{\eta _{2} \cos \theta _{i} + \eta _{1} \cos \theta _{t}}} (8-146a)
\Gamma_{\bot } = \frac{n_{1} \cos \theta _{i} – n_{2} \cos \theta _{t}}{n_{1} \cos \theta _{i} + n_{2} \cos \theta _{t}} = \frac{\cos 60^{\circ } – 2 \times 0.901}{\cos 60^{\circ } + 2 \times 0.901} = – 0.57Substituting k_{r} = k_{i} = 2 and \theta _{r} = \theta _{i} = 60^{\circ } into Eq. (8-143b), we have
\pmb{E}^{r} = \pmb{a}_{y} E^{r}_{o} e^{-j\pmb{k}_{r} \pmb{\cdot} \pmb{r}} = \pmb{a}_{y} E^{r}_{o} e^{-j [(k_{r} \sin \theta _{r}) x – ( k_{r}\cos \theta _{r} )z]} (8-143b)
\pmb{E}^{r} = \pmb{a}_{y} ( – 0.57 ) 10 e^{-j[(2\sin 60^{\circ }) x – ( 2\cos 60^{\circ })z]} = \pmb{a}_{y}( – 5.7) e^{- j (\sqrt{3} x – z )} \\ \mathscr{E}^{r} = Re [ \pmb{E}^{r} e ^{j\omega t}] = \pmb{a}_{z} ( – 5.7) \cos (\omega t – \sqrt{3} x + z )(c) In the region z < 0 , the total electric field intensity is expressed in phasor form as
\pmb{E} = \pmb{E}^{i} + \pmb{E}^{r} = \pmb{a}_{y} e^{- j \sqrt{3} x} [ 10 e^{- jz} – 5.7 e^{+jz}] \\ \quad \quad \quad \quad \quad \quad = \pmb{a}_{y} e^{- j \sqrt{3} x} [ (15.7 – 5.7) e^{- jz} – 5.7 e^{+jz}] \\ \\ \quad \quad \quad \quad \quad \quad = \pmb{a}_{y} e^{- j \sqrt{3} x} [ 15.7 e^{- jz} – 11.4 \cos z]The total electric field intensity is written in instantaneous form as
\mathscr{E} = Re [ \pmb{E} e ^{\omega t}] \\ \quad = \pmb{a}_{y} \left[15.7 \cos (\omega t – \sqrt{3} x – z ) – 11.4 \cos (z) \cos (\omega t – \sqrt{3} x)\right]