Tagged: beam-width, fdtd, photonics
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July 1, 2022 at 7:01 pm
shankargmenon
SubscriberHi,
I was trying to simulate some gaussian beams and noticed that the beam sizes are different than what is expected analytically. In that attached files, I have simulated a gaussian beam with 0.14NA, its FDTD simulated 1/e^2 diameter seems to be 2.17um (for 1360nm). However for a 0.14NA lens, for plane wave input with completel filling of the lens, the expected beam diameter is 11.85um and assuming gaussian input beam it will be 6.18um (assuming focal length of 40mm and beam size of 11.2). So I am lost here why the simulated beam is so off. Could you let me know what could be going wrong? My assumption is it is some sort of 2*pi factor error.
Here are the simulated beams and a fit to its line cut. I have also attached the link to the .fsp file for clarity.
https://drive.google.com/file/d/1x3QD9VNfwBIgdQjlY0C7Hvh6acChrTta/view?usp=sharing
Thanks,
Shankar
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July 4, 2022 at 8:10 pm
Guilin Sun
Ansys EmployeeThere is some confusion here for the NA beam and the scalar Gaussian beam. NA bean is diffracted beam, thinking the fields at the focal plane of a microscope objective. It is Airy pattern with side lobes. It is
It is not Gaussian shape.
The simulated result suffers some discrete errors so the simulated results may not be exactly the same as theoretical value. In addition, the simulated result is full vectorial even through the excitation source has only one non-zero component. Please note that the scalar Gaussian beam is approximate, which assumes only one field component, eg, the polarization (Ex for example) in source excitation.
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July 5, 2022 at 3:44 pm
shankargmenon
SubscriberHi, Thanks for the reply, but I believe there is some misunderstanding. The terminology might have been bit confusing due to gaussian source object and gaussian input beams.
On the first part I am talking about a plane wave incident on the lens producing diffraction limited beam (non gaussian). I was already using the diffraction limited equation that you have pasted in your reply. From the equation that you pasted for the airy pattern, the first dark ring ( first minima) should be where the J1(k a sin(theta)) goes to zero, which corresponds to a beam diameter of the diffracted spot to be 1.22*lambda/NA. For 1360nm input beam with an NA of 0.14 this corresponds to 11.85um. But the FDTD simulated version for fully vectorial calculation gives 2.17um, which is confusing!!! This should be what is simulated by lumerical.
On the second part I show that this is not explained by assuming a gaussian input beam instead of a plane wave. If the input to a lens is gaussian instead of plane wave, the diffraction limited beam (which is non gaussian) is expected to be 6.18um based on the same diffraction limit arguments. So I am confused why the simple fully vectorial FDTD simulation doesnt match any of these.
Thanks
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