About T-matrix and S-matrix, and how to obtain the electromagnetic field distribution of the diffraction structure under the specified amplitude ?

[xungu1]

I noticed that your program uses the enhanced T-matrix method to calculate the diffraction efficiency when calculating the multi-layer structure, and I also noticed that your program seems to provide a calculation method for the S-matrix. I currently have two doubts : The first is if I want to get the S-matrix at the interface of each interface of the multi-layer structure, how do I operate ? Can I use the calculation results of T-matrix to get the S-matrix at the interface of each interface ? If so, how should I operate ? The other is how do I operate if I want to get the effect of a specific amplitude on a certain layer of structure ? I have noticed that the S-matrix can be used as a calculation when there are inputs on both sides of the interface. I want to use this S-matrix to calculate the electromagnetic field energy distribution of the layer under different illumination on both sides at the same time, or can the same operation be achieved in another way ?

[yonghakim]

SMM option in meent was disabled. I will try to make it available again (or make it sure if it's already available) but cannot say when.

Regarding two questions,

• S matrix
  • SMM and TMM are about how the waves of in and out are connected at the interfaces. S matrix is not defined in TMM. You can utilize other information of Enhanced TMM in meent to calculate S matrix. If you need S matrix, I recommend you TORCWA which is open source python RCWA tool. This uses classical SMM.
• Both sides
  • Meent doesn't support this. If what you mean is to calculate two situations (one situation with incident light from the top and one other situation with incident light from the bottom) with single simulation, not one situation with two light sources, then I suggest to look reticolo(matlab) or TORCWA(python). They are both open source so you can take a look. TORCWA uses classical SMM.

[yonghakim]

$S_2$ seems the light source you are looking for - reverse incidence. It is 0 in Moharam's.

[xungu1]

As you say，if SI is delta function, then (14) is same as (41).I think that the reverse incidence can be achieved by flipping the stack, but in fact, whether I pass through the reverse incidence or flip the stack, I cannot bypass the light source that requires the incident light source to be a specified transmission amplitude.What I fail to understand is, if I want to use the specified transmission amplitude as the amplitude of the reverse incident light source, how do I set SI ? According to the data I have consulted, the diffraction efficiency of a certain order of the grating can be defined as the product of the transmission amplitude, that is, | t ^ 2 | = η, where t is the transmission amplitude and η is the diffraction efficiency. So now if I want to set the transmission amplitude to the amplitude of the reverse incident light source, how do I set it.

[xungu1]

I have tried to modify the delta function to 2, but the fact is that this is only the result multiplied by 2.
Could you please provide your contact information conveniently ? Such as Facebook / twitter / whatsapp or other contact information, so that we can contact you in a timely manner.

[yonghakim]

| t ^ 2 | = η, where t is the transmission amplitude and η is the diffraction efficiency.

This is not true. See (67) in meent paper. There is more term.

I have tried to modify the delta function to 2, but the fact is that this is only the result multiplied by 2.

I never tried other source other than delta function so I can't say it's right or wrong.
But it seems reasonable to me that you got a doubled result with a doubled input. What kind of result were you expecting?

[xungu1]

Thank you very much for your previous answers ! I simulated aluminum in the visible light band. The real and imaginary parts of the refractive index of aluminum at different wavelengths were used as input. When viewing the transmission and reflection, it was found that the reflectivity would exceed one at some wavelengths. Is this a normal phenomenon ?
Shouldn 't I take the imaginary part as input ? I set the duty cycle of the aluminum grating is 80 %, the period is 350 nm, the wavelength is from 400-750 nm, and the depth is 35 nm.

[yonghakim]

The sum of the diffraction efficiencies is one when the refractive indices are real.
Whether to use the imaginary part or not, I suggest to follow other precedent works in the same domain.

[xungu1]

I think so too.I 'm also going to try some previous examples.Thank you very much！