DIY ND filter

While repairing old Luna Pro light meter I've noticed that the 6EV filter that gets inserted in the light path for high scale is flaking off - its surface looks like tiny droplets of
some substance etched away filtering material leaving dirty spots on filter's glass. IPA does not clean it. Anyway, since finding good pare filter of this size is not possible,
I had to come up with alternative.

I managed to get about right attenuation by printing wide gray scale wedge on a piece of transparency film, and cutting out small square
section that happen to be required density. It actual took stack of two pieces overlapping each other, but inserting/removing them blocks light 6 stops worth, which is what
the stock filter did. But quality of the surface, while not critical, is not great, even printed at 600 dpi.

Then I had an idea I wanted to run by collective wisdom here. I have variety of color gelatin theatrical projector filters (like these:
https://www.bhphotovideo.com/c/product/1016197-REG/colourlite_mafp_213_vivid_colors_filter_pack.html ). Since we know white light roughly
consists of equal amount of red, green and blue light, by the same thinking, if I overlap red, green and blue filters of about equal density, I get ABOUT neutral "gray" filter.

Is this accurate assumption? I could adjust overall density by having two or more layers of each color. What do you think?

Granted, I can probably just find and buy gray filter online, but the best I found are way too thick, minimum set of two large ones (like 6x6 inches while I need just
8mm x 8mm square, and I'm reluctant to waste at least ~$15 to throw away practically all of it buy the tiny piece. For a light meter printed wedge works just fine, but
I'm thinking laser jet's toner won't last long on transparency. Gelatin is far more durable.

If you have any other ideas, please share. The filter in Luna Pro cannot be more than about 0.3mm thick.

336v said:

But quality of the surface, while not critical, is not great, even printed at 600 dpi.

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Does not matter IMO. This is not an imaging configuration. As long as the global attenuation is right, and the area is approximately uniform, you should be OK.

336v said:

if I overlap red, green and blue filters of about equal density, I get ABOUT neutral "gray" filter.

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Don't overthink the neutrality. When intercomparing some of my light meters (Selenium; CdS, Si) I noted offsets between the meters (of course, nothing is perfect) that would change according to the color of the object being metered; even warm gray versus cold gray. So...

336v said:

If you have any other ideas, please share.

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Why not make if from B/W film? That is how Stouffer wedges are made, and nobody complains about their durability. First iteration will give you a rough idea of the exposure needed, next time take a pic (MF) of a slight gradient at about the right exposure, then select on the negative the spot within the gradient with the exact amount of density; don't worry about the gradient within the small area that you use.

336v said:

Is this accurate assumption?

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No.

But if you want to make neutral density, you can get fairly close with a neutral-base B&W film (something like Fomapan 100 in sheet film format for instance) and develop it to get neutral-toned silver layer. If you follow this link you'll find a post that links to a developer that will do this. Getting the desired density will be a process of experimentation.

336v said:

Then I had an idea I wanted to run by collective wisdom here. I have variety of color gelatin theatrical projector filters (like these:
https://www.bhphotovideo.com/c/product/1016197-REG/colourlite_mafp_213_vivid_colors_filter_pack.html ). Since we know white light roughly
consists of equal amount of red, green and blue light, by the same thinking, if I overlap red, green and blue filters of about equal density, I get ABOUT neutral "gray" filter.

Click to expand...

But light does not consist of equal amounts of red, green, and blue. The Sun's output is greater in green than blue or red. Our eyes have variable sensitivity to different wavelengths, and the same is true for film. And different color filters block light differently.

There is a more fundamental problem with your hypothesis: Even if you actually had filters that are properly red, green, and blue as non-overlapping segments of the spectrum, the result of stacking them together would be pure black, as every single wavelength would be blocked by at least one of the filters. If you are able to stack a red + green + blue filter and still see anything at all, that just tells you that the filters are, in a sense, failing at their job. It means that the red filter must be allowing some blue + green to pass through, and the green filter must be allowing some red + blue to pass through, and the blue filter is allowing some red + green. Basically, it's as if you're adding up the flaws of all the filters in some unpredictable way. That's very much the opposite of ND.

dcy said:

The Sun's output is greater in green than blue or red.

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Sort of; it's more in the blue-cyan border region. It drops off very steeply to the left, and more gradually to the right of this peak. Green is still pretty powerful though. You may be thinking about the spectral sensitivity of the human eye, which peaks strongly in green.

dcy said:

Even if you actually had filters that are properly red, green, and blue as non-overlapping segments of the spectrum, the result of stacking them together would be pure black, as every single wavelength would be blocked by at least one of the filters.

Click to expand...

Assuming perfect filters(*). If you use weak filters, the approach in principle could work, but to get a neutral response you need to pretty carefully balance the filters against each other - not just their plateaus, but also the slopes so you don't get stray peaks or dips in the transmission curve. Maybe there are filter sets commercially available that meet this requirement. However, I doubt it's a very feasible, cost-effective route to go for this particular application.

(*) the common method is actually subtractive filtering...like in most color enlargers. So CMY, not RGB.

koraks said:

Sort of; it's more in the blue-cyan border region. It drops off very steeply to the left, and more gradually to the right of this peak. Green is still pretty powerful though. You may be thinking about the spectral sensitivity of the human eye, which peaks strongly in green.

