Question of reflection and transmission of TEM wave in normal incidenc

1. May 23, 2013

yungman

Suppose TEM wave in +z normal to a boundary on xy plane at z=0. We know E & H are tangential to the boundary. Let $\vec E_i=\hat x E$, be the incident wave towards the boundary therefore incident H is $\vec H_i =\hat z\times \hat x \frac {E_i}{η}=\hat y H$. I have a few questions:

1) How do I know why the reflected E is $\hat x E_r$? How do I determine the direction of $\hat E_r=\hat x$? Why it's not $-\hat x$?

2)From boundary condition, tangential E is continuous cross boundary. Is this the reason the transmitted E is $\hat x E_t$ where it follows the direction of $\vec E_i$?

3) Books always use $\vec E_i,\;\vec E_r,\hbox{ and }\;\vec E_t$ to derive H for incident $\vec H_i$, reflected $\vec H_r$, and transmitted $\vec H_t$ respectively. Is that the reason why $\vec H_r=(-\hat z \times \hat x) H_r$? where $-\hat z$ from the direction of the reflected wave(-z) and $\hat x$ is the direction of the $\vec E_r$?

4) The transmitted $\vec H_t=\hat z \times \hat x H_t$ because the transmitted EM wave in medium 2 is in +z direction and the $\vec E_t$ in +x direction?

5) Is this kind of boundary condition holds even in Oblique incident where the tangential $E_r$ always the same direction as tangential $E_i$? And tangential $H_r$ is always opposite direction to tangential $H_i$?

Thanks

Last edited: May 23, 2013
2. May 23, 2013

yungman

Anyone?
I am quite sure 2) to 5) are correct. The main question is

1) How do I know why the reflected E is $\hat x E_r$? How do I determine the direction of $\hat E_r=\hat x$? Why it's not $-\hat x$?

The book just gave this and I want to know why.

Thanks

Last edited: May 23, 2013
3. May 23, 2013

Zag

Hello yungman,

I believe that the answer to your first question will really depend on the boundary conditions of the interface. Apparently you are making calculations for a case in which the reflected wave has the electric field pointing in the same direction as the electric field of the incident wave at $z=0$. That is

$\vec{E}_{incident}(z=0) = E_{i}\hat{x}$

and

$\vec{E}_{reflected}(z=0) = E_{r}\hat{x}$

However, let's analyse the case of the interface between the vacuum and a conductor. We know that the electric field is always perpendicular to a conductor's surface - that is to say that for the present problem, the total field $\vec{E} = \vec{E}_{incident} + \vec{E}_{reflected}$ has to be perpendicular to the a vector normal do the surface at $z=0$. Let $\hat{n}$ be a unitary vector normal to the surface, so that

$\vec{E}\cdot\hat{n} |_{z=0} = 0$

Since the electric field of the incident wave points in the $\hat{x}$ directon, it is tangent to the conductor's surface. Now, from the rules for the addition of vectors, we know that the only way to make the the total field satisfy the condition above is by choosing

$\vec{E}_{reflected} = -\vec{E}_{indicent}$

Which means that now the reflected electric field points in the direction $-\hat{x}$ at $z=0$, and we have a different situation. :)

Edit:

Okay, thinking a little bit more about it, I advise you to read a little bit about the Fresnel Coefficients. For normal incidence, the relationship between the amplitudes of reflected electric field and the incident electric field is given by

$E_{r} = \frac{n_{1} - n_{2}}{n_{1} + n_{2}}E_{i}$

The Fresnel Coefficient being

$r_{12} = \frac{n_{1} - n_{2}}{n_{1} + n_{2}}$

And now you can see very explicitly that the sign of the amplitude of the reflected wave will really depend on the relation between the refractive indices of the media that create the interface. But be careful, this expression I wrote for the Fresnel Coefficients seems carry many other sign conventions associated to the directions chosen for the wave vectors of the incident and reflected wave (at leat for cases in which the incidence is not normal). So I strongly advise you to read the literature and not just rely on the expression I wrote above.

I hope this helps! ;)

Best,
Zag

Last edited: May 23, 2013
4. May 23, 2013

yungman

Thanks for the detail reply. I have this in mind and agree in what you said. I intentionally not bringing this up to let others to comment on this. This is what I read is from Ulaby, I read it carefully and he did not say anything on this. It is shown in the drawing of Ulaby. What I know from tx line theory and read from this that:

$$\Gamma= \frac {η_1-η_2}{η_1+η_2}$$

Which is like in Tx line that $\Gamma= \frac {Z_L-Z_0}{Z_L+Z_0}$ where if $Z_L<Z_0$ the reflection is -ve. So meaning if $η_1>η_2$, the tangential component of the reflected E field is in the same direction. If $η_2>η_1$, the reflect wave has to be -ve direction of the incident tangential direction.

This is really the heart of my question and I want to hear double confirmation on this as this is very important.

Last edited: May 23, 2013
5. May 24, 2013

yungman

I read the book together with Griffiths. I think I found my answer. At the boundary z=0,
$$E_{0r}=\frac {\eta_2-\eta_1}{\eta_2+\eta_1} E_{0i}=\Gamma E_{0i}$$
So $\vec E_{0r}$ can be in -ve x if $\eta_2<\eta_1$

So the direction is embedded in the $E_{0r}$.