Effects of coherence length on optical interference filter

AI Thread Summary
The discussion centers on the functioning of an optical interference filter designed to block solar heat between 800nm and 1100nm, which comprises 200 layers of coatings. A key point raised is the apparent contradiction between the filter's thickness, which exceeds the coherence length of solar light, and the ability of the filter to still produce interference effects. It is clarified that the relevant coherence length for interference filters is the longitudinal coherence length, which is significantly longer than previously estimated, allowing for effective interference even with thicker coatings. The conversation also highlights that each individual layer's thickness must be less than the coherence length for the filter to operate correctly. Overall, the coherence properties of sunlight and the design of multilayer coatings enable the filter to function effectively despite its thickness.
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I have an optical interference filter that is used to block (reflect) solar light heat between 800nm and 1100nm. That filter is made by using 200 layers of quarter wave high and low index coatings on plastic film. The total thickness of the coating is 25 micrometers or even 50 micrometers.

I was told that interference effect can only happen when the thickness of coating is less than coherence length of light. How can my filter work when the total thickness of interference coating is much larger than the coherence length of solar light that is only about one micrometer?

I can understand locally the adjacent coating layers can still have interference effect but how can the light beam reflected by the top coating layer interfere with light beam reflected by bottom coating layer since the optical path difference between these two beams is far larger than coherence length of solar light?
 
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shiweiliu said:
I was told that interference effect can only happen when the thickness of coating is less than coherence length of light. How can my filter work when the total thickness of interference coating is much larger than the coherence length of solar light that is only about one micrometer?
The coherence length of sunlight is significantly larger than ##1 \mu\text{m}##.
From this reference: Spatial coherence of sunlight... I quote an excerpt (from pg. 98):
1736061207419.png

So for a solar wavelength of ##0.5 \mu\text{m}## the coherence length ##L_\bot## amounts to ##40 \mu\text{m}##.
 
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renormalize said:
The coherence length of sunlight is significantly larger than ##1 \mu\text{m}##.
From this reference: Spatial coherence of sunlight... I quote an excerpt (from pg. 98):
View attachment 355358
So for a solar wavelength of ##0.5 \mu\text{m}## the coherence length ##L_\bot## amounts to ##40 \mu\text{m}##.
Oh. I estimated it according to formula: wavelength^2/Delta wavelength. Wavelength is about 500nm and Delta wavelength is about 300nm. It is about one micrometer. Something is wrong in my method?
 
defining the coherence length L⊥ as the slit separation for which theamplitude of η⊥ decays to 1∕2, we find that L⊥ 80λ fordirect sunshine
It looks like they are defining something that they call for "a coherence length", but I am not sure that is the same as coherence length defined as "the maximum optical path difference to still observe interference".
For unfiltered sunlight the interference would disappear withing 1um or less path difference. Using a narrow bandpass filter (or spectral device) it is possible to get interference with much longer path difference.
However, there is a limit for thermal sources that I have seen in a physics books, which is connected to the maximum coherence time not exceeding 10-8s or 2..3m of path difference.
 
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renormalize said:
The coherence length of sunlight is significantly larger than ##1 \mu\text{m}##.
From this reference: Spatial coherence of sunlight... I quote an excerpt (from pg. 98):
View attachment 355358
So for a solar wavelength of ##0.5 \mu\text{m}## the coherence length ##L_\bot## amounts to ##40 \mu\text{m}##.
That is spatial coherence or the transverse coherence length. It tells you how point-like the light source is or what the maximal separation of slits may be to observe interference in a double slit experiment.

For interference filters, the longitudinal coherence length (coherence time times the speed of light) is the relevant quantity. The coherence time is essentially given by the Fourier transform of the power spectral density of your light field. Simply speaking: a broad spectrum results in a short coherence time. Therefore, one may improve coherence time by spectral filtering. Depending on what is going on in the atmosphere, sunlight filtered by the atmosphere is usually considered to have a coherence time on the order of one to few femtoseconds:

https://www.nature.com/articles/s41598-022-08693-0

A coherence time of 1 femtosecond corresponds to a longitudinal coherence length of about 300 nm.
 
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Intersting article in Nature, but it doesn't give the answer to the topic starter question, asking why a thick multilayer coating is working while sunlight has a very short coherence length.
 
The Fabry-Perot effect (present in interference filters) is often taught as multiple reflections that occur that then might need fairly long coherence lengths to be completely effective. You might find this Insights article that I authored a few years ago of interest, where the interference effects can occur at a single dielectric interface, and where multiple reflections are not the fundamental reason why the interference occurs. The multiple layers with the multiple interfaces do play a role, but as this article explains, interference occurs from two mutually coherent sinusoidal sources incident onto a single dielectric interface from opposite directions.

See https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/
 
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I need to add one more item to the above=at each dielectric interface in the multi-layer system there is a right-going wave and also a left-going wave (amplitude) on the left side of the interface, and also a left-going wave and a right-going wave on the right side of the interface. There is a phase relation depending on the path distance of each layer that also comes into play, (between the right-going waves at adjacent interfaces, etc. and between the left-going waves at adjacent interfaces). This is how the multi-layer problem is solved rather than trying to compute all possible multiple reflections.

