• Ultraviolet Photography
  •  

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

Filters
80 replies to this topic

#1 JMC

    Member

  • Members+G
  • 1,094 posts
  • Location: London, UK

Posted 12 September 2020 - 13:16

A few weeks ago, I asked for assistance in trying to answer the question of what is the filter transmission of Hoya U-340 and U-360 in the part of the spectrum where the Hoya datasheets don't provide any data (original post here - https://www.ultravio...dpost__p__38106). Many of you kindly agreed to help out and contribute to the buying of custom made 2 filters - Hoya U-340 and U-360, both at 0.5mm thick, so I could try and answer that question.

The filters have arrived, and I have started to do some measurements on them. In this thread I will update my progress. I'll provide information on what I am doing and the method I'm using to measure transmission. I'll provide some graphs at relatively low resolution, so you can see what I am up to. Anyone who has contributed can have copies of any raw data I generate and the Excel files the graphs come from.

I also welcome advice and comment from you all regarding my approach. My end goal is to provide you with a set of data which can fill the gaps in the transmission data that Hoya provide. I hope to do this over the next few weeks.

Thank you for your help in making this happen.
Jonathan M. Crowther

http://jmcscientificconsulting.com

#2 colinbm

    Member

  • Members+G
  • 1,988 posts
  • Location: Australia

Posted 12 September 2020 - 13:43

Good news Jonathan
Did the supplier provide their own graph too ?

#3 JMC

    Member

  • Members+G
  • 1,094 posts
  • Location: London, UK

Posted 12 September 2020 - 14:02

View Postcolinbm, on 12 September 2020 - 13:43, said:

Good news Jonathan
Did the supplier provide their own graph too ?
Well yes, sort of. I asked for a spectra for each one, and got a printout only, and only of the region around the main transmission peak. In hindsight I should have been more specific.

So it won't be much use, apart from allowing me to check whether mine and theirs match up in that part of the spectrum.
Jonathan M. Crowther

http://jmcscientificconsulting.com

#4 colinbm

    Member

  • Members+G
  • 1,988 posts
  • Location: Australia

Posted 12 September 2020 - 14:06

Thanks Jonathan

#5 Stefano

    Member

  • Members(+)
  • 971 posts
  • Location: Italy

Posted 12 September 2020 - 15:40

I can't wait to see the results. I always wondered what the transmission is in those regions, especially compared to Schott glass. Thank you for finally giving us an answer.

#6 JMC

    Member

  • Members+G
  • 1,094 posts
  • Location: London, UK

Posted 12 September 2020 - 15:54

Test 1 - simple transmission measurement

First assessment of the filters. This is simple transmission measurement. Light source is an Ocean Optics DH-2000-BAL using both lamps (deuterium and halogen), left to warm up for 30mins before any measurements. A 600 micron extreme solarisation fiber then couples this to an Ocean Optics 74-UV collimating lens at 90 degrees to the test filter surface. Below the filter is another 74-UV collimator at 90 degrees to the filter. Then another 600 micron fiber leading to a Ocean Optics FX spectrometer, also left for 30mins between switching on and measurements. Measurements and calibration was done in a darkened room (although there was ambient light from laptop screen). Data captured between 250nm and 800nm.

The Transmission measurement was set up in the software with 2000 scans of 1.4ms and a Boxcar width of 1. After setup the filter was run and a baseline scan was also run straight afterwards. Recalibration and setup of the experiment was done before each filter.

First the scans shown on a linear scale and including the peaks.
Attached Image: 2000scans both filters and baseline.jpg

Now, the scans concentrating on the low % transmission region.
Attached Image: 2000scans low trans area.jpg

And finally the data shown using a Log scale.
Attached Image: 2000scans log scale.jpg

Few things to note;

These are low res images just for update purposes. Any I share with those who contributed will be original Excel files and graphs. This was just a quick test to have a look at them, so do not take this as being the final definitive data - more work will be done.

There is transmission in the 450nm to 600nm region for both filters, but even at 0.5mm thickness the degree of transmission is low. The shape of these peaks is similar between the 2 filters, but not identical.

Baseline after the U-360 looked good, but the one done after the U-340 scan was a little high.

I seem to have a slightly dodgy pixel in my sensor at around 553nm, as I consistently get a dip there.

My spectrometer is a solid state one, and like all of this design does suffer a bit from stray light effects. This manifests itself as an increase in readings to the left of areas which have high transmission. As both of these filters have a high IR transmission, I expect the readings for the filters in the 400-700 region to be slightly effected by this increasing the values shown here. I have a way around this I think - using a UV/IR cut filter in the path of the light source when I do the 100% transmission calibration, and I will be trying this.
Jonathan M. Crowther

http://jmcscientificconsulting.com

#7 Stefano

    Member

  • Members(+)
  • 971 posts
  • Location: Italy

Posted 12 September 2020 - 16:52

I know these are only initial tests, but I would have expected more transmission in the green for Hoya U-340, and less for U-360. Hoya U-340 seems to peak at 0.25% in the green, and that becomes only 0.000625% at 1 mm (OD 5.2). That's a comparable OD to the infrared one in a well-filtered stack. Of course the peak (minimun) OD matters up to a certain point, what matters more is the area under the graph, followed by the camera sensitivity and so on. You can force leaks in a OD 5+ filter, but I doubt they can affect a UV image. If the data is accurate (keeping in mind that this is just an initial test) I would say 1 mm thick U-340 is safe to be used in a UV stack. Anything above OD 5, even with a large bandwith (large area under the graph, as mentioned before) should not be visible in normal circumstances.

