Question about the FFP - Spatial filtering example

I am learning the example in the https://kb.lumerical.com/solvers_far_field_projections_far_field_filtering.html
In the script, solver_far_field_filter.lsf, I have some questions.

  1. In line 25&32:

set(“angle”,0);
set(“angle”,90);

Why these two dipole phi angle can represent the p and s polarization? When setting the angle to 90, the main electric field is still not Ez. We know that in this situation, the Ez can be regarded as s polarization.

R_s = ( (n1cos(theta1) - n2cos(theta2))/(n1cos(theta1) - n2cos(theta2)) )^2;
R_p = ( (n1^2n2cos(theta2) - n2^2n1cos(theta1))/(n1^2n2cos(theta2) + n2^2n1cos(theta1)) )^2;

I checked some theory books, the R_s should be

R_s=( (n1cos(theta1) - n2cos(theta2))/(n1cos(theta1) + n2cos(theta2)) )^2;
and R_p should be
R_p=( (n2cos(theta1) - n1cos(theta2))/(n2cos(theta1) + n1cos(theta2)) )^2;

So why it is not consistent with the theory book?
Thanks if anyone could help me.

  1. In line 61

theta1 = n2*sin(theta2)/n1;

I think theta1 should be sin(theta1).

Hello @makeqi,

  1. These angles do not represent S, P polarization they represent the orientation of two dipoles. Dipoles will radiate at many polarization’s simultaneously, and so can be to efficiently find the T and R for a complete set of incidence angles. They have very little directivity; however, which is why the emission pattern is run through a spatial filter in this example.

  2. R_s is incorrect, and you have given the appropriate correction. The R_p values which you quote are equivalent since you can cancel a factor of n1*n2 in all terms.

  3. Again you are correct this a typo, the correct form should be. theta1 = asin(n2*sin(theta2)/n1);

Thanks for posting these questions. I’ll log this and look at fixing this example in the near future.

Regards,

@trobertson
Thanks for your reply.

  1. I have got that the orientation of the two dipoles does not represent S, P polarization. But I have checked that when the angle is set to 90 degree, the electric field generated by this dipole is Ex, Ey and the Ez component is 0. But in this situation, if we want to get the S polarization far field( since the script want to get the far field of S and P polarization and add them together. In line 42: # collect data from both polarizations in the far field), we should have Ez componet. But Ez is 0 in this situation.

2&3: If I am correct, there comes a strange question, why is the simulation result consistent with the result got from the theory using the script? Since we now know that the formula used in the script to calculate the result is not correct.

Regards

Hello @makeqi,

Both Ex and Ez correspond to S polarization components i.e. normal to the plane of incidence for ky propagating light. I am not sure what monitor you are looking at but if you visualize the field data of the above monitor at the plane of incidence you will see Ex and Ey components. There will not be any Ez components since this is out of the plane of the dipole. That being said Ex and Ez are equivalent for this geometry so you could look at simulations in the yz plane if you like.

The far field monitors need to collect all of the field information to perform the required projection. Knowing that Ez is exactly zero is an important piece of information for recreating the far field radiation pattern.

If you look at what the script calculates it plots the Rp value which was correct, and it uses the incorrect form for theta1. I made the correction in the script and plotted the correct and previously used incorrect value below.

As you can see there id very little variation between the two values which is likely why this typo was not caught before.

Best Regards,

Thanks for your reply. @trobertson

Both Ex and Ez correspond to S polarization components i.e. normal to the plane of incidence for ky propagating light.

The plane of incidence is a plane and S polarization means that the electric field is perpendicular to the plane of incidence. So I think only one electric field component can be the S polarization. In the situation of the example on KB, the plane of incidence is XY, so I think only the Ez component can be regarded as S polarization. In the picture I uploaded above, I use a 2D-Z normal monitor. So I still don’t think that the simulation can get the reflectivity of S polarization. Am I right?
image

There will not be any Ez components since this is out of the plane of the dipole.

Another question is about the plane of the dipole. I am still not clear about which electric component can be generated by a dipole. I made a test like this:
image
In this situation, the dipole is along y-axis(phase = 0, theta = 90, phi = 90). I use a monitor of 2D-Z normal to get the electric field. Ex and Ey component are not zero, Ez is zero.
image
In this situation, the the dipole is along x-axis(phase = 0, theta = 90, phi = 0). I use a monitor of 2D-Z normal to get the electric field. I can get Ex and Ey component are not zero, Ez is zero.
image
In this situation, the the dipole is along x-axis(phase = 0, theta = 0, phi = 0). I use a monitor of 2D-Z normal to get the electric field. I can get Ex and Ey component are zero, Ez is not zero.

I don’t quite understand why it is like this. In each situation, what’s the plane of the dipole?
In the first and second situations, two component are not zero. But there is only one non-zero component Ez in the last situation. What is the reason? Could you give me the detailed explanation of the electric field generated by the dipole source? I have read the information about the dipole source on KB but I still can not get the answer by myself. Thanks a lot.

PS:
one small question:
Why it is equivalent using “set(“phi”,45)” and “set(“angle”,45)” when setting the dipole?

@trobertson
Could you help me to solve the questions above?

Hello @makeqi,

S polarized light is in the plane of the interface between the dielectric and air. The Ey component is P polarized, you can choose any plane that intersects the interface and it is equivalent. It is often convenient to choose the one where the polarization is purely Ez or Ex, but it really does not matter.

I would ask that you refer to the wkipedia page on dipole antennas for information on the radiation pattern and electric field generated by a dipole.

Regards,

@trobertson Thank you for your suggestion about the dipole antennas. The second question has been solved. But for the first question, this is a 2D simulation. The injection plane is XY. So I think only Ez can be considered as S polarized light. Right?

In the simulation, it runs two simulations and both of them get the far field of P polarized light. Am I right?

Hello @makeqi,

In the situation you describe you are correct. The confusion here, I believe, is that you referenced many simulations and the incidence plane was not always XY. The plane of incidence depends on the E field component you are considering and therefore the orientation of the dipole. By changing theta of the source you change the plane of incidence, but any change in theta is equivalent you are just changing the plane of incidence for this simple example.

In general, 3D/specular interface, the incidence plane basis is easy to construct since you know the k vector is drawn from the source - to the point of interest; furthermore, it must contain the surface normal of the interface. You can then decompose the radiation into components in plane (P or TM) with those normal to it (S or TE).

I hope this helps.

Regards,

Hi @trobertson
I think in this simulation, the injection plane is always XY. I have a opinion as following: for a 2D simulation in the XY plane, we can not get any information about light propagating in the Z direction, because it is uniform in the Z direction. The plane of incidence depends on the propagation direction of light and the surface normal of the interface. In this 2D situation, the propagation direction is always in the XY plane, and the surface normal is in the Y direction. So the plane of incidence is always XY.
By changing the theta of the source, we only change the electrical field component.
If I set theta = 0, the electric field component on the interface is Ez only. We only get the information of S polarization.
If I set theta = 90, the electric field component on the interface is Ex and Ey. We only get the information of P polarization.
If I set theta = 45, the electric field component on the interface is Ex, Ey and Ez. We get the information of S and P polarization.

Are all my statement right?

@trobertson Could you have a look at my question, please.
Regards.