Effects of coherence length on optical interference filter

  • #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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  • #51
@Gleb1964 The term is very insignificant, but if you go the other way and write ## (m+1 )(\lambda-\Delta \lambda)=2nd \cos{\theta} ##, you then get ## \Delta \lambda=\lambda^2/(2nd \cos{\theta}+\lambda) ##.

My suggestion is to leave off the ## \pm \lambda ## term in the denominator. It simplifies matters, although it may not appreciably affect any numerical result, but I think it will then agree with what you might find in the literature.

and I saw your edit above. Very good. :)
 
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  • #52
The difference of ## \pm \lambda ## term is getting very insignificant at large m, but it may be significant at low order. Otherwise the formula with ## \pm \lambda ## is exact and approximal without ## \pm \lambda ## term.
 
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  • #53
renormalize said:
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?
The experiment is a demonstration of "a very thick coating" where the path length between interfered beams are many times exceed formal "coherence length" of unfiltered thermal source. Halogen lamp used in experiment can be approximated by a Black Body radiation with the temperature about 2600K.

The white light is consist of many independent emission acts, so called wave trains.
Every wave train can only interfere with itself and they do not interfere with each other. If the path shift exceed the length of the wave train, it is not capable to produce an interference.

We are meaning the coherence as ability to produce stationary interference picture on detector (screen), where detector is performing unilinear respond to the squared amplitude of electromagnetic wave. Strictly saying interference always happened only at detector (or screen), not at interference layers of coating, but sometimes we can speak about interference at layer, meaning that result interference would be at detector (!).

The coherence length of independent wave trains are much-much longer compare to the coherence length of combined white light. Would be wave trains infinity long they would be absolutely monochromatic and would not be able to transmit any energy. Because the wave trains are limited in length they have finit bandwidth.

How long can be wave trains for unfiltered sunlight is disputable, I don't have a solid digit, but I know it is in an order of tens of cm and may be up to meters going into NIR range.

Calculation of broadband coating can be done by splitting the spectral range in many wavelength and calculating them separately, handling any of wavelength as with "infinity long coherence" compare to the coating thickness.
 
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  • #54
Summary: Starting from the solar visible-spectrum characterized by a coherence-length ##l_\text{coh}<1\mu\mathrm{m}##, I model sunlight transmission through a glass window of thickness ##\tau##, focusing on the total spectrum-integrated power ##P_T## delivered to a solar cell behind the glass. I find that ##P_T## depends strongly on ##\tau\,##: for ##\tau\lesssim l_\text{coh}## interference effects are pronounced and ##\tau## can be chosen for nearly perfect power delivery; in contrast, for ##\tau\gtrsim l_\text{coh}##, interference is negligible and ##P_T## approaches a constant smaller value as ##\tau\rightarrow\infty##. This suggests that ##\tau\sim l_\text{coh}## marks the onset of useful interference in wideband layered filters.
Gleb1964 said:
How long can be wave trains for unfiltered sunlight is disputable, I don't have a solid digit, but I know it is in an order of tens of cm and may be up to meters going into NIR range.
Calculation of broadband coating can be done by splitting the spectral range in many wavelength and calculating them separately, handling any of wavelength as with "infinity long coherence" compare to the coating thickness.
Charles Link said:
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.
Although true, imo what the above statements overlook is the necessity of engineering a wideband interference filter to maximize its effectiveness over the full spectrum. This is where I think the broad-spectrum, short coherence-length ##l_\text{coh}## comes into play. To illustrate this, and inspired by @Gleb1964's single-layer film measurements in post #45, below I consider a free-standing glass slab of high-refractive-index ##n##, intended to serve as an efficient window to pass the complete visible-spectrum to a solar cell. (I analyze such windows down to an impractically-thin ##1\text{nm}\,##, so this is really a learning exercise aimed at highlighting the significance of ##l_\text{coh}##).
Let's start by estimating the full-spectrum longitudinal coherence-length. Taking the wavelength-range of visible light to be ##380\mathrm{\mathrm{n}m}\equiv\lambda_{1}\leq\lambda\leq\lambda_{2}\equiv780\mathrm{\mathrm{n}m}##, its easy to see that ##\lambda_{2}-\lambda_{1}\equiv\Delta\lambda\sim\lambda\,## implying that the small-difference approximation breaks down. Therefore, here I'll calculate ##l_\text{coh}## from a formula that's more appropriate to a wideband spectrum (http://electron6.phys.utk.edu/optics421/modules/m5/coherence.htm) :$$l_{coh}\equiv\frac{\lambda_{1}\lambda_{2}}{2\pi\left(\lambda_{2}-\lambda_{1}\right)}=118\mathrm{nm}\tag{1}$$The window is illuminated by atmospheric-filtered sunlight as it reaches ground level, which I approximate as ##T=6500°\text{K}## black-body radiation ##B##:
$$B\left(v\right)=\frac{Nv^{3}}{\exp\left(\frac{hcv}{kT}\right)-1}\tag{2}$$where ##v\equiv1/\lambda## is the radiation wavenumber and ##N## is a normalization constant chosen to make the area under the spectral-curve unity: ##\intop_{1/\lambda_{2}}^{1/\lambda_{1}}dv\,B\left(v\right)=1##:
1737000960101.png

