Optimal thickness for 2-layer AR coating

[96Firebird]

Long shot that anyone here knows how to do this, but might as well ask...

Here is what I am doing. Coating a glass substrate with refractive index (n value) of 1.82 @ 1300nm. Outside layer is SiO2, which has an n value of 1.53 @ 1300nm. Inside layer is Ta2O5 which has an n value of 1.78 @ 1300nm. I want to obtain an optimal thickness value for each coating material so that it's minimum reflectance is at 1300nm.

Thanks for any help...

 

[Sho'Nuff]

I believe that each layer has to have an optical path length (sometimes referred to as optical thickness) that is 1/4 of the wavelength of interest.

See page 5-11

http://www.gen-opt.com/lamda_admin/Lamda_Edit/UploadFile/2011222112937198.pdf

See also http://journals.tubitak.gov.tr/physics/issues/fiz-02-26-5/fiz-26-5-3-0108-1.pdf

 

[96Firebird]

I tried that, and it gives me a thickness of 212.4nm for the SiO2, and 182.6nm for the Ta2O5. However, when I use this calculator: http://www.filmetrics.com/reflectance-calculator , and mess with the thicknesses, I am able to achieve a lower reflectance using other thicknesses. For the numbers above, I get a reflectance of .0224. Right now I am down to .01666 for the reflectance at 227nm for SiO2 and 310nm for the Ta2O5. I'm not sure if that calculator is assuming the wrong refractive index values for 1300nm (since you need to put the n value in for 632.8nm for each material).

 

[Daedalus685]

When only working with a single wavelength the value is always going to be a quarter wave at the optimal index.

The materials you've chosen are very close to the optimal index of root(1.83) = 1.35 so you'd get decent transmission at 1300nm with a quarter wave.

I'm assuming you are making a laser window? The 1/4 wave advice will center your transmission on 1300 and give you pretty good results for a laser.

I used to use McLeod for optimizing layer thicknesses when making more complex coatings.

http://www.thinfilmcenter.com/essential.html

If you want to know how to calculate things like transmission % in theory you'll have to draw out your system and figure it out for each boundary.

R = ((n2-n1)/(n2+n1))^2

I can help you calculate that if you want. Its very impractical after two layers and if you want more than a single wavelength it is impractical at 2 by hand .

The numbers you have should be correct for this application. The calculator is likely just not set up for not being centered at visible and who knows what assumptions it is making about the index changing.

Though you're really close to a half wave layer there now... Trying to remember why that would behave like that....

This is why I'd only hand bomb a single layer. The optimizations get tricky. I'll get my head on straight after work and see if I can actually properly think about it. I haven't worked in coating in about a year and I haven't thought about outside of computer optimizations in a long time... Sorry.

 

[silverpig]

Semi-related: are you evaporating your coatings on your substrates?

 

[96Firebird]

Daedalus, thanks for the insight. First question, how did you decide that the materials are a good fit for the optical index of 1.35? I did the calculation somewhere but can't find it now. I have so many hand calculations spread all over the place I've lost track.

As far as the application, I can't really say much about it right now. silverpig, we are not evaporating.

So I must calculate the reflectance at each boundary layer? How do I then calculate the final reflectance value for the whole system?

Basically, the stacks will look like such:

Air (n=1.00)
SiO2 (n=1.53)
Ta2O5 (n=1.78)
Glass (n=1.82)

Optimal thickness values are unknown, but I guess should just be quarter-wave? So for SiO2, t = 325nm/1.53 = 212.4nm, and for Ta2O5, t = 325nm/1.78 = 182.6nm.

So far you guys have been helpful, didn't think I'd get much of a response here. Thanks.

 

[Daedalus685]

Daedalus, thanks for the insight. First question, how did you decide that the materials are a good fit for the optical index of 1.35? I did the calculation somewhere but can't find it now. I have so many hand calculations spread all over the place I've lost track.

For single layers the optimum index is the root of the product of the index of the medium and the substrate. In air this is just root(nglass) In Hecht optics the derivation is on page 428 (4th edition). It is the reduction of the characteristic matrix for the n that results in zero reflectance.

For double layers you want (n2/n1)^2 to be the index of the substrate while n1 and n2 also being as small as possible (such that if n1 =1 it reduces to teh single layer expression above). I'm sorry to say that my brain fart earlier means that i was thinking 1.35 instead of 1.83 so you are not that ideal .

As far as the application, I can't really say much about it right now. silverpig, we are not evaporating.

