Effects of coherence length on optical interference filter

[avlok]

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?

 

[Charles Link]

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.

 

[Gleb1964]

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.

 

[Charles Link]

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).

 

[Gleb1964]

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.

 

[Charles Link]

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

 

[renormalize]

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:

I now examine how the highlighted restriction impacts the design of the coating:

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:

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).

 

[Gleb1964]

@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.

 

[Charles Link]

@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).

 

[Gleb1964]

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.

Theoretical peak spacing should be like this:

Here’s the measured spectrum:

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

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).

 

[renormalize]

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):

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?

 

[renormalize]

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?

 

[Charles Link]

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.

 

[Charles Link]

@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).

 

[Gleb1964]

$m \lambda = 2nd \cos(\theta)$
$(m-1)( \lambda+\Delta \lambda) = 2nd \cos(\theta)$

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.

 

[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.