Next I write down the expression for the power transmission-coefficient ##T## of the glass slab:$$T\left(\tau,n,v\right)=\frac{8n^{2}}{n^{4}+6n^{2}+1-\left(n^{2}-1\right)^{2}\cos\left(4\pi\,n\,\tau\,v\right)}\tag{3}$$This comes from solving JD Jackson, Classical Electrodynamics, problem 7.2 (complete solutions can easily be found online). Finally, I multiply (3) by (2) and integrate over ##v## to find the total power ##P_T## across the spectrum that is transmitted through the window:$$P_{T}\left(\tau,n\right)=\intop_{1/\lambda_{2}}^{1/\lambda_{1}}dv\,B\left(v\right)\,T\left(\tau,n,v\right)\tag{4}$$Choosing ##n=1.70## and repeatedly evaluating integral (4) numerically for increasing thickness ##\tau## yields the resulting graph:
1737005933801.png

What this tells me is that the total range of possible slab thicknesses ##\tau## is divided roughly at ##\tau\sim l_{coh}## into two subranges that can be summarized by:$$
P_{T}\left(\tau\right)=\begin{cases}
\sim100\% & \tau\lesssim l_{\text{coh}}\text{ (high-inteference subrange)}\\
\sim P_{T}\left(\infty\right) & \tau\gtrsim l_{\text{coh}}\text{ (low-inteference subrange)}
\end{cases}
\tag{5}$$I do acknowledge what you've both been emphasizing: one can choose any specific wavelength from a bright, incoherent source, like the sun or an incandescent bulb, and always find coherent wave-trains of more than sufficient length to get through any etalon. But based on the above analysis, I fail to see how this fact has any bearing on the design and performance of wideband multilayer interference filters intended to deliver maximum intensity across the band. Instead, I think that the optimal configuration for such a filter is achieved by striving to keep the optical thickness of each layer less than the broad-spectrum coherence-length; i.e., well under a micron. That's why I continue to maintain that the submicron-scale coherence-length of sunlight is quite relevant for the synthesis and operation of wideband thin-film filters.
 
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  • #55
@renormalize Excellent. You showed it conclusively. Perhaps what @Gleb1964 and I have been stuck on is that you can in fact generate a comb filter with a very thick platelet, but this comb filter, as you have shown, would have no practical use. It would be incapable of blocking or passing (i.e. generating interference that doesn't average out over) a wide ## \Delta \lambda ##. [Edit: This seems to be in hindsight what this coherence length number of one micron for sunlight is all about. It isn't trying to put a limit on how coherent the individual components that make up the light source are.] [Edit 2 and I agree with your expressions for B an T ].

