Reuse of Pyrogallol developer

Silverhalide Emulsions / Chemistry.
Ed Wesly
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Re: Reuse of Pyrogallol developer

Post by Ed Wesly »

Here's where the use of using a rehalogenating bleach to de-fog plates got started (1988): http://edweslystudio.com/Publications/recycle.pdf
"We're the flowers in the dustbin" Sex Pistols
jrburns47
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Re: Reuse of Pyrogallol developer

Post by jrburns47 »

Yes, read your interesting writeup on pgs 43-52 of your informative 58 page Darkroom writeup on your website. Thank you. There seem to be different opinions about the best solutions and methods. I wonder if anyone has had actual experience and success trying your approach or Jeff B’s approach on old Agfa 8E75HD or Ilford SP696T plates. I was a little confused by Jeff’s comments here (forwarded to me by John Wagner):
https://holographyforum.org/forum/viewt ... ged#p40366

The solution is similar but different and also discusses a “reactivation” solution which may or may not be necessary.

Would soaking in a 1% TEA solution work just as well and thus bypass other “defogging” or “dark reaction” approaches?

I’m asking since I have a fair number of old Ilford SP696T and 8E75HD plates.

Any guidance based on actual experience will be greatly appreciated😊!!!
jrburns47
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Re: Reuse of Pyrogallol developer

Post by jrburns47 »

Dumb question, to which I believe I’ve known the answer for decades, but one needing an answer please.

If I’m reading 25mw CW at the recording plate surface, with a Newport wand detector 818-ST with no attenuator, with a normally reliable Newport 840-C light meter, and I make a one second exposure, how much energy is exposing the material in microwatts per cm2 per second? If you prefer to answer in uJ/cm2, (which I believe is the same amount of energy but in a single pulse), no problem.

How many ergs/cm2/sec?
Thanks.
lobaz
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Re: Reuse of Pyrogallol developer

Post by lobaz »

The size of the 818-ST detector should be 1x1 cm2. That is, if light completely fills the sensor area (unlike measuring a thin collimated beam) and you read 25 mW, it means the irradiance is 25 mW/cm2.

As 1 W = 1 J/s, it follows it equals to 25 mJ/s/cm2.
For the exposure time 1 s, the plate gets 25 mJ/cm2.

As 1 W = 10^7 erg/s, then 25 mW/cm2 = 25 x 10^-3 x 10^7 erg/s/cm2 = 25 x 10^4 erg/s/cm2 = 250 k erg/s/cm2.

(There is still a somewhat open question whether the detector should be parallel to the plate or perpendicular to the incoming light, but that's a different story.)
BobH
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Re: Reuse of Pyrogallol developer

Post by BobH »

Use the detector perpendicular to the beams when measuring to calculate beam ratio. Beam ratio is independent of the orientation of a recording plate, as it determines the fringe visibility in space. Use it parallel to the recording plane when calculating exposure if you like. Since exposure is usually bracketed to find what works best for a given situation, and because there is usually shadowing of the detector when used tilted at 45 degrees or more typical in holography, I use it perpendicular to the beams for that as well (for consistency and convenience).
Last edited by BobH on Sun Jan 23, 2022 9:53 am, edited 1 time in total.
jrburns47
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Re: Reuse of Pyrogallol developer

Post by jrburns47 »

Thanks Petr. Glad I’m not losing my mind! (Yet😂)…
Although I’ve gotten decent results on very old 10E75 (35-40 years old) and on 8E75HD expiration date 6/2001), I had been completely unable to get any exposure at all on old (25-30 years old) Ilford SP696T - just an all over even emulsion coloration after processing. All above 4x5” plates. Finally, yesterday, got an ok result with what I suspected might be a bad batch of the SP696T by giving the 4x5” plate an 83” exposure! That’s approximately 20,000,000 ergs/s/cm2. Superb result on 8E75HD plate w/ 1,000,000 ergs/s/cm2. About 100,000 ergs for great result on 10E75.

I’m testing a 500mw Cobolt 660nm laser. Beam splitter is two HWPs & PBS cube made for 660nm. Power of laser confirmed on both legs of beam via the 840-C meter with 818-ST wand detector w/ attenuator in place. If anything, the meter readings may be a little low. The Cobolt Monitor software says I’m getting 525mw but meter gives between 500-505mw. Both OB & RB are going through SPFs to first surface collimators.

Now I’m going to try to perfect the SP696T exposure to see if I can get a result as great as the 8E75HD.

So far, no attempt to “defog” or presensitize anything. I will have to for testing some very old German HRT BB640 4x5s.

Have no idea how fresh materials might react to this laser.

The 10E75 4x5s were all processed with Pyrochrome processing as were the initial batch of failed tests with SP696T and the only two 8E75 (pre HD & probably 35-40+ years old) and the first 8E75HD (there but dim). I tried a D-19/fix “control” plate with the SP696T and still absolutely nothing.