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It is always possible that I'm wrong, but... I don't think I am... According to Wien's law, the peak wavelength of a blackbody at 5,772 K is 502 nm which is in the green part of the spectrum.

koraks said:

No.

But if you want to make neutral density, you can get fairly close with a neutral-base B&W film (something like Fomapan 100 in sheet film format for instance) and develop it to get neutral-toned silver layer. If you follow this link you'll find a post that links to a developer that will do this. Getting the desired density will be a process of experimentation.

Click to expand...

that's a good way I think and only needs little experimentation. Three exposures,hopefully catching densities above and below target values,then connected to a simple curve in excel will nail the required exposure sufficiently precise. Just make sure not to change too many variables.

dcy said:

It is always possible that I'm wrong, but... I don't think I am... According to Wien's law, the peak wavelength of a blackbody at 5,772 K is 502 nm which is in the green part of the spectrum.

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The atmosphere will attenuate quite a bit in the visible range. here is a link to the spectrum. https://www.sciencedirect.com/topics/physics-and-astronomy/solar-spectra

Ken Wilkens said:

The atmosphere will attenuate quite a bit in the visible range. here is a link to the spectrum. https://www.sciencedirect.com/topics/physics-and-astronomy/solar-spectra

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Attenuates it in the direction of shifting the peak further to longer wavelengths. The norm is for opacity to shift a spectrum to longer wavelengths, not shorter ones (caveat for absorption lines). Here's the plot from your link:



Notice that the peak has shifted slightly toward the right, and is clearly at a wavelength longer than 500 nm. Blue is 450-495 nm. Green is 495-570 nm. Whether we look at the location of the peak or the integral under the curve, this plot says that there's more green than blue hitting the Earth's surface.

EDIT: Once again, it is always possible that I'm wrong and I want to be cautious here... But... I think I'm right about this.

To follow-up on my previous comment: Notice that in the plot the ratio of the irradiance at Earth's surface vs irradiance at the top of the atmosphere decreases for shorter wavelengths, at least for lambda < 800 nm. Compare the height of the two curves as lambda decreases. This is the Rayleigh scattering that also makes the sky blue.

dcy said:

It is always possible that I'm wrong, but... I don't think I am... According to Wien's law, the peak wavelength of a blackbody at 5,772 K is 502 nm which is in the green part of the spectrum.

Click to expand...

502 sounds about right. That's cyan. 525-550 is green. If you want to call 500nm 'green', I can sort of get behind that. We're on the same page on the spectrum. Of course, the whole solar spectrum thing isn't very relevant to the topic at hand.

dcy said:

It is always possible that I'm wrong, but... I don't think I am... According to Wien's law, the peak wavelength of a blackbody at 5,772 K is 502 nm which is in the green part of the spectrum.

Click to expand...

• It’s not Wien’s it’s Planck’s
• Where the maximum is depends on whether one is plotting flux per unit wavelength or per unit frequency; in the latter case (still for 5770K) the maximum is at the frequency corresponding to 922nm, almost twice as much. There is no reason other than habit or tradition to prefer one over the other. Or to state that the sun peaks in the green.

See e.g. https://www.scielo.br/j/rbef/a/58MBHJ7TF4mhynrfNr4rCHR/

Instead of reinventing the wheel, can't a Wratten 96 filter do the job as long as you chose the right density?

koraks said:

502 sounds about right. That's cyan. 525-550 is green. If you want to call 500nm 'green', I can sort of get behind that. We're on the same page on the spectrum. Of course, the whole solar spectrum thing isn't very relevant to the topic at hand.

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Yeah. Agree 100% Colors don't have well-defined boundaries. Wikipedia says that green is 495-570 nm but it also says that cyan is 490–520 nm, so ... ¯\_(ツ)_/¯ ... Clearly "cyan" is the most precise & accurate way to describe it.

I was thinking back to an exercise we did in grad school --- "We see red stars and blue stars, why don't we see green stars?" ---. The punchline was the the Sun is a "green" star. Clearly in that context we were not thinking about anything more subtle than R/G/B.

bernard_L said:

• It’s not Wien’s it’s Planck’s
• Where the maximum is depends on whether one is plotting flux per unit wavelength or per unit frequency; in the latter case (still for 5770K) the maximum is at the frequency corresponding to 922nm, almost twice as much. There is no reason other than habit or tradition to prefer one over the other. Or to state that the sun peaks in the green.

See e.g. https://www.scielo.br/j/rbef/a/58MBHJ7TF4mhynrfNr4rCHR/

Click to expand...

Wikipedia has a nice plot on this topic:


Using Wien's was not wrong; it was the sort of imprecision you'd expect from casual low-stakes conversation. It's a reasonable stab at identifying where the Sun's output is. If we were being precise, we should instead be computing the total irradiance within each frequency band. Then we'd get into a debate about whether it's cheating if one color is associated with a wider wave band than another, and whether cyan counts as green.

dcy said:

Then we'd get into a debate

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I think we need to get back to the debate of ND filters for the application of OP. Thanks.