The other item that we might still need to answer is whether a stack of layers will work where the thickness of the stack greatly exceeds the coherence length, provided the thickness of each layer is less than the coherence length. I think the answer to that is yes. Edit: and I see @renormalize in post 10 above is in agreement with this.[Edit: I do see post 42 that doesn't seem to be real clear on this, and may be in disagreement=the highlighted "provided that the thickness is less than the coherence length"].

and reading the OP again, they ask how does it work if the stack is much thicker than the coherence length? We seem to be in agreement on this that it does work, even though the thickness of the stack greatly exceeds the coherence length. There doesn't seem to be any major restriction here.

Edit: I need to add something to the above. The H alpha type etalon, mentioned by @Gleb1964 , post 12 above , even though it is a comb filter blocks perhaps 95 % or more of the broadband sunlight. So this comb filter does have a practical use, but it does not transmit a broad passband and block a broad passband, which is what I think the OP is trying to do.
 
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  • #56
a follow-on to the above: @renormalize I have mentioned it in post 7 and also in posts 37 and 41, but now that we have read your research on this topic, I'm hoping you can take the time to read about something I came up with about 15 years ago on this topic. To give a little history of this, as a graduate student in college, I was the teaching assistant for the upper level undergraduate Optics course for a couple semesters. We learned all about the Fabry-Perot effect and the method for treating the multi-layer films, which is not by trying to consider multiple reflections, but rather by having a left-going and a right-going wave on both sides of each dielectric interface. The solution for the several equations and several unknowns is straightforward, but it seemed to be missing something which I couldn't put my finger on. I encountered a similar thing a number of years later at the workplace with reflections from r-f cable connections due to slight impedance mismatches in the cables, again the Fabry-Perot effect was occurring, but again, something seemed to be missing. I finally spotted the missing piece in 2008-2009 when I worked through the problem of the interference of two sinusoidal sources (having some arbitrary phase ## \phi ## between them) incident from opposite directions onto a single interface, using the Fresnel coefficients to calculate the result. I was very surprised to find how consistent the results were, and that they conserved energy, and were really the same kind of interference that occurs in the Michelson interferometer.

Please see
https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/
and
https://www.physicsforums.com/threads/if-maxwells-equations-are-linear.969743/#post-6159689

I would enjoy your feedback on these two write-ups that I think are relevant to this topic. You have enough background in this area that I think you might find them fairly easy reading. Thanks. :)
 
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  • #57
renormalize said:
Next I write down the expression for the power transmission-coefficient ##T## of the glass slab:$$T\left(\tau,n,v\right)=\frac{8n^{2}}{n^{4}+6n^{2}+1-\left(n^{2}-1\right)^{2}\cos\left(4\pi\,n\,\tau\,v\right)}\tag{3}$$This comes from solving JD Jackson, Classical Electrodynamics, problem 7.2 (complete solutions can easily be found online).
Thank for your excellent and educative post!

It appears for me that our positions are not so different.
Expression (3) accounted for infinity orders of multibeam interference in a slab and doesn't limit a slab thickness. Efficiently, (3) should assume that at any monochrome wavelength/frequency any order of interference is perfectly coherent .

parallel glass slab multibeam interference.jpg


Accounting for high order reflections in a glass slab is a similar approach as accounting for multiple layers of a multilayer coating with a similar assumption that all reflections are coherent. Not accounting for the coherent reflections would produce significantly different result.
 
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  • #58
Gleb1964 said:
It appears for me that our positions are not so different.
Perhaps we should just agree to disagree!
Recall that you said:
Gleb1964 said:
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.
I still assert that this statement is wrong for wideband interference filters. Based on my post #54, the coherence-length of sunlight, ##l_\text{coh}<1 \mu\text{m}##, is indeed relevant to a broad-spectrum filter since ##l_\text{coh}## is (approximately) the maximum layer-thickness possible if you seek to maximize power transmission. If ##l_\text{coh}## were irrelevant, you could just as well construct the ##\text{SiO}_2/\text{ZrO}_2/\text{SiO}_2/\text{ZrO}_2## stack-filter of post #42 out of layers that are millimeters thick, rather than ##\sim 100## nanometers, without sacrificing performance.
 
