Effects of coherence length on optical interference filter

[Charles Link]

@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. :)

 

[Gleb1964]

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.

 

[Gleb1964]

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.

 

[renormalize]

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.

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.

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: $\intop_{1/\lambda_{2}}^{1/\lambda_{1}}dv\,B\left(v\right)=1$:

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:

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.

 

[Charles Link]

@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.

 

[Charles Link]

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. :)

 

[Gleb1964]

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 .

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.

 

[renormalize]

It appears for me that our positions are not so different.

Perhaps we should just agree to disagree!
Recall that you 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.

 

[Charles Link]

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.

 

[Charles Link]

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.

 

[Tom.G]

Following may, or may not, be relevant (out of my field ); see if it helps.

For Optical Coherence Tomography (OTC), perhaps take a look at pg. 4 in:

https://www.edmundoptics.com/ViewDocument/OCT_Whitepaper_18.pdf

 

[Charles Link]

@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.

 

[Charles Link]

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/

 

[renormalize]

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:

 

[Charles Link]

@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.

 

[Charles Link]

@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.

 

[Charles Link]

Even though it's been a couple months, I'd like to comment on @renormalize post 64. It may be obvious to some that you can and do get interference from two sinusoidal sources incident on a single interface, I was surprised myself when I discovered about fifteen years ago that that is the case. Yes the transfer matrix works for that, and the case at normal incidence is solved the same way as the Michelson interferometer case with angles of incidence at 45 degrees, but I think it may be worthwhile for instructors to point this out in teaching the subject. (It is really even simpler than using the transfer matrix. You simply employ the Fresnel coefficients).

It was 40+ years ago that I was a teaching assistant for an upper level undergraduate optics course, and we learned all about the mathematical method for multi-layer systems with the Fabry-Perot effect, but it wasn't until about 25 years later that I recognized you do get an interference that conserves energy from two sinusoidal sources incident on a single interface from opposite directions, with the result dependent upon the relative phase of the two sources.