The breakthrough yesterday happened once it occurred to me that I might be too low by at least an order of magnitude on exposure for the SP696T and 8E75HD. I also changed to different processing - moved to Nick’s #5 catechol dev versions for both the 8E75HD and SP696T and FeEDTA bleach. Also moved to graded alcohol dry since I don’t like the imperfect drying of photoflo final rinse.
Later today will post an annotated photo of the 4x5” “quadrant” exposed and processed 8E75HD. I will also repost the first successful 10E75 test with corrected exposure amount annotations.
BobH
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Re: Reuse of Pyrogallol developer

Post by BobH »

Squeege after the fotoflo?
Din
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Re: Reuse of Pyrogallol developer

Post by Din »

jrburns47 wrote: Sun Jan 23, 2022 12:44 am Dumb question, to which I believe I’ve known the answer for decades, but one needing an answer please.

If I’m reading 25mw CW at the recording plate surface, with a Newport wand detector 818-ST with no attenuator, with a normally reliable Newport 840-C light meter, and I make a one second exposure, how much energy is exposing the material in microwatts per cm2 per second? If you prefer to answer in uJ/cm2, (which I believe is the same amount of energy but in a single pulse), no problem.

How many ergs/cm2/sec?
Thanks.
You're confusing your units, Jody. Energy is not measured in microwatts per cm², irradiance is. Consider another form of energy, such as heat (not temperature, which is a measure of heat). If there is a certain amount of heat energy in some volume, such as a room, then the amount of heat energy is measured in Joules, or Ergs or BTU (British Thermal Units). If the heat leaks outside the room, then the amount of heat that leaves in 1 second is the power exerted by the heat, and power is measured in watts, or ergs per second or horsepower (BTU per second). However, the amount of heat leaking from the room may not be as useful as the heat leaving the room per second per unit area, so we have watts per m², or ergs per second per cm², or horsepower per feet². This is known as irradiance, or energy per unit time per unit area.

In holography, the molecules in the medium (silver halide, dcg, polymer etc) needs to undergo a physical change - a reduction of silver halide to silver or a cross linking of gelatin etc, on exposure to light. This kind of light-activated chemical change is known as an actinic reaction. In order for the molecules to undergo an actinic reaction, they need a certain amount of energy per unit area, ie an energy necessary to modulate the medium to some value for some efficiency. This figure is the sensitivity of the plate, given in energy units per unit area, eg mJ/cm², but, as Petr pointed out, most meter heads have an aperture of 1 cm², so, as long as you fill the aperture, you don't need to worry about the 'unit area' part. The meters themselves integrate, so the number displayed is the energy/per unit time, or power.

So, if, for example, your plate has a stated sensitivity of 100 millijoules/cm², then, if you read 25 mW/cm², you need 100/25 = 4 seconds of exposure. In other words, your source, the laser, is providing 25 mJ of energy every second, and you have to 'pour in' 100 mJ of energy for the actinic reaction to take place. The exposure is:

t = S/p (t = exposure time, S = sensitivity, p = measured power)

One other caveat is that you don't want to make the exposure too short. This depends on the exposure system, but, if your exposure system is meausured in seconds (such as whisking a baffle away), then, if your exposure is one second, the uncertainty will be too large. For example, if your plate needs a 1 second exposure, and your measurement error is 25%, you'll give your plate 25% more energy than necessary.
lobaz wrote: Sun Jan 23, 2022 6:26 am
(There is still a somewhat open question whether the detector should be parallel to the plate or perpendicular to the incoming light, but that's a different story.)
My feeling is that the detector has to be perpendicular to the incoming light, because that's the direction of the actinic radiation. That is, you're measuring the value of the energy imparted to the medium in that particular direction, ie the value of the poynting vector. If you measure parallel to the plate, you're measuring the value of the cos of the Poynting vector and so giving the plate less energy than it requires.
jrburns47
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Re: Reuse of Pyrogallol developer

Post by jrburns47 »

Thanks Bob. Yes, I always read the beams perpendicular to direction of propagation and add the result and calculate ratio based on those readings. I seem to remember that, as the angle of the reference beam increases, the actual light entering the emulsion decreases since more is reflected from the surface. I don’t factor that into my total exposure or beam ratio calculations.
Din
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Re: Reuse of Pyrogallol developer

Post by Din »

BobH wrote: Sun Jan 23, 2022 9:50 am Beam ratio is independent of the orientation of a recording plate, as it determines the fringe visibility in space.
Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate.

The efficiency due to beam ratio, r, is given by:

η = 2√r/(1+r)

[I = I(1) + I(2) + 2{√I(1)I(2)}cos(phase), the ratio is r = I(1)/I(2), so now, we have I = (1+r)I(2)[1 + 2√r/(1+r) cos(phase)] and the 2√r/(1+r) is the amplitude of the cosinusoidal variation]

As you can see, if r = 1, η = 1, but if r = 0.25, η = 0.8
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