  • #59
It may be worth mentioning that in the case of a Fabry-Perot slab where there are two interfaces, the two methods of solution, the infinite (geometric series) vs. the algebraic with waves traveling to the left and right at the interfaces, are directly analogous to solving an electronic feedback amplifier by having the feedback loop create an infinite (geometric) series of terms at the input vs. solving for the output algebraically with a fraction of the output (even with a phase term) being added to the input.

First method, infinite series: ## E_{1r}=E_{incident} \tau_1 (1+e^{2i \phi} \rho^2 + e^{4 i \phi} \rho^4+...)=E_{incident} \tau_1/(1-e^{2 i \phi} \rho^2) ##.

##E_{2r}=E_{1r} e^{i \phi} ##

(## E_{1r} ## and ## E_{2r} ## are both inside the material)

(## E_{2r} ## is analogous to ## V_{out} ## and ## E_{incident} \tau_1 ## is analogous to ## V_{in} ##.
## e^{i \phi} ## corresponds to the open loop gain ## a_v ##, and ## \rho^2 e^{i \phi } ## is the feedback coefficient ## \beta ##. The mathematics is very analogous for the electronics and optics cases. For the feedback amplifier we have ## V_{out}=V_{in} \frac{a_v}{1-a_v \beta} ##).

## E_{transmitted}=E_{2r} \tau_{21} ##

Now for the algebraic method:
## E_{1r}=E_{incident} \tau_1+\rho E_{1l}=E_{incident} \tau_1+\rho E_{2l} e^{i \phi} =E_{incident} \tau_1+\rho (\rho E_{1r} e^{i \phi})e^{i \phi} ##
so that
## E_{1r}(1-\rho^2 e^{2 i \phi})=E_{incident} \tau_1 ##, in agreement with the infinite series method.

It may be worth contemplating which method best describes the actual physics that occurs.
 
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  • #60
One or two additional comments about post 59 and the single slab two interface case: it should be noted that the interference occurs exclusively at the first interface. There we have two sources incident on the interface from opposite directions. The interference dos not occur at the second interface where we can even do the computation with the energy reflection coefficient ## R ## which works when there is a single source incident on the interface. [Edit: It should be noted the energy reflection coefficient will not tell you whether or not there is a ## \pi ## phase change upon reflection]. It is at the first interface that we need to use the Fresnel coefficients because we have sources incident on the interface from opposite directions, and thereby the ## R ## is no longer a good number. See also the links in post 56 above.
 
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  • #62
@Tom.G Interesting, but it doesn't cover the Fabry-Perot interference. It was good to get some feedback from someone in any case. Right now it seems we must have a very limited number of viewers who are reading through the posts in any detail.

Edit: I do think students should find this thread as a very good way to learn all about the Fabry-Perot effect. I'm really surprised that we don't seem to be hearing from any students.
 
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  • #63
I want to take a couple of minutes one more time to show IMO what the basis of the Fabry-Perot effect is all about. This is very relevant to any calculations you would do with it involving thin films.

If you take a single interface with air on one side and a material of index ## n ## on the other side such that you get a 50-50 split in energy for any electromagnetic wave incident on it at normal incidence where one half the energy is transmitted and one half reflected, you really get an interesting case when you consider two waves incident on this interface from opposite directions. They interfere with each other and you no longer get a 50-50 split. You need the Fresnel coefficients to calculate the result because they remain as good numbers even though the energy reflection coefficient ## R ## is no longer a good number when two waves are present.

