- #1

DemoniWaari

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## Homework Statement

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 [itex] \varepsilon_r = 16 [/itex] where [itex] \varepsilon_r [/itex] is the relative permittivity.

## Homework Equations

[itex] \varepsilon = \varepsilon_r \varepsilon_0 [/itex]

[itex] \eta = \sqrt{\frac{\mu}{\varepsilon}}[/itex] Where [itex]\mu = \mu_0 [/itex] Because of the dielectric material.

[itex] \vec E(\vec r) = \vec E_0 e^{-j \vec k \cdot \vec r}[/itex]

[itex] \vec H(\vec r) = \frac{1}{\eta} \vec E_{0p} e^{-j \vec k \cdot \vec r}[/itex] Where [itex] \vec E_{0p} [/itex] is perpendicular to [itex] \vec E_0 [/itex] and has the same magnitude.

[itex] \vec S(\vec r) = \frac{1}{2} \vec E(\vec r) \times \vec H(\vec r)^{*}[/itex] The complex poyinting vector.

## The Attempt at a Solution

[itex] \vec E_+(\vec r) = \vec E_0 e^{-j \vec k_1 \cdot \vec r}[/itex] wavefront in the air.

And

[itex] \vec E_-(\vec r) = \vec E_0 e^{-j \vec k_2 \cdot \vec r}[/itex] wavefront in the material

And [itex] \eta_1 = \sqrt{\varepsilon_0 \mu_0}[/itex] is the wave impedance.

Thus we get

[itex] \vec H_+(\vec r) = \frac{1}{\eta_1} \vec E_{0p} e^{-j \vec k_1 \cdot \vec r}[/itex]

And

[itex] \vec H_-(\vec r) = \frac{1}{\eta_2} \vec E_{0p} e^{-j \vec k_2 \cdot \vec r}[/itex]

Now the complex poyinting vectors are

[itex] \vec S_+(\vec r) = \frac{1}{2} \vec E_+(\vec r) \times \vec H_+(\vec r)^{*}[/itex]

We're not interested in the directions so we can just check the magnitudes thus we get.

[itex] \vec S_+(\vec r) = \frac{1}{2\eta_1} |E_0|^2[/itex]

And same thing for the other wavefront...

[itex] \vec S_-(\vec r) = \frac{1}{2\eta_2} |E_0|^2[/itex]

So the passing fraction is

[itex]\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[/itex]

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

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