Applications are more fun if secret

So I must calculate the reflectance at each boundary layer? How do I then calculate the final reflectance value for the whole system?

Basically, the stacks will look like such:

Air (n=1.00)
SiO2 (n=1.53)
Ta2O5 (n=1.78)
Glass (n=1.82)

Optimal thickness values are unknown, but I guess should just be quarter-wave? So for SiO2, t = 325nm/1.53 = 212.4nm, and for Ta2O5, t = 325nm/1.78 = 182.6nm.

So far you guys have been helpful, didn't think I'd get much of a response here. Thanks.

You have to assume the quarter wave to make it possible to compute by hand (possible is not the right word but I'm not about to type the equation for even a single layer). Thus you get perfect destructive interference and the total reflectance is just the remainder at the end.

For two layer situations you can get more complicated interactions and thus have less reflection in the situations where reality keeps your index from ideal. I can't tell you how to do it by hand as it requires very complicated matrix math just to compute a single value, let alone optimize it, that I haven't done by hand in 8 years if ever. For that you need an optimization program like McLeod. For almost every situation a quarter wave or a quarter wave double are good enough for single wave transmission if you have a close enough index. You aren't going to do anything by hand unless you use quarter wave or hate your life entirely

For a single wave coating there is no reason to use more than one layer unless you cant find the right materials to get the correct n. In your case a single layer would need n=1.35 which is remarkably close to mag fluoride. MgF2 is tricky to coat on thicker than a few hundred nm though. Adding more layers is to broaden the minimum and allow the use of more common materials. A perfect index quarter wave will always be by definition perfect transmission at the desired wave.

 

[Daedalus685]

Something to consider when you are playing with the reflectance calculator online (which I have to say is very cool, it doesn't optimize but even being able to plot that for free is amazing even if it isn't likely perfectly accurate for your application).

Notice that with your materials and a quarter wave layer each you actually get a worse minimum than with only a single layer of silicone dioxide. What you gain is that the minimum is far broader. But for only one wavelength this benefit is moot.

I don't know your application (if you must use those materials or notm or if you need a wider pass) but plug in a single layer of MgF2 @ 240nm thick. Very easy to coat that kind of layer and the performance around 1300nm will be unbeatable on SF5 or SF6 glasses. Also MgF2 is very inexpensive and can be similar durability to SiO2 if done at decent temperature. Just something to consider.

 

[96Firebird]

Daedalus, thanks for the help. Unfortunately, we cannot coat mag fluoride. But I agree, it is a great match for the glass.

Another day, more changes. Now we are using a different glass, this time it has a refractive index of 1.98 @ 1100nm. We did some coatings of SiO2 and realized the coating actually has a lower refractive index than handbook values, which helps us out here. Our measured refractive index for the SiO2 is 1.43, and sqrt(1.98) = 1.40. So we are pretty close. We are going to try coating 227nm of SiO2. Found the thickness by using quarter-wave (1300nm/4 = 325nm) divided by the refractive index of our SiO2 (1.43).

Our current spectrometer maxes out at ~850nm, so we will be using the minimum reflectance at around 435nm to check the sample. Hope we can get some good results.

 

[Daedalus685]

Daedalus, thanks for the help. Unfortunately, we cannot coat mag fluoride. But I agree, it is a great match for the glass.

Another day, more changes. Now we are using a different glass, this time it has a refractive index of 1.98 @ 1100nm. We did some coatings of SiO2 and realized the coating actually has a lower refractive index than handbook values, which helps us out here. Our measured refractive index for the SiO2 is 1.43, and sqrt(1.98) = 1.40. So we are pretty close. We are going to try coating 227nm of SiO2. Found the thickness by using quarter-wave (1300nm/4 = 325nm) divided by the refractive index of our SiO2 (1.43).

Our current spectrometer maxes out at ~850nm, so we will be using the minimum reflectance at around 435nm to check the sample. Hope we can get some good results.

Depending on your deposition rates and gas levels the SiO2 will have quite a wide variety of physical and optical properties. Pure SiO2 crystal will have very close to the book value. However, if you use plasma assistance or don't have enough O2 gas you will end up with a layer that is of varying density with contamination of other Si oxides (We used to coat with SiO regularly for very hard IR on SI windows). So don't expect perfection . Unless you are using a multi watt laser the tenths of percent difference wont really matter for single band. Measuring near IR is a pain. Commercial FTIR spectrophotometers don't go that short and visible designs don't go that long .

Sounds like you're well on the way!