#8 JMC

    Member

  • Members+G
  • 1,094 posts
  • Location: London, UK

Posted 12 September 2020 - 17:56

Well Stefano, that's the thing about testing something that's not been done before - they don't always give you what you expect. That's the joy of science :grin:

Something to be careful of though is assuming 1mm is the same as 2x 0.5mm filters (and I think 0.25% for 0.5mm would become 0.00000625% for 1mm based on that assumption). You need to take into account reflections from the surface(s) - I'm measuring filter transmission, not internal transmittance. Needless to say we are looking at extremely small numbers here for any filter with a meaningful thickness.

All,

This is somewhere I need a bit of reality check. Looking at the Hoya datasheets I got the following for Reflection factors;
U-340 - reflection factor = 0.907
U-360 - reflection factor = 0.883

Question is for a given filter, does this need to be applied twice - once for each air/filter interface the light would experience when passing through the filter?
Jonathan M. Crowther

http://jmcscientificconsulting.com

#9 Stefano

    Member

  • Members(+)
  • 971 posts
  • Location: Italy

Posted 12 September 2020 - 18:10

View PostJMC, on 12 September 2020 - 17:56, said:

I think 0.25% for 0.5mm would become 0.00000625% for 1mm based on that assumption
0.25% is 0.0025 on a scale from 0 to 1 (1/400). Squaring it gives you 0.00000625 (1/160,000), but converted in percentage becomes 0.000625%. Anyway it is not that important, what matters is that those numbers are quite small.

#10 Stefano

    Member

  • Members(+)
  • 971 posts
  • Location: Italy

Posted 12 September 2020 - 18:15

View PostJMC, on 12 September 2020 - 17:56, said:

Question is for a given filter, does this need to be applied twice - once for each air/filter interface the light would experience when passing through the filter?
You have to apply it once converting Ti to T (and viceversa). Each air-glass interface reflects 4-5% back (there is also a formula to calculate this).

Edited by Stefano, 12 September 2020 - 18:18.


#11 JMC

    Member

  • Members+G
  • 1,094 posts
  • Location: London, UK

Posted 12 September 2020 - 18:27

Stefano, yep you're absolutely right. I forgot to apply the final x100 step when I did the math to get back to a percentage. As you say though, extremely small numbers.

Ok, thanks about the Ti to T.
Jonathan M. Crowther

http://jmcscientificconsulting.com

#12 dabateman

    Da Bateman

  • Members+G
  • 2,042 posts
  • Location: Maryland

Posted 12 September 2020 - 22:15

Jonathan,
Thank you for taking this on and starting the study. I would also, if possible be interested for you to compare with your 0.5mm ug11 filter. I know you bought it as ug5, were told it was ug5. But I am 99.999% certain its ug11.


#13 Andy Perrin

    Member

  • Members+G
  • 3,341 posts
  • Location: United States

Posted 13 September 2020 - 00:45

Jonathan- regarding the reflection factor, there is an exact formula for it in terms of the individual reflection coefficient of each surface. And you can calculate the surface reflection coefficients from refractive index. Which is a slight function of wavelength (you can fit a line to the values given in the data sheet). If you want I can walk you through how to extract the Ti based on the measured T and the above description?

#14 Stefano

    Member

  • Members(+)
  • 971 posts
  • Location: Italy

Posted 13 September 2020 - 01:16

That's the formula I was talking about above: https://www.google.c...d=1599959498674

I hope the link works as I want. If it doesn't direct you there, go to "Theory" and then "Reflection".

Edit: I think it should work. Andy, are you talking about this using the refractive indexes or already the coefficients?

Edited by Stefano, 13 September 2020 - 01:23.


#15 Andy Perrin

    Member

  • Members+G
  • 3,341 posts
  • Location: United States

Posted 13 September 2020 - 01:47

I’m talking about calculating the single-reflection coefficients (not to be confused with the OVERALL reflection coefficient given by Hoya) based on the refractive indexes, which themselves need to be calculated as functions of wavelength. Then the single reflection coefficients get used along with the absorption to calculate the overall reflection coefficient accounting for ALL the partial reflections simultaneously.

Stefano if you are interested in the details google “transfer matrix method” and look for the papers by Katsidis and Siapkas in Applied Optics 2002

Edited by Andy Perrin, 13 September 2020 - 01:57.


