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UltravioletPhotography

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


JMC

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What I’m saying is that the shortcut is only marginally easier than the exact calculation so why wouldn’t you just do it right? I have no idea how much it will matter in this case without actually carrying it out both ways. Sure there are other sources of error but let’s minimize the ones under our control.

Sounds like a good idea.

To minimise method errors I will lean back, waiting for a proper simplified formula. :smile:

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All,

 

Just to be clear, the original aim of this work was to try and find a route to answer the question about the missing data in the Hoya data sheets for the U-340 and U-360. My goal at the moment, is to get as good and reliable set of transmission data as I can, so that that question can be answered. With regards to converting the measured transmission to internal transmission (or transmittance - I'm never sure which is right) and back again for different filter thicknesses, I shall certainly need help with the spreadsheet for the conversion if it is more complicated than the simple single number conversion factor. I can bolt things together and make stuff, but maths is not my forte.

 

Stefano, at the moment this is my priority, any side experiments will need to parked for now. I do not have any compact fluorescent tubes, nor do I have 1mm U-340 unfortunately. I am currently about 10h into the work for this, as I've been trying a few things to improve it, and I expect to easily have another 10h ahead of me to get the rest of the data, so as you can imagine I have my work cut out for me.

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Stefano, at the moment this is my priority, any side experiments will need to parked for now. I do not have any compact fluorescent tubes, nor do I have 1mm U-340 unfortunately. I am currently about 10h into the work for this, as I've been trying a few things to improve it, and I expect to easily have another 10h ahead of me to get the rest of the data, so as you can imagine I have my work cut out for me.

I understand. Don't worry, take your time.
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Regarding the formula, I worked it out. To get the actual transmission from the internal transmission, the formula is,

post-94-0-47055800-1600014411.png

where r is the reflection coefficient due to a single bouce, which can be found from

post-94-0-12796600-1600014499.png

with N as the refractive index. (I think we can approximate at this level and just use the real part of the refractive index, because otherwise this really does turn into a mess.)

 

To get internal transmission from the actual transmission, you have to solve the above formula for Ti, which is slightly messy because it's quadratic, but it's doable, and the solution is,

post-94-0-50428500-1600014617.png.

----

 

As a check you can see that if you plug r=0 into the top formula, it reduces to T=Ti, as expected.

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All, bit of an update. Last night I rethought my plan a bit, and decided to try using a 1mm S8612 to help isolate the visible part of the spectra for the measurements.

 

This morning has been quite productive. I ran the U-340 and U-360 and also my Mystery filter (which is very likely to be UG11). I now have he following sets of data;

 

250nm to about 420nm using the deuterium lamp from the light source only.

420nm to 640nm using both deuterium and halogen lights, and also having a Schott S8612 1mm in the way to help reduce long wavelength stray light effects.

640-800nm using the halogen light.

 

I though it better to us the S8612 than th UV/IR cut filters, as the s8612 is not dichroic, and doesn't have any sharp cutoffs. The plan now is for me to go through all the data, and pull it together in a format to share.

 

Andy, that formula fill me with dread. Does the square root go all the way to the end? I believe there is only 1 refractive index for each filter (1.568 for U-340 and 1.663 for U-360 according to the datasheets) so I presume it's just that that needs to be used?

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The square root covers up to and including the 4T^2 term ....

 

You can extract the term ((1-r)/r)^2 and make it a constant to simplify the expression, as the refractive index would be specific for each filter, true?

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Yes, there are two terms under the square root. It could be simplified a little further.

 

If we only have one refractive index then that’s all we can work with. Schott certainly provides two, and if you have an Abbe number and a refractive index then it’s enough to find the variation with wavelength. Don’t be afraid of long formulas, as I tell my students, it’s just some extra typing. And you can break it down and do things like compute (1-r)/r first to simplify it.

 

Alternatively, in Excel you could use the Solver on the top formula instead.

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Thanks for the info on the maths. I shall endeavour to get something to work in Excel, once the spectra are done (speaking of which, I should hopefully be able to share them later today).....

 

Ulf - only a couple of basic micrometers. My imperial one is more accurate than my metric one, but I can double check the thicknesses.

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I have some graphs to share, as of course we all like to see some data.

 

A couple of graphs for each filter. Each graph has 6 lines on it. At the short wavelength end in blue is data collected just using the deuterium lamp on the light source. In the middle in green is data collected with both lamps and a 1mm S8612 to cut out the IR and try and reduce the effects of stray light in the spectrometer. In red at the long wavelength end is data collected with just the halogen lamp from the light source. This approach was done to try and minimise stray light effects, especially in the region of interest in the visible part of the spectrum.

 

In each region of the graph there are 2 lines - transmission through the filter, and a baseline with the lamp off to give an idea of what the baseline variability was.