It just takes a little algebra to show that if the two waves are in phase with each other at the interface, all of the emerging energy will go to the right, because there is a ## \pi ## phase change for the reflected E field that is incident on the higher index material, causing destructive interference with the transmitted wave that comes from the right.

If there is a ## \pi ## phase difference at the interface between the two incident waves, then all the emerging energy goes to the left.

I think this may not be the way this is usually presented in the textbooks, but it really isn't at all in the category of personal theory. I'm just applying the well-established Fresnel coefficients to the simple case of two sources incident from opposite directions onto a single interface. I urge you to try it for yourselves. So far, I've received almost zero feedback on this concept, but I think trying the calculation for yourselves may well be worth the while. To me, it really explains what is going on with the Fabry-Perot effect.

See https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/
 
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  • #64
Charles Link said:
I'm just applying the well-established Fresnel coefficients to the simple case of two sources incident from opposite directions onto a single interface. I urge you to try it for yourselves. So far, I've received almost zero feedback on this concept, but I think trying the calculation for yourselves may well be worth the while. To me, it really explains what is going on with the Fabry-Perot effect.
@Charles Link , I've refrained from commenting on your Insight article because I don't see the novelty of your approach. Analyzing two oppositely-directed waves at single dielectric interfaces is the "bog standard" transfer-matrix method that's used to design stack filters. Here's a random example from online:
Temperature dependence of refractive indices...
An excerpt from that article:
1737829490063.png

1737829569971.png
 
  • #65
@renormalize I learned the above method many years ago in graduate school. It was only in the last several years though that I recognized it works even for a single interface.

If you look at it with two or more interfaces, it looks like a system of multiple reflections with phase differences, etc. Perhaps it is obvious to many that it works as well with a single interface. It wasn't obvious to me. For a single interface, you need to introduce a second source, and I really was surprised to find the energy is indeed conserved with the result that depends on the relative phases of the two sources.

Edit: Looking it over more carefully, in graduate school I did not learn the matrix method, which when I computed a couple terms just now, it takes a little extra algebra. (We treated the multi-layer system though with a left-going wave and a right going wave on each side of every interface). One can write e.g. ## E_{2r}=\rho_{12}E_{2l}+\tau_{12}E_{1r} ##, where with the matrix method I found some Fresnel coefficients wind up in the denominator for the expressions. The matrix method is a very mechanical algebraic process, and it really doesn't provide for much insight into the underlying physics.
 
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  • #66
@renormalize I looked over the write-up above somewhat carefully from post 64 and I believe (4) is in error.
In nomenclature you should be able to follow, I get the following:

## E_{1r}=E_{2r}/\tau_{12}-\rho_{12} E_{2l}/\tau_{12} ## and

##E_{1l}=\rho_{21}E_{2r}/\tau_{12}+(\rho_{21}^2/\tau_{12}+\tau_{12})E_{2l} ##.

This gives ## (M_{21}/M_{11})^2=\rho_{21}^2 ##, but that is not the energy reflection coefficient of the system. The energy reflection coefficient of the system is ## R=(E_{1l}/E_{1r})^2 ##.

The expression ## R=\rho_{21}^2 ## only holds for a single source incident on an isolated interface. The person who wrote this up seems to have blundered with their formula (4) above. I'm not infallible, but on this one I'm pretty sure I am right, and the "book" is wrong.[Edit: See below=the book did get it right]. Even though the Fresnel coefficient ## \rho_{21} ##, (which has ## R=\rho_{21}^2 ## when there is a single source incident on the isolated interface), remains a good number for multiple sources and multiple layers, this ## R ## is no longer a good number, but instead needs to be determined by what the whole system is doing.

Edit: I looked it over some more and wondered if perhaps I made a mistake, because their ## M ## is for the whole system. and I think I must retract the above, because I see how they get (4). Since the final interface has no left going wave on the right side, they set the right going wave to some arbitrary number, and the left-going wave to zero. They then have ## R=(E_{1l}/E_{1r})^2=(M_{21}/M_{11})^2 ##. My mistake.
 
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