#16 Stefano

    Member

  • Members(+)
  • 971 posts
  • Location: Italy

Posted 13 September 2020 - 02:43

So you will calculate the refractive indexes of the glass for every wavelength (like every nanometer), and use them to precisely calculate reflection losses using the formula I linked to obtain the internal transmission?

I found the paper you cited. I don't understand why matrixes are involved, probably I will study these things at university.

#17 Andy Perrin

    Member

  • Members+G
  • 3,341 posts
  • Location: United States

Posted 13 September 2020 - 04:14

Quote

So you will calculate the refractive indexes of the glass for every wavelength (like every nanometer), and use them to precisely calculate reflection losses using the formula I linked to obtain the internal transmission?
Yup that's the general idea.

Quote

I don't understand why matrixes are involved, probably I will study these things at university.

Stefano, the matrices are because you decompose the phenomenon into three steps, 1) reflections off the first surface, 2) transmission through the interior of whatever made it inside, 3) reflections/transmission off the rear surface, BUT all three of these are coupled together, so they need to be solved simulaneously. The matrix handles the coupling for you in a convenient way. You could also do at as a series (single reflections + double reflections and absorption + triple reflections and absorption + ...) , but the matrix method does the equivalent of summing the resulting power series without even needing to write it out. The wiki formula, T=2R/(1+R) is a simplified formula because it does not account for absorption inside. Actual formula is naturally somewhat longer but it reduces to that one when absorption is zero.

#18 JMC

    Member

  • Members+G
  • 1,094 posts
  • Location: London, UK

Posted 13 September 2020 - 08:33

Morning everyone. I see there has been some more discussion on this.

Quick update - no graphs this time. I did a test last night using a Baader UV/IR cut to try and isolate the transmission in the visible band. The approach looks promising, but I want to try it in a bit more controlled manner.

David - I am wary of including that mystery filter I any formal test, as it does not seem to be as described. I agree it looks like UG11, but there is no proof. However I also agree it would make an interesting comparison. I'll run it along with the others, so that I have an equivalent set of data for the 3x 0.5mm thick ones that I have, and then figure out how to deal with it later.

Andy - you've thrown me a bit now. I was expecting a simple conversion using the reflection factor from the datasheets - is that not the case then? If not then I will absolutely need assistance.

My plan for next steps;

I'm going to break down the lighting into 3 zones, to help minimise the effects of stray light in the spectrometer. Firstly, use the deuterium lamp alone (for the region 430nm and below). Secondly, both deuterium and halogen lamps and the Baader UV/IR cut filter for the region between about 420nm and 700nm. Finally the halogen lamp alone for 700nm and above.

Once I have a decent set of data then I can worry about the maths.....
Jonathan M. Crowther

http://jmcscientificconsulting.com

#19 UlfW

    Ulf W

  • Members+G
  • 1,167 posts
  • Location: Sweden, Malmö

Posted 13 September 2020 - 09:04

The transfer matrix method is excellent for reaching a more exact result.
We'll also need the refractive index, by wavelength, for the glass materials tested, to go all the way.

I wonder how big the deviation from the real value we will see by just using the simplified formula, as the size of the higher order sub-terms decrease rather quickly.
At high attenuation areas that would be even more true, as the internally reflected light dies away even faster, leaving just the media transition terms.
I think my reasoning here is sound. If not, please feel free to correct me, anyone.

When looking at Jonathan's graphs at the high transmission areas and using the Ti-values calculated from the Schott calculator for T, (Ti x reflection factor), they agree rather well.
The the peaks values I get are 88.6% and 83.6% for U-340 and U-360.

There might also be a practical reason that there is little data from the filter glass manufacturer in these areas.
It might be that the deep OD varies significantly from melt to melt, but normally are kept high.
The green leak (not the weggy :grin: ), might have been a temporary ting.

It is a challenge to measure low transmission filters with array spectrometers as the results are affected by offset drift and internal crosstalk.
I think we should wait and not draw any conclusions until Jonathan has refined his measurements further.


I must embarrassed confess that I have forgotten how to practically use matrix calculations as I have not needed them for any of my professional tasks.
I have not inverted a matrix for about 35 years.

It would be interesting to know how to best apply this practically to the current problem without access to a Mathlab licence.
When I studied there existed no personal computers.

Edited by UlfW, 13 September 2020 - 09:09.

Ulf Wilhelmson
Curious and trying to see the invisible.

#20 Andy Perrin

    Member

  • Members+G
  • 3,341 posts
  • Location: United States

Posted 13 September 2020 - 14:24

Ulf- there is only a single layer here so the matrix method is not very involved in this situation and one can eventually simplify it down to a single formula for the transmission without any MATLAB whatsoever. The high absorption areas will be the least affected by multiple bounces, you were correct. But we are interested in the transparent parts here. Most of these data sheets give at least two values of refractive index so one can just fit a line to them to get the dispersion.