April 20, 2022 at 9:38 ammschnellSubscriber
Hi, I have a few questions about the simulation of tightly-focused beams.
1) How does the pupil function of a Gaussian beam relate to the far field projection? I inject fields with a Gaussian beam and a constant pupil function (NA=0.8). I collect the beam with a monitor and perform a far field calculation. Interestingly, the observed pattern is not constant, but seems to be a cosine(theta) shape. Shouldn't it be constant like the pupil function?April 20, 2022 at 9:10 pmApril 20, 2022 at 10:52 pmGuilin SunAnsys EmployeeThis is a very interesting topic, with different concepts.
A1: Pupil function is defined in k-space. It tries to make the farfield within the given NA uniform. If you disable the Pupil function, the farfeild intensity will be Gaussian-like.
try to use set("use custom pupil function",false); and check the simulation result.
High NA beam will be like the Airy diffraction pattern.
As you can see, the pupil function is uniform in k-space:
The actual source injection is in the near field on the source plane, so I believe internally it has been transformed to nearfield. After injection, simulation, you get the farfield. During this process, errors are introduced, mostly due to truncation to finite size. Therefore the simulated result is not as ideal as the pupil function.
A2: It might be due to this note:
I used this script
source_theta = 36; # azimuthal angle
source_phi = 0; # polar angle
NA = 0.4; # maximum NA of the beam
n = 501; # number of points in each direction of k-space
and then compared to pure NA=0.4 beam, the result is very similar:
A3: I do not know the details. From the KB page, it should just use the regular angled injection, since the angles are set not in pupil function but in the source settings.
April 25, 2022 at 8:40 ammschnellSubscriberThanks for getting back to this so fast.
A1: I calculated with the custom pupil function turned off and the result is nearly the same.
From what I understand about a Gaussian beam source is that each element of the pupil function is weighted by a ÔÇ£strength factorÔÇØ sqrt(cos(theta)) (├çapo─ƒlu, ─░. R., Taflove, A. & Backman, V. Computation of tightly-focused laser beams in the FDTD method. Opt. Express, OE 21, 87ÔÇô101 (2013)). This factor relates the field amplitude of the ray at the entrance pupil against the ray after the lens (facing the sample). I guess this explains the non-uniform patterns seen in the far field projection.
I think the strength factor is not considered in the far field projection because no lens is assumed in the far field projection, but only the rays coming out from the sample are calculated. I suppose that if I needed to calculate the field at the exit pupil of the collecting lens, I would need to divide by the strength factor again. Not 100% sure if this is right.
A2: Ok, I will use the rotation option, thanks for pointing this out.
April 25, 2022 at 3:22 pmGuilin SunAnsys EmployeeYou are right that at very large NA this can be more un-uniform as expected. When you want to integrate the farfield strength you will need to correct this angle-dependency to get the actual power.
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