Edit: For the coherence length calculation, it would appear that all you would need to get interference with such a multi-layer system is a coherence length that is perhaps 2 or 3 times the optical thickness of each layer.
 
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shiweiliu said:
I have an optical interference filter that is used to block (reflect) solar light heat between 800nm and 1100nm. That filter is made by using 200 layers of quarter wave high and low index coatings on plastic film. The total thickness of the coating is 25 micrometers or even 50 micrometers.

I was told that interference effect can only happen when the thickness of coating is less than coherence length of light. How can my filter work when the total thickness of interference coating is much larger than the coherence length of solar light that is only about one micrometer?

I can understand locally the adjacent coating layers can still have interference effect but how can the light beam reflected by the top coating layer interfere with light beam reflected by bottom coating layer since the optical path difference between these two beams is far larger than coherence length of solar light?
Multilayer dielectric coatings can be designed to have a (nearly) arbitrary spectral reflectivity- ultrabroadband is a simple example:

https://www.laser2000.com/en/mirrors/82436-ultra-broadband-dielectric-mirror.html

How do multilayer dielectric coatings work? Any decent optics book will have a thorough discussion. Apps can design the specifics based on your desired reflectivity profile, and companies can fabricate your designs

https://www.rp-photonics.com/rp_coating.html
https://www.chroma.com/capabilities/manufacturing

Note- the thickness of each individual layer of a multilayer coating is indeed less than the coherence length of the light source.
 
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  • #10
Andy Resnick said:
Note- the thickness of each individual layer of a multilayer coating is indeed less than the coherence length of the light source.
This is the key! In my previous post above I incorrectly cited the spatial/transverse coherence length of sunlight, but the crucial parameter for multilayer interference-filters is the temporal/longitudinal coherence length, which is indeed ##\sim 1\mu m## as stated by the OP. As long as each individual layer thickness is less than this value, a multilayer filter will work for any overall stack height. Here's a reference that discusses designing filters by this criterion: Optical model for multilayer structures....
From the introduction:
"Optical models for thin-film multilayer structures are used in several important applications such as design and analysis of optical coatings and thin-film solar cells. The objective of these optical models is to predict reflectance, transmittance or absorption depth profile as a function of wavelength. For accurate calculation of these parameters it is crucial that the interference between multiple internal reflections is taken into account correctly. Thus far this has been especially challenging for thin-film solar cells which typically have an absorber layer thickness on the order of the coherence length of the incident sunlight. Layers with an optical thickness less than the coherence length of the incident light cause observable interference fringes and are commonly referred to as coherent layers. Much thicker layers, that do not give rise to observable fringes, are referred to as incoherent layers. Partly coherent layers represent the intermediate case."
 
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  • #11
Very good, and I pretty much agree with the previous two posts, but there IMO is something missing from the standard explanations of multiple reflections and/or the case of multiple layers with a left-going and right going wave amplitude on each side of the interfaces. It took a number of years for me to see the part that I originally couldn't put my finger on, and that is mentioned in post 7: The second interface for the Fabry-Perot effect is needed to make multiple reflections, but its real purpose is to provide for the left-going wave incident on the first interface=that way you have two waves from opposite directions incident on the first interface and interference results from this simplest of systems. ( I was surprised myself when I discovered that interference results from the emerging waves from two sinusoidal sources incident onto a single interface from opposite directions that completely conserves energy). Edit: Note that the resulting energy redistribution from these two sources will change as the phase of the left-going wave is changed=the distance to the second interface is indeed important.

Otherwise, yes, the standard explanations written up in the literature will get you the results that you need, but this fundamental process is often ignored. With it, both the Fabry-Perot and Michelson interferometers function from the same fundamental concepts, and the two aren't as different as the standard explanations could lead one to believe=in both of them there are two waves incident from different directions onto a dielectric interface, and the emerging wave amplitudes interfere, with the result that the energy redistribution depends on the relative phases of the two incident waves. (Note that the dielectric interface splits each wave into two components=a reflected and a transmitted).
 
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  • #12
The sunlight comprise emission events from many atoms with a temporal coherence on a level at least 10-11s or longer. That mean the filtered sunlight coherence length may exceed > 1cm or more. The fact that the thick multilayer interference coating if working evident the the light coherence of elementary atoms bursts are much longer then the thickness of the coating.

Another example I can take is the Lunt Fabry–Pérot interferometer used for the observation of the Sun in H-alfa.

Lunt Fabry-Perot etalon.jpg


From the picture above it is easy to calculate that the path length between two parallel mirrors is 525 waves or 344µm. The width of the narrow peaks suggest that reflective coefficient is about 0.9 or so. Taking into account that the light is traveling the gap multiple times, > 20..30 times to decay, that gives the coherence length well above 2..3 cm to make it works.

Taken from Britannica Propagation and coherence
Light from the Sun or from a lightbulb comes in many tiny bursts lasting about a millionth of a millionth of a second and having a coherence length of about one centimetre. The discrete radiant energy emitted by an atom as it changes its internal energy can have a coherence length several hundred times longer (one to 10 metres) unless the radiating atom is disturbed by a collision.

