Fourier-transform Infrared Spectroscopy Thin Film Thickness

[Tian En]

I had an experiment involving FTIR (incoherent light source) to measure the thickness of plastic film through making use of interference fringes. I don't understand how the interference fringes can occur over a frequency range (in term of wave number ν for FTIR) and also how it can occur at the detector side, not at the side facing the source. Below is what I already know.

Thin Film Interference
I learned that thin film occurs due to the path difference of light. For example, when light ray hit a optically denser plastic surface, it splits into 2 rays. The first ray, immediate reflected light from the plastic surface, undergoes λ/2 phase shift, due to refractive index larger than the air. The second ray, having penetrated the layer and undergo internal reflection, has traveled extra distance of 2 thickness. The thickness of ghe material determines the corrsponding frequency for constructive or destructive interference. For this to work, λ needs to be comparable to the thickness. The interference happens at the surface facing the light source and not the other surface.

Michelson Interferometer (red laser coherent light source)
I also learned that moving the mirror by 1/2*d can shift one of the ray by d. I can count the number n of complete (dark to dark transition or bright to bright transition) cycles to determine λ = 2d/n.

 

[pixel]

To answer your second question, some fraction of the second ray that you refer to refracts into the air on the (back) side facing the detector. And, after the internal reflection that you refer to, some fraction of the ray will reflect off of the first surface and also end up refracting into the air on the back side Those two rays that emerge on that side can interfere with each other as one has traveled an extra distance of two thicknesses.

I'm not sure what you are asking in your first question. Are you using a range of frequencies?

 

[Tian En]

Are you using a range of frequencies?

Yes, I use broadband IR source, which goes through michelson interferometer. The equipment varies the movable mirror distance to obtain an interferogram which then undergo Fourier transform to get transmittance vs wave number graph. The graph look like this:

From wave nunber 2500 to the right, the interference fringes are more pominent. On top of the extra distance of 2 thickness due to internal reflection, is there phase shift of λ/2 when the ray bounce off the interface during internal reflection (2 bounces make it a phase shift of λ)? Thank you.

 

[pixel]

On top of the extra distance of 2 thickness due to internal reflection, is there phase shift of π/2 when the ray bounce off the interface during internal reflection (2 bounces make it a phase shift of π)? Thank you.

I think you mean λ/2. The mnemonic that I learned years ago is: "low to high, phase change π," where low/high refers to the index of refraction. So the internal reflection does not have the phase shift.

 

[Tian En]

I think you mean λ/2. The mnemonic that I learned years ago is: "low to high, phase change π," where low/high refers to the index of refraction. So the internal reflection does not have the phase shift.

Oops, it should be λ/2. Thanks. I have understood how interference occurs at the outgoing ray. I still don't get the interference fringes over the transmittance vs wave number graph. Not sure how that phenomena happen.

 

[sophiecentaur]

Oops, it should be λ/2. Thanks. I have understood how interference occurs at the outgoing ray. I still don't get the interference fringes over the transmittance vs wave number graph. Not sure how that phenomena happen.

To clear this up. If the OP is talking about phase shift then the answer is an angle (for all frequencies) and not a path difference, which is wavelength dependent. Transmission depends on the path differences being near whole wavelengths. There can be many internal reflections, of course. It may be worth while thinking in terms of transmission lines where the incident power is 'accepted' or even 'pulled' through the surface because of the Impedance presented at the surface. The internal 'cavity' resonates and changes the presented impedance from a severe mis-match to a good match.

 

[pixel]

I still don't get the interference fringes over the transmittance vs wave number graph. Not sure how that phenomena happen.

Constructive interference occurs when OPD (the optical path length difference) = λ, 2λ, 3λ, etc. So you get one at λ = OPD, another one at λ = OPD/2, another one at λ = OPD/3, etc.

I can't make out the numbers by the peaks on your graph, but you have a peak at roughly 750 cm^-1. This would correspond to a wavelength of 13,333 nm. There is another peak at roughly 1500 cm^-1. This corresponds to 6,666 nm, or half of 13,333. The peak at 3000 cm^-1 is at a wavelength of 3,333, or 1/4 of 13,333.

Edited: Now I'm a little confused as to what your graph is showing as these are not peaks but transmission minima.

 

[Tian En]

Constructive interference occurs when OPD (the optical path length difference) = λ, 2λ, 3λ, etc. So you get one at λ = OPD, another one at λ = OPD/2, another one at λ = OPD/3, etc.

I can't make out the numbers by the peaks on your graph, but you have a peak at roughly 750 cm^-1. This would correspond to a wavelength of 13,333 nm. There is another peak at roughly 1500 cm^-1. This corresponds to 6,666 nm, or half of 13,333. The peak at 3000 cm^-1 is at a wavelength of 3,333, or 1/4 of 13,333.

Edited: Now I'm a little confused as to what your graph is showing as these are not peaks but transmission minima.

I guess I have understood what's happening there. When there is no phase shift of λ/2, and the ray just traveled extra distance of 2 thickness, considering the troughs, I will have such formula 2t = (m+1/2)λ for destructive interference, considering wave number v (in cm^-1) as 1/λ_air, given medium refractive index n = 1.5, I have the relationship t = (m+1/2)*1/(2*n*v_air). Graph fitting and analysis yield thickness t = 1.13*10^-5m. Thank you.

 

[pixel]

Okay, if you are considering destructive interference, then OPD = λ/2, λ, 3λ/2, etc. and for a given thickness, you get transmission minima at these wavelengths and the corresponding wavenumbers.

You forgot the n in the first equation of your last post (the OPD = n2t) but have it in the second one.

 

[Tian En]

How do I find error bar for my FTIR graph peaks?