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Given is a source $S$ which produces an electromagnetic wave $E(x,y,z)$. The source is in vacuum. At z=0 there is an interface between vacuum and a perfect dielectric with $\epsilon$.

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The electric field can be calculated in the space where we have vacuum, by numerically evaluating an Integral (If you think it is necessary I can add the formula here, but it is a 2D integral over the free space Green's function of the wave equation of electromagnetism)

The question is now what happens at the interface? I know that a part of the wave will be reflected and a part will be transmitted. But how do I get this information in my case (I have no plane wave coming in!).

For the transmitted field my idea would be as follows: I know $E(x,y,0+)$. With the interface condition $E(z=0+)_{||}=E(z=0-)_{||}$ it follows directly $E_{x,y}(x,y,0-)=E_{x,y}(x,y,0+)$. And from the interface condition $D_z(z=0+)=D_z(z=0-)$ follows $E(x,y,0-)=\frac{1}{\epsilon}E(x,y,0-)$. But how do I get the reflected field?

A few more technical information: I have a formula to propagate the field from the source S to the interface. I evaluate the formula at a discrete set of points. For each point I need the information which amount is transmitted and which is reflected. The reflected part is propagated again into the vacuum space using the propagation formula and the transmitted part is propagated into the dielectric space using the propagation formula. If I do this I have constructed the whole solution of my problem.

Has anyone an idea how to get the reflected part? Also completely new ideas to solve the problem are welcome.

EDIT:

A possible Ansatz could be to assume that locally at a point $x_0$ at the surface we have

$\begin{pmatrix}E_x\\E_y\\E_z\end{pmatrix}=\begin{pmatrix}\tilde{E}_x\\\tilde{E}_y\\\tilde{E}_z\end{pmatrix}e^{ik\cdot x_0}=\begin{pmatrix}\tilde{E}_x\\\tilde{E}_y\\\tilde{E}_z\end{pmatrix}e^{i n\omega |x_0| \cos{\alpha}}$

but in this parameterization I have now 4 parameters ($\tilde{E}_x,\tilde{E}_y,\tilde{E}_z,\alpha$ ???

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  • $\begingroup$ Is it possible to use Image Method to construct the solution? But I don't know how to set it right to match the boundary conditions... $\endgroup$ – K_inverse Sep 20 '18 at 10:37
  • $\begingroup$ @K_inverse I don't know. Maybe one also needs the interface conditions for the B-field? And then one can see which reflected field one has to add to fulfill the interface conditions also for the B field? $\endgroup$ – Jan SE Sep 20 '18 at 11:23
  • $\begingroup$ This probably isn't the answer you're looking for, but the "fully Kosher" approach to this problem is not to use the free-space Green's function, but rather to solve for the transmission/reflection problem at the level of a point source, yielding the Green's function for your actual configuration, and then integrating over your source. $\endgroup$ – Emilio Pisanty Sep 20 '18 at 22:19
  • $\begingroup$ @EmilioPisanty I couldn't find something about the Fully Kosher approach. Do you maybe have a reference? $\endgroup$ – Jan SE Sep 21 '18 at 8:42
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The electromagnetic waves also have a magnetic field component that must also satisfy boundary conditions. The electric and magnetic field components are at right angles to each other and the wave-vector$^{*}$. The electric and magnetic field component magnitudes are different by a constant that depends on the speed of light in the medium they are propagating.

These additional pieces of information allow you to set up another set of equations that can be solved along with what you already write down in your question.

$^{*}$ Assuming that you are dealing with a transverse radiation field. If you cannot assume this then the problem is more difficult, but in principle the same.

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  • $\begingroup$ Thanks for the answer. I don't think that E and B field are at right angles to each other in this case. This statement holds only for plane waves, but in this case we do not have a plane wave. Also I am wondering how you want to extract out of these informations the reflected part? Furthermore note, that in this case the k-vector is only locally defined and not globally like in the case of a plane wave. $E(x,y,z)e^{ik(x,y,z)\cdot x}$ $\endgroup$ – Jan SE Sep 20 '18 at 13:08
  • $\begingroup$ @JanSE E and B are perpendicular for transverse electromagnetic waves, of which plane waves (or the superposition of plane waves) are a subset. Are you considering the "near-field" of a transmitter or something like that? Of course $\vec{k}$ is a function of position, there will be no "general solution" for the reflected wave, it will depend on position along the interface. $\endgroup$ – Rob Jeffries Sep 20 '18 at 16:11
  • $\begingroup$ yes the near field is included and it can play a role in what I am doing. When I take the Maxwell eq. and fourier transform them the fourier transformed field stand perpendicular on each other, but not the fields $E(x,y,z)$ and $B(x,y,z)$. Do you agree? $\endgroup$ – Jan SE Sep 20 '18 at 18:15
  • $\begingroup$ I agree with you that the reflected wave will depend on the position along the interface. Right now I am computing with my code the field from S which hits the interface, e.g. I have at each interface point 3 complex values for the E field. I now need to know what amount of this field gets reflected? At the moment I don't hav any idea how to do it... P.S. with a little bit more effort I could also numerically calculate the B-field at each point of the interface, but I don't know how this should help me... $\endgroup$ – Jan SE Sep 20 '18 at 18:18
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You can expand your wave in plane waves. Then you find the transmitted wave for each of these plane waves. Sum these up and you are done.

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  • $\begingroup$ Can you please give a bit more details? I just have the numerical values at the interface at a discrete set of points. I don't see how I should do the expansion. $\endgroup$ – Jan SE Sep 21 '18 at 9:02
  • $\begingroup$ @Jan SE you may want be more explicit. Do you know where the source is located? $\endgroup$ – my2cts Sep 21 '18 at 9:08
  • $\begingroup$ You can assume that the source is a circular aperture. The field which hits the surface is the diffracted field, including near fields,e.g. the full solution. The aperture is very close to the interface such that near field effect will play a role. $\endgroup$ – Jan SE Sep 21 '18 at 9:13

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