Transmission through thin (0.5mm) Hoya U-340 and U-360 filters

[Andy Perrin]

I tested my formulas from the other page out on this (with the n and k given in the Hoya datasheet above) and I get perfect agreement to 3 digits:

In case there was any doubts about my formulas!

[Stefano]

The satisfaction when things work!

[Stefano]

Did you write a formula for P by doing T formula/Ti formula and plugging in the r values (obtained with the other formula)?

 

In theory you could write a nasty formula for P, that would not look nice but would be correct.

[Andy Perrin]

It's just the first formula in the other post with Ti on the other side:

 

the d_ref = 2.5mm according to the datasheet, and they give the absorption coefficient k. And yes to plugging in the r values from the other formula.

[Stefano]

Oh it was that simple! I didn't pay too much attention.

[Stefano]

Is k expressed in meters? So that 10^-3 means that 1 mm of glass has a ~36.8% transmission?

 

It doesn't make sense at 500 nm for example, where k is close to 10^-3. I know I am missing something.

[Andy Perrin]

Stefano, I see what it is - also I slightly goofed above but it didn't affect the results.

The k, which they list as absorption coefficient, must actually be the imaginary part of the refractive index instead. In that case, the absorption coefficient is 4πk/lambda = 27000 m^-1 or so at 500nm, so therefore the Ti is roughly 0.

 

If you put Ti=0, the formula still gives correct results here in that case! Basically it was what Ulf was saying upthread about the second bounce not mattering if the absorption is really large. That won't be true at every wavelength, it depends which part of the filter spectrum we are looking at. In the passband it will matter more.

[Stefano]

I have been thinking about the second bounce not mattering if the absorption is high... but I think it does. That's how I understood it: light goes through the first air-glass interface, and we lose a bit (typically 5%). Then it is absorbed to some extent and then, when it goes through the second glass-air interface, we lose a bit again (still ~5% of the light that made it through the first interface). In this case, assuming two bounces only, we have 1-0.95² of reflection losses, which is almost 10%. This is a fixed number. No matter the absorption (Ti), we always have to do *0.95 at the beginning and *0.95 at the end. If we remove an additional 5% from a small number (high absorption) we remove little, but in proportion is still 5%.

 

Did I miss something (again)?

 

I suggest to be extra precise with the high OD parts, like the blue and green bumps, because these regions are very sensitive to errors. If Jonathan's filters are 0.5 mm thick (we discovered that that's not exactly true, but let's imagine for simplicity that they are), by bringing the thickness up to 2 mm we do a fourth power on the transmission, and this amplifies any error.

[Andy Perrin]

If we remove an additional 5% from a small number (high absorption) we remove little, but in proportion is still 5%.

 

Did I miss something (again)?

Yes! In the high absorption region, you don't lose 5% in the middle, it's close to 100% in the middle. Almost none of the light survives to make bounce 2.

[Stefano]

Yes! In the high absorption region, you don't lose 5% in the middle, it's close to 100% in the middle. Almost none of the light survives to make bounce 2.

Yes, I get that almost no light survives to make bounce 2, but let me make a few examples: let's assume a 5% loss at each interface (9.75% total reflection losses). If Ti=1 (or 100%), we have T=90.25% That's 100-9.75, easy. Note (it will be important later) that T=Ti*0.9025.

 

Let's assume Ti=1%. We start with 100%. Light enters the glass, and we have 95%. This gets reduced to 0.95% because of Ti, and then, at the second bounce, we remove 5%, so we have 0.95*0.95=0.9025%. That's still Ti*0.9025. Ti can be 0.001% (OD 5), but still T will be 0.0009025%. The additional 5% removed at the second bounce is very little (in the very last example, it is 0.0000975%, which is nothing compared to the initial 100%), but it is still 5% of the value our "intensity" has before exiting the glass. 5% less means adding 0.0222 at the OD (log(base 10) of 1/0.95), and converting the thickness from 0.5 mm to 2 mm this gets multiplied by four, becoming almost 0.09. It may still not seem a lot, but it weights as much as the 5% of the first reflection.

 

Hope it wasn't too contorted.

[Andy Perrin]

Stefano, 95% of the signal was lost before it got to the other end, then it switches direction (small loss) and loses another 95% going back the other way. So there is not much of a 2nd bounce, or a 3rd or 4th. I'm getting tired of explaining now, so I'll leave you to wrestle with it yourself after this.

[JMC]

More data to share, and I think this finishes off the experiment.

 

Using the reflection factors from the data sheets, I calculated a transmission for 2.5mm thick Hoya U-340 and U-360 based on my measurements and plotted it against the values in the Hoya datasheets for the filters at that thickness.

 

U-340 2.5mm thick comparison

 

U-360 2.5mm thick comparison

 

Overall there is a pretty good match between predicted data and that from the datasheet.

 

While not strictly part of this work, I also did a similar comparison for the mystery filter which is most likely Schott UG11. The datasheet provides information on that at 1mm thickness, so that is what I am showing here.

 

After this, I am more certain of this being UG11.....

 

Ok folks, I hope you are happy with this. It seemed to work out quite well, although I must admit it was more work than I'd thought to get data I was happy with. Overall the black hole in the visible range for the U-340 and U-360 filters has been explored, and as a bonus even managed to compare them with a Schott filter.

 

Anyone who contributed can have copies of the Excel files to do with as you see fit. I guess I can send them through the mail system for the forum (not tried with excel before), if not and you want a copy just let me have your email address if we've not already shared emails. Thank you for your help.

[Stefano]

Thanks Jonathan for your work. Your data does match pretty well with the Hoya one.

 

Just a little question, it may be trivial but I want to be sure: did you use the actual thicknesses of your filters to calculate those curves for 2.5 mm thick filters? As we know they aren't exactly 0.5 mm thick.

[JMC]

Hi Stefano. Yes, I used my measured thicknesses for the calculations.

[Stefano]

Thanks Jonathan

[colinbm]

Thanks Johathan

Have you done diabatic charts Please ?

[JMC]

Colin, as far as I am aware I can't do diabatic graphs in Excel. It can do logarithmic ones though. I'll include those in the Excel files I share.

[colinbm]

Thanks Jonathan

Is Ulf still going to do tests ?

[Stefano]

Anyway I think I understood the reflection thing. The odd reflections (3^rd, 5^th, 7^th and so on) contribute to the output. Summing some terms agrees with the simplified Wiki formula. I didn't understand the working initially.

[JMC]

Yes Colin, I'll be sending the U-340 and U-360 to Ulf for him to have a look at too.

[colinbm]

Thanks John

So the graphs could be altered after Ulf has tested them ?

[JMC]

Colin, I think we'd have to see mine and Ulfs as different experiments (and therefore as different sets of results), as we do not have identical equipment, however I'd expect any differences will be minor. If there are any significant differences, we can work through that if it happens.

[Andy Perrin]

I have previously written diabatic plotting software so if Jonathan sends me the excel files I can easily replot them for us?

[Stefano]

Or can we import the data in the Schott filter calculator program and use its diabatic plot?

 

(I'm not suggesting it is a better idea, it is just an alternative)

[colinbm]

That will be useful Andy, then it gets the data into a format we are used to seeing, thanks.