# Wavefront's power losses due to perpendicular collision with dielectric material.

1. Mar 1, 2012

### DemoniWaari

1. The problem statement, all variables and given/known data
How large proportion of wavefront's power penetrates dielectric material's surface in a perpendicular collision from air.

The only parameter that I have is $\varepsilon_r = 16$ where $\varepsilon_r$ is the relative permittivity.

2. Relevant equations
$\varepsilon = \varepsilon_r \varepsilon_0$
$\eta = \sqrt{\frac{\mu}{\varepsilon}}$ Where $\mu = \mu_0$ Because of the dielectric material.
$\vec E(\vec r) = \vec E_0 e^{-j \vec k \cdot \vec r}$
$\vec H(\vec r) = \frac{1}{\eta} \vec E_{0p} e^{-j \vec k \cdot \vec r}$ Where $\vec E_{0p}$ is perpendicular to $\vec E_0$ and has the same magnitude.
$\vec S(\vec r) = \frac{1}{2} \vec E(\vec r) \times \vec H(\vec r)^{*}$ The complex poyinting vector.

3. The attempt at a solution

$\vec E_+(\vec r) = \vec E_0 e^{-j \vec k_1 \cdot \vec r}$ wavefront in the air.
And
$\vec E_-(\vec r) = \vec E_0 e^{-j \vec k_2 \cdot \vec r}$ wavefront in the material
And $\eta_1 = \sqrt{\varepsilon_0 \mu_0}$ is the wave impedance.
Thus we get
$\vec H_+(\vec r) = \frac{1}{\eta_1} \vec E_{0p} e^{-j \vec k_1 \cdot \vec r}$
And
$\vec H_-(\vec r) = \frac{1}{\eta_2} \vec E_{0p} e^{-j \vec k_2 \cdot \vec r}$
Now the complex poyinting vectors are
$\vec S_+(\vec r) = \frac{1}{2} \vec E_+(\vec r) \times \vec H_+(\vec r)^{*}$
We're not interested in the directions so we can just check the magnitudes thus we get.
$\vec S_+(\vec r) = \frac{1}{2\eta_1} |E_0|^2$
And same thing for the other wavefront...
$\vec S_-(\vec r) = \frac{1}{2\eta_2} |E_0|^2$

So the passing fraction is
$\frac{ \vec S_-(\vec r)}{\vec S_+(\vec r)} = \frac{\frac{1}{\eta_2}}{\frac{1}{\eta_1}} = \frac{\eta_1}{\eta_2} = \frac{\sqrt{\frac{\mu_0}{\varepsilon_0}}} {\sqrt{ \frac{\mu_0}{\varepsilon_r \varepsilon_0} }} = \sqrt{\epsilon_r}=4$

So interestingly output is four times higher than input... So I have a problem here =(

Last edited: Mar 1, 2012
2. Mar 1, 2012

### fluidistic

I suggest to use [itex.] rather than [tex.] or the latex will be displayed on a new line for each expression.
When you close the [itex.], use [/itex.] rather than [\itex.]. This is the reason why it doesn't work.
Without the . inside the []'s.

3. Mar 1, 2012

### DemoniWaari

OH YES!
Thank you kind sir.

4. Mar 1, 2012

### rude man

Penetrated power = incident power minus reflected power.