nopacman said:
I'm quite sure it is. In any case, would you be so kind as to present any evidence to the contrary?
Please keep in mind I'm not talking about the ratio of intensities, but the electric field ratio.
The transversal components of the electric and magnetic fields have to be continuous. Assuming normal incidence to the x = 0 plane, and that the electric field vector is polarized along y-direction, then the magnetic field-vector is polarized along the z-direction, we have:
Medium 1:
<br />
E_1 = E_0 e^{i k_1 x} + E_r e^{-i k_1 x}<br />
<br />
H_1 = \frac{n_1}{Z_0} \left( E_0 e^{i k_1 x} - E_r e^{-i k_1 x} \right), \ Z_0 \equiv \sqrt{\frac{\mu_0}{\varepsilon_0}}<br />
Medium 2:
<br />
E_2 = E_t e^{i k_2 x}<br />
<br />
H_2 = \frac{n_2}{Z_0} E_t e^{i k_2 x}<br />
where:
<br />
k_i = \frac{n_i \, \omega}{c}, \; n_i = \frac{1}{\sqrt{\varepsilon_i \, \mu_i}}, \ (i = 1, 2)<br />
The continuity conditions give:
<br />
\begin{array}{l}<br />
E_0 + E_r = E_t \\<br />
<br />
\frac{n_1}{Z_0} \left(E_0 - E_r \right) = \frac{n_2}{Z_0} E_t<br />
\end{array}<br />
and, solving for E_r/E_0 and E_t/E_0, gives:
<br />
\begin{array}{l}<br />
\frac{E_t}{E_0} - \frac{E_r}{E_0} = 1 \\<br />
\frac{n_2}{n_1} \frac{E_t}{E_0} + \frac{E_r}{E_0} = 1<br />
\end{array}<br />
<br />
\begin{array}{l}<br />
\left(1 + \frac{n_2}{n_1} \right) \frac{E_t}{E_0} = 2 \\<br />
<br />
\left(1 + \frac{n_2}{n_1} \right) \frac{E_r}{E_0} = 1 - \frac{n_2}{n_1}<br />
\end{array}<br />
But, the power flux is given by the Poynting vector, which, in harmonic time dependence is given by:
<br />
\mathbf{S} = \frac{1}{2} \mathrm{Re}\left(\mathbf{E} \times \mathbf{H}^{\ast} \right)<br />
In our case, the cross product \hat{\mathbf{y}} \times \hat{\mathbf{z}} = \hat{\mathbf{x}} is directed along the x-direction and the projections are:
Medium 1:
<br />
S_1 = \frac{n_1}{2 \, Z_0} \mathrm{Re} \left[ \left(E_0 e^{i k_1 x} + E_r e^{-i k_1 x} \right) \left(E^{\ast}_0 e^{-i k_1 x} - E^{\ast}_r e^{i k_1 x} \right) \right]_{x = 0} = \frac{n_1}{2 Z_0} \left( |E_0|^2 - |E_r|^2 \right)<br />
Medium 2:
<br />
S_2 = \frac{n_2}{2 \, Z_0} \mathrm{Re} \left[ E_t e^{i k_2 x} E^{\ast}_t e^{-i k_2 x} \right]_{x = 0} = \frac{n_2}{2 Z_0} |E_t|^2<br />
Keeping in mind that these amplitudes are all real, and the above continuity equations, it is easy to see that:
<br />
S_1 = S_2<br />
which means that the Poynting vector is continuous on the boundary. This is an expression of the Law of Conservation of energy. The incident power density is:
<br />
S_{inc} = \frac{n_1}{2 Z_0} E^2_0<br />
The reflected power density is (notice the flux is in the negative x-direction):
<br />
S_r = \frac{n_1}{2 Z_0} E^2_r<br />
and the transmitted power density is:
<br />
S_t = \frac{n_2}{2 Z_0} E^2_t<br />
The ratios:
<br />
T \equiv \frac{S_t}{S_{inc}} = \frac{n_2}{n_1} \left(\frac{E_t}{E_0}\right)^2 = \frac{2 \frac{n_2}{n_1}}{\left( 1 + \frac{n_2}{n_1} \right)^2}<br />
and
<br />
R \equiv \frac{S_r}{S_{inc}} = \frac{E^2_r}{E^2_0} = \frac{\left(1 - \frac{n_2}{n_1} \right)^2}{\left(1 + \frac{n_2}{n_1} \right)^2}<br />
are
defined as the transmission and reflection coefficients, respectively. Notice that each is less than 1, and that:
<br />
T + R = 1<br />