Would be the "intrinsic" coherence length just above a single layer thickness the entire multilayer coating would not work! The fact the the multilayer coating is working with high efficiency evident that the "intrinsic" coherence is much-much longer than the entire coating thickness.
 
  • #13
renormalize said:
.. the crucial parameter for multilayer interference-filters is the temporal/longitudinal coherence length, which is indeed ##\sim 1\mu m## as stated by the OP. As long as each individual layer thickness is less than this value, a multilayer filter will work for any overall stack height.
Misleading sentence.
I want to emphasise once more that the coherence length of broadband white light has nothing with the ability of the thick multilayer coating to work.
The fact that multilayer coating does work declare that the intrinsic coherence length of the elementary emission events combining the white light is indeed many times longer than the thickness of the coating otherwise it would not work.
 
  • #14
Gleb1964 said:
From the picture above it is easy to calculate that the path length between two parallel mirrors is
The calculation for the distance between mirrors in a Fabry-Perot cavity is ## d=\frac{\lambda^2}{2 \, \Delta \lambda} ## (where ## \Delta \lambda ## is the spacing between peaks in the spectral transmission curve) which is one half of what you computed. (Comes from ## m \lambda_1=2nd ##, and ## (m-1)\lambda_2=2nd ##).

Meanwhile a google seems to suggest that the coherence length of the H alpha line is a few millimeters, and I think it is likely that it has a longer coherence length than the blackbody type portion of the solar spectrum.

I'm inclined to believe that the coherence length necessary for the layered interference filter to work will be about 3 or 4 times the width of each layer (rather than depending upon the width of the entire stack), but we don't have an authoritative source on that yet that anyone has referenced.
 
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  • #15
Gleb1964 said:
The fact that multilayer coating does work declare that the intrinsic coherence length of the elementary emission events combining the white light is indeed many times longer than the thickness of the coating otherwise it would not work.
Can you provide a reference which quantifies this claim involving total coating-thickness instead of individual layer thicknesses? Like, e.g.:
"A multilayer interference filter of total thickness ##T## will function with ##X##% efficiency under light with coherence-length ##L## provided that ##T## satisfies ##T\leq\Lambda L##, where ##\Lambda=(\text{specified})##."
 
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  • #16
One or two additional comments on post 12: I'm assuming this is the experimental result of the transmission spectrum run with a diffraction grating type spectrometer, where most likely an incadescent source (typically about a 2500 K blackbody) was used, with one run with the Fabry-Perot etalon, and the other without it. The optical path length 2d=344 microns was computed, and that would indicate the 2500 K blackbody source must have a coherence length a good deal greater than that.

I don't know that we are getting good numbers for the coherence length of solar light. In an early post it was stated to be around one micron in length, but a google indicates it to be in the 50 micron range. The fact that the OP's filter works leaves this as an open item, but it would be nice if we could determine if the coherence length needs to be of the order of a couple of layers of the interference filter or the width of the entire stack. I am going with the former.
 
  • #17
Charles Link said:
I don't know that we are getting good numbers for the coherence length of solar light. In an early post it was stated to be around one micron in length, but a google indicates it to be in the 50 micron range.
As I mentioned in post #10, there are two coherence-lengths for sunshine: spatial/transverse (##\sim 50\mu\text{m}##) and temporal/longitudinal. It is this second quantity that is relevant for multilayer interference filters. From the discussion on pg. 2 of The coherence time of sunlight..., depending upon its definition, the temporal coherence time of sunlight is quoted in the range ##0.58-3\text{fs}##. Using the vacuum speed-of-light ##0.3\mu\text{m}/\text{fs}##, this corresponds to a longitudinal coherence length of ##0.17-0.9\mu\text{m}##.
 
  • #18
renormalize said:
corresponds to a longitudinal coherence length
I doubt that the number is less than one micron. If that were the case, even the single layer of a stack of thin film layers would fail to operate.

I must admit, I lack expertise in this area in regards to the individual photons or groups of them in the same boson mode, and what they are defining as a coherence length. Meanwhile though, I did spend a lot of time working the interferometer concepts, (Fabry-Perot and Michelson), and that's something that I think others might find useful. See e.g. posts 7, 8, and 11.
 
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  • #19
Charles Link said:
I doubt that the number is less than one micron. If that were the case, even the single layer of a stack of thin film layers would fail to operate.
That's not true.
A layer that's thicker than the light coherence-length still functions optically to a calculable degree; i.e., it is partially coherent. Have you read the reference I gave in post #10? From its conclusion:
"In summary, we showed that the net-radiation method can be used for coherent calculations of multilayer structures. We demonstrated that by averaging only a few calculations with equidistant values of ϕ, the coherence of any layer can be removed partly or completely. This makes this relatively simple method a powerful optical design tool for multilayer structures consisting of coherent, partly coherent and incoherent layers, such as thin-film solar cells. We illustrated this method by optical simulation of an a-Si-H solar cell with various degrees of absorber layer coherence."
 
