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yungman

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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

##[\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

##(\vec E_0+\vec E_0''-\vec E_0')\times\hat n=0## (7.37c) for tangential

##\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

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

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

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