 

The 0.5mm U-340

post-148-0-15571900-1600163221.jpg

 

And the 0.5mm U-360

post-148-0-77198100-1600163222.jpg

 

And especially for David, my mystery filter which the data would suggest is 0.5mm UG11.

post-148-0-42020400-1600163218.jpg

 

Assuming it is UG11, then it is showing less transmission in the 550nm region that the U-340 which may address the question you had Stefano (note the magnified graph at the bottom is a different y axis scale than the Hoya ones). Keep in mind though, these values for transmission are low even for such a thin filter. At 2mm thickness or above, I doubt these leaks would be visible unless you were using a 500 or 550nm LED and long exposure.

 

I think these are as good as I can get them while trying to reduce the stray light effects. I'll give the maths a go now to try and get some data for internal transmittance so we can have a look at different thicknesses. I may be some time.....

 

Ulf - I'll get the thicknesses checked too.

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Good job Jonathan. I wonder if the ends of the three regions "connect" together (maybe you are still working on this). U-340 does leak green twice as much as U-360 and the Mystery filter at 0.5 mm (peak transmission) and this difference increases at larger thicknesses, especially since the numbers are small. So far the graphs look good.
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Ok, thicknesses. Not quite as expected, so thanks for reminding me to check them Ulf;

 

UG11 mystery filter - 0.55mm

U-340 - 0.60mm

U-360 - 0.59mm

 

All slightly thicker than expected. But I checked with both micrometers, and they both gave the same readings. It's strange, the UG11 felt thinner as I was testing it, but I'm surprised I can tell a difference of 5 microns by touch.

 

Colin - even without doing more maths, the key message is any leaks in the 450nm to 600nm region for these glasses are tiny, really tiny, for any meaningful thickness of filter (1mm and above). If you're using monochromatic light sources, and combinations of these with other filters, then yes of course a leak could be seen, but for sunlight, or camera flash they wont be a problem. Is the transmission zero - no. Is it effectively zero when you have a couple of mm of the glass - yes.

 

Stefano - I'm at the limitations of what I can do with the setup I have (and the level of knowledge I have). If I had a double monochromator spectrometer then stray light leaks would be less of an issue, but I don't. From my perspective, I wouldn't attempt to join the ends up. Each full wavelength range 'scan' is a combination of 3 individual scans, each done with a setup to try and make the best of it. However at the edges I'm not surprised they don't line up perfectly because of the issues I've talked about above. Perhaps if I had other band pass filters to allow me to isolate the regions around where one ends and the other starts, I could get it better, but I have used the kit that I had.

 

What'll be interesting to see is whether once all the maths is done the final calculated transmission curve matches a real life filter.....

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Uv photography with a 500nm Led,

Challenge accepted!

 

I may actually have one. I only have 1mm thickness filters.

1mm ug11

1mm ug1

2mm ug1

2mm u340

2mm Zwb1

1.5mm zwb2

 

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Thanks Andy. For my own peace of mind, I'm going to try the simple, single digit reflection factor first, and then see what that looks like before doing the more comprehensive maths.
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Jonathan see what I just posted! It has the three digit reflection coefficients already done for you.

Ooh thanks Andy. The datasheet I had just had a single number for the reflection coefficient. I shall use these then.

 

What would you recommend - Do I use the 400nm one for up to 400nm, and then the 500nm up to 500nm etc, or is it best to assume changes in the middle between the ranges? My guess is the latter (changing in the middle between the ranges) but just want to check. Will have a look at them tomorrow, and get one for the U-360 too.

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JMC, if you are going to use the tabulated values then I would just do what you said and use constant interpolation in each range of values switching at midpoints. If we are going to go the whole hog, then I will curve fit a formula for N for you and then we can use that in the other formulas I worked out yesterday. Honestly, based on what I see here, in the range shown probably constant or linear interpolation of the P values is going to be good enough. It's only at the third digit that we see changes.
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[a bit off-topic, but still related]

I noticed Hoya graphs are smooth, so they collected data at a finer resolution (like every nm) and used that to plot their graphs. I don't think they have data in the visible spectrum, especially if they measured their filters at 2.5 mm, but they should have finer data in the UV and IR regions, they just don't publish it. If what I said is correct, can we ask Hoya to provide us their data?

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Cheers Andy. Will do that. Thanks for the link for the Hoya datasheet. I did a google search and the first that came up was by Sydor - that had just the single number. I should have looked further down the page. I've now downloaded both the proper ones from Hoya and will do it tomorrow with these numbers. Like you I don't think doing it that way will be a problem, given the small changes as a function of wavelength.

 

Stefano - if you'd like to contact Hoya, please feel free to do so, and let us know if you hear anything.

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