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  • #20
renormalize said:
That's not true.
A layer that's thicker than the light coherence-length still functions optically to a calculable degree; i.e., it is partially coherent. Have you read the reference I gave in post #10? From its conclusion:
"In summary, we showed that the net-radiation method can be used for coherent calculations of multilayer structures. We demonstrated that by averaging only a few calculations with equidistant values of ϕ, the coherence of any layer can be removed partly or completely. This makes this relatively simple method a powerful optical design tool for multilayer structures consisting of coherent, partly coherent and incoherent layers, such as thin-film solar cells. We illustrated this method by optical simulation of an a-Si-H solar cell with various degrees of absorber layer coherence."
I just read through it=thank you. E. Hecht is listed in the article near equation (10) as saying the coherence length for sunlight is about one micron. His Optics book (Hecht and Zajac) is a good one and we'll have to go with that. From what I could get from the article, it seems they are saying the distance of interest is the thickness of the individual layers. I don't know that they came out and stated it explicitly, but it seems that is what they are calculating.

Meanwhile though, I do think the coherence length does need to be nearly one micron for these filters to work=otherwise you start getting in a region where the coherence length is shorter than the wavelength where you are trying to generate interference. e.g. a coherence length of 0.2 microns would not work. (Note the visible light at wavelength of 600 nm is 0.6 microns)
 
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  • #21
While the coherence length of unfiltered sunlight is less than one micron, it is possible to enhance the coherence by using a bandpass filter. However, the coherence length can only be improved up to a certain limit, which is determined by the lifetime of the emitters.
 
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  • #22
Gleb1964 said:
While the coherence length of unfiltered sunlight is less than one micron, it is possible to enhance the coherence by using a bandpass filter. However, the coherence length can only be improved up to a certain limit, which is determined by the lifetime of the emitters.
You seem to be changing your opinion from that of post 13, but that is ok. One of the purposes of the discussion is to try to learn something. :)
 
  • #23
renormalize said:
Can you provide a reference which quantifies this claim involving total coating-thickness instead of individual layer thicknesses? Like, e.g.:
"A multilayer interference filter of total thickness ##T## will function with ##X##% efficiency under light with coherence-length ##L## provided that ##T## satisfies ##T\leq\Lambda L##, where ##\Lambda=(\text{specified})##."
May I refer to the echelle spectrometers used for star and Sun observation?
In multilayer coating interference is occurred between reflections from multiple layers, in diffraction gratings interference is occurred between light diffracted from different grooves across grating aperture. Echelle gratings are used with large angle of incidence that makes quite large path difference across the aperture.
How large the size of echelle gratings that used for stars/sun?
There are plenty references about 300..400 mm size.
The echelle spectrometer for Giant Magellan Telescope (GMT) is planning to us echelle with 0.3 x1.25 m size (here is the link The optical design of the G-CLEF Spectrograph ),:
G-CLEF Spectrograph Optical Layout.jpg


That would be above 1m of the optical path different across the aperture. I understand that they assume that the light from stars would be coherent otherwise it would be not possible to realize the planned spectral resolution of the spectrometer.
 
  • #24
Gleb1964 said:
While the coherence length of unfiltered sunlight is less than one micron, it is possible to enhance the coherence by using a bandpass filter. However, the coherence length can only be improved up to a certain limit, which is determined by the lifetime of the emitters.
Can you expand on this? In general terms, what type of bandpass filter design(s) would be suitable for producing filtered sunlight with a longitudinal coherence-length greater than ##1\mu\text{m}##?
 
  • #25
I am simply repeating what I have read in physics textbooks (which I revisit from time to time to refresh my memory).

White light consists of numerous acts of elementary emission. Undisturbed atoms can emit light for durations of about 10−6..10-8 seconds, producing electromagnetic waves that consist of up to 106..108 cycles before the phase changes randomly. Collisions between atoms can shorten both the time and length of these undisturbed emissions. Interference can occur only if the path difference does not exceed the length of an individual elementary emission act.
Thus, the ability to produce interference is not inherently limited to 1 µm for white light; its intrinsic coherence length is much longer. However, without filtering, the interference cannot be revealed. For example the human eye can distinguish colors and observe tens of Newton fringes even in daylight, much more than suggest 1um coherence length. Before I have used other examples to explain this subject.
 
  • #26
Gleb1964 said:
Thus, the ability to produce interference is not inherently limited to 1 µm for white light; its intrinsic coherence length is much longer. However, without filtering, the interference cannot be revealed.
I've listed my references that state ambient sunlight has a longitudinal coherence-length of ##\sim 1\mu\text{m}## or less. Yet you claim "its intrinsic coherence length is much longer". Can you please cite scholarly references that explicitly support this specific claim and quantify "much longer" with actual numbers? If you can't, this seems like a personal theory.
 
  • #27
I found one "link" on the subject that may be of interest. Perhaps I can help @Gleb1964 on this item.

See https://mpl.mpg.de/fileadmin/user_upload/Chekhova_Research_Group/Lecture_5_2.pdf

They have the formula ## l_{coh} \approx \frac{\lambda^2}{2 \, \Delta \lambda} ##.

This would suggest that if we run the sunlight through a diffraction grating spectrometer, and sample a very small ## \Delta \lambda ## using narrow slits that this light would then have a very long coherence length=far beyond one micron (by a factor of 1000 or more).

