Help with vector calculus in reflection and transmission of plane wave

[yungman]

This is not a homework, this is concerning reflection and transmission of electromagnetic wave ( plane wave) at a flat planar boundary between two media. But the work in question is pure vector calculus. I ultimately want to proof if $\vec E_I=\hat y E_I$ then $\vec E_R$ and $\vec E_T$ are $\hat y$ direction also. I have a lot of difficulty in this as it is very long. I have the first road block that I need someone to check my work.

As shown in the figure, the $\vec E_I,\;\vec E_R,\;\hbox { and }\;\vec E_T$ are all in xz plane and the boundary is the xy plane at z=0.

At z=0, $\vec E_I|_{z=0}=\vec E_{0I},\;\vec E_R|_{z=0}=\vec E_{0R},\;\hbox { and }\;\vec E_T|_{z=0}=\vec E_{0T}$

[PLAIN]http://i40.tinypic.com/kaj5x.jpg[/PLAIN]

We let:
$\vec E_{0I}=\hat y E_{0I}$.
$\vec E_{0R}=\hat x E_{0R_x}+\hat y E_{0R_y} + \hat z E_{0R_z}$
$\vec E_{0T}=\hat x E_{0T_x}+\hat y E_{0T_y} + \hat z E_{0T_z}$
$\hat k_I=\hat x\sin\theta_I+\hat z \cos\theta_I$, $\hat k_R=\hat x\sin\theta_R-\hat z \cos\theta_R$ and $\hat k_T=\hat x\sin\theta_T+\hat z \cos\theta_T$.

This is the boundary equations of J D Jackson p304 (7.37) that is being used here
$[\epsilon(\vec E_0+\vec E_0'')-\epsilon'\vec E_0']\cdot \hat n=0$ (7.37a) for normal E.
$[\hat k \times \vec E_0+\hat k''\times\vec E_0''-\hat k'\times\vec E_0']\cdot \hat n=0$ (7.37b) for normal B.
$(\vec E_0+\vec E_0''-\vec E_0')\times\hat n=0$ (7.37c) for tangential E.
$\left[\frac 1 {\mu}(\hat k \times \vec E_0+\hat k''\times\vec E_0'')-\frac 1 {\mu'}(\hat k'\times\vec E_0')\right]\times\hat n=0$ (7.37d) for tangential B.
Where $\hat k=\hat k_I,\;\hat k'=\hat k_T,\;\hat k''=\hat k_R$
$\vec E=\vec E_I,\;\vec E'=\vec E_T,\;\vec E''=\vec E_R,\;\hat n=\hat z$


(7.37a)$\Rightarrow\;[\epsilon_1(\vec E_{0I}+\vec E_{0R})-\epsilon_2\vec E_{0T}]\cdot \hat z=0$ (A).
(7.37b)$\Rightarrow\;[\hat k_I \times \vec E_{0I}-\hat k_R \times\vec E_{0R}-\hat k_T \times\vec E_{0T}]\cdot \hat z=0$ (B).
(7.37c)$\Rightarrow\;(\vec E_{0I}+\vec E_{0R}-\vec E_{0T})\times\hat z=0$ (C).
(7.37d)$\Rightarrow\;\left[\frac 1 {\mu_1}(\hat k_I \times \vec E_{0I}-\hat k_R\times\vec E_{0R})-\frac 1 {\mu_2}(\hat k_T\times\vec E_{0T})\right]\times\hat z=0$ (D).

From (C) $[\vec E_{0I}+\vec E_{0R}-\vec E_{0T}]\times\hat z=0\;\Rightarrow\;\hat y E_{0I_y}\times \hat z + (\hat x E_{0R_x}+\hat y E_{0R_y} + \hat z E_{0R_z})\times \hat z + (\hat x E_{0T_x}+\hat y E_{0T_y} + \hat z E_{0T_z})\times \hat z=0$
$\Rightarrow\; \hat x E_{0I_y}-\hat y E_{0R_y}+\hat x E_{0R_y}+\hat y E_{0T_x}-\hat x E_{0T_y}=0$

Therefore $E_{0I_y}+E_{0R_y}-E_{0T_y}=0\;\hbox { and }\;E_{0R_x}=E_{0T_x}$ (E)


From (B) $[\hat k_I \times \vec E_{0I}+\hat k_R \times\vec E_{0R}-\hat k_T \times\vec E_{0T}]\cdot \hat z=0$
Also according to Snell's Law, $\theta_I=\theta_R$. Let $\theta_1=\theta_I=\theta_R$ and $\theta_2=\theta_T$ here.

\begin{align}\Rightarrow\; &[(\hat x \sin\theta_1+\hat z \cos\theta_1)\times \hat y E_{0I_y}+(\hat x \sin\theta_1-\hat z \cos\theta_1)\times(\hat x E_{0R_x}+\hat y E_{0R_y} + \hat z E_{0R_z})\\
&-(\hat x \sin\theta_2+\hat z \cos\theta_2)\times(\hat x E_{0T_x}+\hat y E_{0T_y} + \hat z E_{0T_z})]\cdot \hat z=0
\end{align} Because of the $\cdot \hat z$ at the end, only the $\hat x \times \hat y$ terms in the equation remain:
$\Rightarrow\; E_{0I_y} \sin\theta_1 +E_{0R_y}\sin\theta_1 - E_{0T_y}\sin\theta_2=0$.(F)

If you compare (E) to (F)
It cannot be both true as $\theta_1$ is not equal to $\theta_2$. Can anyone check my work, I have check 3 times already and I cannot see the problem.

Thanks

 

[yungman]

Anyone?

I have another question: The wave function is really defined as:
$$\vec E_R = \vec E_{0R} e^{-jk_R(x\sin\theta_1 - z\cos\theta_1)}=\hat k_R E_{0R} e^{-jk_R(x\sin\theta_1 - z\cos\theta_1)}$$
which the amplitude vary sinusoidally along the path of $\vec k_R$. I should still use $\vec E_R(\vec k_R)=\hat x E_{R_x}+\hat y E_{R_y}+\hat z E_{R_z}$. Is this true?