I think @Gleb1964 may have it correct, while for a broad solar spectrum, the ## \Delta \lambda ## is approximately ## \lambda ##, which is consistent with the one micron number for the coherence length. Hopefully this helps.
 
  • #28
renormalize said:
Can you please cite scholarly references that explicitly support this specific claim and quantify "much longer" with specific numbers? If you can't, this seems like a personal theory.

As you wish, here is fragment from Sivuhin Optics T.5
..removed...

Translation of essential part:
An example of this is the radiation from an isolated atom. An excited atom emits a series or, as it is commonly called, a wave packet during the emission time Temission, which typically lasts around 10^{-8} seconds. Such a wave packet (or can be translated as a wave train) contains 10^6..10^8 waves. During this time, the atom "de-excites" and transiting to an unexcited state. Due to various processes, such as collisions with other atoms or impacts from electrons, the atom can return to an excited state and then begin emitting a new wave packet. As a result, a sequence of wave packets is emitted by the atom at short and irregularly changing intervals.
 
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  • #29
Gleb1964 said:
As you wish, here is fragment from Sivuhin Optics T.5
View attachment 355651
PF is an English language forum. Please offer a translation of this text-image or else cite a different reference published in English.
 
  • #30
I have to struggle to find a similar explanations, may be here is saying something similar 5.3.1 Coherence of Light Sources. Unfortunately not about thermal light sources.

But I have translated the essential part of text above. I hope it helps.
 
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  • #31
Gleb1964 said:
Translation of essential part:
Thank you for translating. That quote discusses atomic emission but says nothing regarding solar light, which is the subject of the original post. So I ask again: can you or can you not provide a reference stating that the longitudinal coherence-length of ambient sunlight as it arrives at the Earth's surface is "much longer" than ##\sim 1\mu\text{m}##?
 
  • #32
renormalize said:
Thank you for translating. That quote discusses atomic emission but says nothing regarding solar light, which is the subject of the original post. So I ask again: can you or can you not provide a reference stating that the longitudinal coherence-length of ambient sunlight as it arrives at the Earth's surface is "much longer" than ##\sim 1\mu\text{m}##?
I shouldn't try to answer completely for @Gleb1964 , but have you seen my post 27? I think that might answer at least part of this question. We should try to avoid personal theories whenever possible, but this is one topic where the mathematics can get somewhat complex, e.g. in the link of post 27, and is subject to interpretation. It can even be difficult to predict whether a stack of layers of an interference filter will work or not, but the proof is in the pudding.
 
  • #33
Charles Link said:
I shouldn't try to answer completely for @Gleb1964 , but have you seen my post 27? I think that might answer at least part of this question. We should try to avoid personal theories whenever possible, but this is one topic where the mathematics can get somewhat complex, e.g. in the link of post 27, and is subject to interpretation. It can even be difficult to predict whether a stack of layers of an interference filter will work or not, but the proof is in the pudding.
Your reference in post #27 is indeed informative, but it only mentions sunlight in the context of Young's experiment measuring spatial/transverse coherence, and even then Young had to project the sun through a pinhole to achieve interference. In contrast, the OP in this thread is asking about a multilayer IR-reflecting (heat-rejecting) film, such as that applied to a window and exposed to raw ambient sunlight. To practically design such a film requires knowing the temporal/longitudinal coherence-length ##L## of raw sunlight. To me, whether ##L\sim 1\mu\text{m}## or ##L\gg 1\mu\text{m}## is a crucial difference that impacts the film design/performance and is hardly just a matter of "interpretation". That's why I ask for a credible source justifying @Gleb1964's claim that ##L\gg 1\mu\text{m}## for ambient sunlight.
 
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  • #34
Something wrong with my post #12?
Lunt Fabry-Perot is routinely used for solar observations in ##H \alpha##, its transmission spectrum is represented and the minimum coherence length > 1cm is derived in #12.
renormalize said:
..claim that ##L\gg 1\mu\text{m}## for ambient sunlight.
 
  • #35
renormalize said:
Your reference in post #27 is indeed informative, but it only mentions sunlight in the context of Young's experiment measuring spatial/transverse coherence,
If you look right around figure 3 on around the 3rd page, it has a formula ## l_{coh} \approx \frac{\lambda^2}{2 \, \Delta \lambda } ## that seems to apply in general. From what I can deduce from this formula, and I mentioned this in post 27, it appears to me that you are both right.

To be able to predict or accurately model the multi-layered interference filter is still a little bit of a puzzle though. If the coherence length of the broadband sunlight is somewhere around one micron, (and I think we are now all in agreement on this), it remains an open item whether the type of filter described by the OP should work in principle, even though we have experimental proof that it does. The source that you linked in post 10 seems to be our best reference right now, but I found the details of their numerical methods a little difficult to follow.

Meanwhile I still could enjoy getting some feedback on posts 7,8, and 11. I mentioned that it took me a number of years to finally recognize the interference is essentially the result of two sinusoidal sources incident from opposite directions on a single interface for the Fabry-Perot case. I first saw the Fabry-Perot effect as an undergraduate student in 1975, and I studied the calculations with the multi-layer thin films as a graduate student in 1979. I encountered the Fabry-Perot effect numerous times at the workplace. There always seemed to me that there was something missing in the standard explanations, which I finally figured out in 2008-2009 by trying some calculations and computing the interference effects (using the Fresnel coefficients) when there are two sinusoidal sources incident from opposite directions onto a single interface. I urge you to try it for yourself and/or read my Insight article listed in post 7. I think you might find it worth your while.
 
  • #36
Hi, my name is Soniya. I'm a [Spam link redacted by the Mentors] expert. What role does the coherence length of a light source play in the performance of optical interference filters? Can someone explain how shorter or longer coherence lengths impact the interference effects in practical applications?
 
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  • #37
If the coherence length is too short, the interference effects won't occur, but the precise details of how short is too short is still an item for discussion.

Here is a copy/paste from post 35 that I could enjoy some feedback on. I'm repeating it here, because we are now on another page, and very few people might otherwise see it:

Meanwhile I still could enjoy getting some feedback on posts 7,8, and 11. I mentioned that it took me a number of years to finally recognize the interference is essentially the result of two sinusoidal sources incident from opposite directions on a single interface for the Fabry-Perot case. I first saw the Fabry-Perot effect as an undergraduate student in 1975, and I studied the calculations with the multi-layer thin films as a graduate student in 1979. I encountered the Fabry-Perot effect numerous times at the workplace. There always seemed to me that there was something missing in the standard explanations, which I finally figured out in 2008-2009 by trying some calculations and computing the interference effects (using the Fresnel coefficients) when there are two sinusoidal sources incident from opposite directions onto a single interface. I urge you to try it for yourself and/or read my Insights article listed in post 7.=See
https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/
I think you might find it worth your while.
 
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  • #38
avlok said:
Hi, my name is Soniya. I'm a [Spam link redacted by the Mentors] expert. What role does the coherence length of a light source play in the performance of optical interference filters? Can someone explain how shorter or longer coherence lengths impact the interference effects in practical applications?
That short coherence length ##1 \mu\text{m}## of unfiltered solar light, that have been intensively disputed here, has about nothing with the performance of the interference filters.

Calculating a multilayer interference filter you need to split wavelength range into arbitrary amount of narrow ranges and handle those
1) like every has infinite coherence length
and
2) ranges are independent from each other (no cross correlation).

At least compare to the thickness of multilayer coating the coherence length of any very narrow intervals can be considered as >> then thickness. At every of intervals all layers are involved into interference.
That would be a practical approach delivering your reasonable result.
 
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  • #39
The signal from sunlight seems to be a rather difficult one to model mathematically and predict exactly what an interference filter will do. It would be far simpler to treat a very coherent monochromatic source incident on a multi-layered filter, but in the case of sunlight, the system is basically swamped in noise.

Edit: Perhaps the question we need to answer is, might there be some order to the noise? The best thing we seem to have right now is when @Gleb1964 mentions there is an intrinsic coherence that is present in the individual sources that make up this noise.

Edit 2: It may be worth mentioning that once at my workplace I did measure the transmission spectrum of a Fabry-Perot type platelet that had a width ## d ## of 1.00 mm with a diffraction grating type monochromator=I used an incadescent lamp (thermal source) to do the spectral runs, and I did get the expected spacing in the peaks in the spectral transmission of ## \Delta \lambda=\frac{\lambda^2}{2 nd} ##, so these thermal sources do work with interference type filters, and there seems to be a fairly long intrinsic coherence length. (On the order of millimeters or more rather than microns). (Some more detail: the slits of the monochromator were very narrow to get the necessary resolution, so that I was working with very low signals, and I used an optical chopper along with a lock-in amplifier to process the signal from the photodiode/preamp so that I was able to get a reasonably good signal. If I remember correctly I also collimated the incident light onto the platelet with an off-axis paraboidal mirror to get the maximum amount of interference to occur, but that is a minor detail).
 
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  • #40
Fourier spectrometers are based on Michelson interferometer where the pathlength of arms variated to produce modulation of signal. Getting resolution of 0.5 cm-1 means that the mirror in the Michelson interferometer is moved by ±1cm range. The optical path is variated twice of the mirror displacement, ±2cm.
NIR and MIR Fourier spectrometers are using broadband thermal sources to cover extended spectral range. There is no questions about ability of thermal light to produce interference with a such path difference.
 
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  • #41
I do think the article that @Gleb1964 translated part of it for us in post 28 above has it right, that there is an intrinsic coherence to the various components making up the sunlight, and that it is indeed the case that these components have much longer coherence lengths than one micron=perhaps on the order of several millimeters or more.

We still haven't completely answered the puzzle of whether we need to look simply at an individual layer of an interference filter or whether we need an intrinsic coherence length that is closer to the width of the entire stack, but this question has been interesting to ponder.

It would be nice to hear again from @renormalize on the subject. Meanwhile, I still would like to get some feedback on posts 7, 8, 11, and also 37 and 39. I spent years off and on looking at the Fabry-Perot effect and what seemed to be an unexplained non-linearity to the system before I pieced it all together. See also https://www.physicsforums.com/threads/if-maxwells-equations-are-linear.969743/#post-6159689
 
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  • #42
I believe our fundamental disagreement stems from the distinction between narrowband vs. wideband optical filters. The Fabry-Perot etalon talked about by @Gleb1964 is an optical cavity whose spatial extent (mms to cms) and reflectivity is carefully engineered to employ interference to pass what is (almost) a single specific wavelength, and so requires coherence-lengths on the order of at least mms to cms to function. In other words, an etalon is an ultra-narrowband filter. When illuminated by the incoherent spectrum of sunlight, it selects-out from that spectrum only those particular wave-trains of sufficient length and precise frequency and blocks the rest. As such, the coherence-length of the transmitted radiation says precisely nothing about the source (sunlight), rather it is simply a characteristic of the etalon itself. This is obvious since if we replace the source by a highly-coherent laser of the appropriate wavelength and intensity, the transmitted light is unchanged. Thus, an observer with access only to the output from the etalon cannot distinguish whether its input is the broadband sun or the narrowband laser. Thus, the light emerging from an etalon bears no relation to the average coherence-length of the broad solar spectrum.
In contrast, the OP's question is aimed at the design of wideband thin-film filters intended to pass sunlight over a broad part of its spectrum. Particular examples are antireflection (AR) coatings that act as UV-IR cutoff filters, passing only visible light for photography, eyeglasses and the like, and AR coatings for solar cells to match the transmitted solar radiation to the power-band of those cells, while reflecting undesirable radiation outside that band. Here is a reference describing such a coating for cadmium-telluride cells: Multilayer Broadband Antireflective Coatings.... In section II the authors state:
1736752068449.png

I now examine how the highlighted restriction impacts the design of the coating:
1736752282042.png
1736753417175.png

1736753101008.png
1736753579455.png

From the wavelength range, I can use the relation ##l_{coh}\approx\frac{\lambda^2}{2 \,\Delta\lambda }## given by @Charles Link to estimate the coherence-length to be ##l_{coh}\approx 625^2/2/\left(850-400\right)=434\text{nm}##. Note that ##l_{coh}## falls into the solar longitudinal-coherence range ##170-900\text{nm}## that I quoted in post #17. This coherence length is to be compared with the ##\text{Optical Thickness}\equiv\text{(Refractive Index)}\times\text{(Physical Thickness)}## of each layer, as well as to the entire 4-layer stack:
\begin{matrix}
\text{Material} & \text{Index} & \text{Thick.(nm)} & \text{Opt. Thick.(nm)} & \text{< 434nm?}\\
\hline \text{SiO}_2 & \text{1.45} & \text{94.12} & \text{136.47} & \text{Yes}& \\
\hline \text{ZrO}_2 & \text{2.13} & \text{133.99} & \text{285.20} & \text{Yes}& \\
\hline \text{SiO}_2 & \text{1.45} & \text{30.40} & \text{44.08} & \text{Yes}& \\
\hline \text{ZrO}_2 & \text{2.13} & \text{18.81} & \text{40.07} & \text{Yes}& \\
\hline & & \text{Total OT:} & \text{505.83} & \text{No}
\end{matrix}
Clearly, even though the optical stack-height exceeds the coherence-length ##l_{coh}\,##, this filter functions because the optical thickness of each individual layer is less than ##l_{coh}\,##, completely consistent with the highlighted statement above. Thus, I claim that the declaration:
Gleb1964 said:
That short coherence length ##1 \mu\text{m}## of unfiltered solar light, that have been intensively disputed here, has about nothing [to do] with the performance of the interference filters.
is simply wrong. @Gleb1964 is focused on "apples" (ultra-narrowband interference filters) when he should be examining "oranges" (wideband interference filters).
 
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  • #43
@renormalize
I propose, we just take more simple case, an uncoated 1 mm glass plate and calculate it transmission spectrum for broadband source, Ok? Because that would be more simple case to see any inconstancy in used argumentation.
 
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  • #44
Gleb1964 said:
@renormalize
I propose, we just take more simple case, an uncoated 1 mm glass plate and calculate it transmission spectrum for broadband source, Ok? Because that would be more simple case to see any inconstancy in used argumentation.
I did something similar to this, many years ago=see post 39. The substrate/platelet was partially silvered on both sides. I don't remember whether I used an incadescent lamp (## T \approx 2500 K ##) or a T=1000 C blackbody for the source. There were spectral peaks spaced at ## \Delta \lambda=\frac{\lambda^2}{2 nd} ##. I don't think we have enough information at this point to make a statement that anyone is incorrect in their analysis of this topic, (referencing to the end of post 42).
 
  • #45
Well, let’s conduct an experiment.

I found I have a plastic film with a thickness of 0.17 mm and excellent thickness uniformity. I use a Fourier spectrometer to measure the reflectance spectrum. In the configuration I’m using, the spectrometer measures the sample in reflection mode, but the transmission spectrum can simply be considered additive to the reflection spectrum.
20250113_164026.jpg
20250113_141329-2.jpg


Theoretical peak spacing should be like this:
peaks spacing theory.jpg


Here’s the measured spectrum:
reflectance spectrum 1400-2600nm.jpg

I have discarded data from 700nm to 1400nm because of aliasing (peaks are not resolved).

calculation peaks 1730nm and 2160nm.jpg

We take several peaks in the range of 1730 nm, where the step between peaks is 5 nm, and estimate the interference order to be 338 and the optical path difference between the reflected beams to be 586.5 microns.
Now, we take several peaks in the range of 2160 nm, where the step between peaks is 8 nm, the interference order is 270, and the optical path difference is 585.6 microns.

We account for a small correction due to beam divergence, corresponding to an equivalent angle of 5 degrees. Then, we adjust the refractive index in the range of 1.715 to 1.718 to obtain the measured film thickness of 170 microns.

The light source is a broadband thermal source (type of halogen lamp with reduced voltage). There is no additional filtering applied rather than spectral throughput of the instrument and spectral respond of the detector (InGaAs detector).
 
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  • #46
Charles Link said:
I did something similar to this, many years ago=see post 39. The substrate/platelet was partially silvered on both sides. I don't remember whether I used an incadescent lamp (## T \approx 2500 K ##) or a T=1000 C blackbody for the source. There were spectral peaks spaced at ## \Delta \lambda=\frac{\lambda^2}{2 nd} ##. I don't think we have enough information at this point to make a statement that anyone is incorrect in their analysis of this topic, (referencing to the end of post 42).
Yes, what you describe is a comb filter (https://www.cloudynights.com/articl...-etalons-and-solar-telescope-technology-r1943):
1736792837061.png

But I am unclear as to how this relates to coherence-length and the design of thin-film optical filters intended to have a broad passband. Can you explain?
 
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  • #47
Gleb1964 said:
Well, let’s conduct an experiment.
Thank you for this excellent input!
Can you do a tl;dr of your post to summarize:
  • what this experiment tells you about the coherence-length of your halogen thermal source?
  • how this informs the design of wideband interference filters?
 
  • #48
renormalize said:
But I am unclear as to how this relates to coherence-length and the design of thin-film optical filters intended to have a broad passband. Can you explain?
We are trying to make an assessment as to whether the thermal source that was used in the measurement of this spectrum has a "coherence length" that is sufficiently long to create the interference that occurs to be able to generate interference peaks on a spectral run where the incident light is broadband. Do we get any interference at all, or does the transmitted energy keep its original spectral shape?

If we had a laser source at one of the peaks or valleys, clearly it would be affected, but we do in fact (experimentally) get some interference with the broadband source at a platelet thickness of 1 mm. I didn't have sufficient resolution in my diffraction grating type monochromator even with very narrow slits to be able to tell whether or not the spectral peaks were as prominent as they might have been at a higher resolution, but I did observe significant spectral peaks with the predicted spacing.

This to me would support the idea of an intrinsic coherence length for the individual components that make up the broadband thermal source. That makes it so that when designing an interference filter, regardless of the spectral extent of the passband, that the filter is likely to work and is not restricted by a coherence length number of 1 micron, which is IMO a little misleading.

Edit: The intrinsic coherence will place an upper limit on how thick this interference platelet could be though and still work. It seems to work ok for thermal sources when the thickness is a millimeter or two, but it may not work for a centimeter. With a laser source that has a coherence length of a meter or more, it would still be affected by the comb spectral transmission curve created e.g. by a platelet of thickness greater than one centimeter.
 
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  • #49
@Gleb1964 For post 45, shouldn't the peak spacing be ## \Delta \lambda=\lambda^2/(2nd) ##?

I'm being a little fussy, but ## m \lambda_1=2nd ## and ## (m-1) \lambda_2=2nd ##.
With a little algebra,
## \Delta \lambda=\lambda_1 \lambda_2/(2nd) \approx \lambda^2/(2nd) ##.
(Instead of ## \lambda^2/(2nd + \lambda) ##). (The difference is minimal in any case. It may not affect the numerical result).

Otherwise a very excellent post.

I also once ran (many years ago) a transmission spectrum (with a diffraction grating type spectrometer/monochromator) on a mica platelet that was ## d=.018 ## mm thick and it showed the same type of spacing of the transmission peaks, i.e. ## \Delta \lambda=\lambda^2/(2nd) ##.

Perhaps it is a very side item, but worth mentioning=If the thickness starts getting much greater than about one tenth of a millimeter, you do need to start making sure the source is sufficiently collimated or the interference will wash out from variations in the optical path distance, where the angular dependence of the optical path difference between the incident and doubly reflected beams is ## \Delta=2 nd \cos(\theta_r) ## if I remember correctly. (The thicker platelet has Newton rings that are more closely spaced in angle).
 
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  • #50
## m \lambda = 2nd \cos(\theta) ##
## (m-1)( \lambda+\Delta \lambda) = 2nd \cos(\theta) ##

From here I have:
## m = \frac{( \lambda+\Delta \lambda)}{\Delta \lambda} ##

## \frac{( \lambda+\Delta \lambda)}{\Delta \lambda} \cdot \lambda = 2nd \cos(\theta) ##

Finaly ## \Delta \lambda = \frac{ \lambda^{2}} {2nd \cos(\theta) - \lambda} ##

Which is exactly matching the data.
* Remark - ## \Delta \lambda ## can be "+" or "-" ,depending what wavelength and order you consider to be the main.
But you are right that the difference in very small.
 
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