Using a 660nm laser to expose legacy AgX Agfa, Ilford, HRT (BB) plates

Silverhalide Emulsions / Chemistry.
Ed Wesly
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Joined: Wed Jan 07, 2015 2:16 pm

Re: Using a 660nm laser to expose legacy AgX Agfa, Ilford, HRT (BB) plates

Post by Ed Wesly »

The curve in the post above is not exactly the true picture, as it is a graph of relative sensitivity, not absolute, unlike this curve, from Agfa Technical Information Bulletin 21.7271(688)LI, the rest of the bulletin available here: http://edweslystudio.com/Materials/AgX/ ... 271688.PDF

Image

The absolute sensitivity never falls to zero, just varies over the spectrum, with that area that is the borderline between blue and green, cyan, where the 488 nm line of the Argon laser reigns supreme, is where it dips to its lowest value. The guidelines were put in place to show that the difference between the sensitivity at the He-Ne red of 633 nm is about an order of magnitude greater than that at 488 nm, almost 10X, or about 3 or 4 photographic stops, which are binary orders of magnitude, which is borne out in practical experience.

The -2 on the ordinate (vertical axis) could be translated as 100 microJoules/cm^2, and the -3 as 1000 microJoules/cm^2 or 1 full milliJoule/cm^2, since the axis is logarithmic. These figures are again borne out in practice, as 100- 200 microJoules/cm^2 is usually the bracketing point when I make test exposures at red wavelengths.

Comparing the 2 graphs, the shape of the curve is the same, but the numbering of the relative's ordinate, which is also logarithmic, starts at zero, and ascends as powers of 10, and once again it can be seen that the sensitivity difference between the 2 lambdas is an order of magnitude off. The drawback to the relative graph is that it gives an impression of absolutely no sensitivity, but the zero in a log axis is 10^0 = 1, so that dip is used as unity, the standard of comparison, not as an exposure suggestion.
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Din
Posts: 458
Joined: Thu Mar 12, 2015 4:47 pm

Re: Using a 660nm laser to expose legacy AgX Agfa, Ilford, HRT (BB) plates

Post by Din »

Ed Wesly wrote: Fri Apr 08, 2022 9:53 am The curve in the post above is not exactly the true picture, as it is a graph of relative sensitivity, not absolute, unlike this curve, from Agfa Technical Information Bulletin 21.7271(688)LI, the rest of the bulletin available here: http://edweslystudio.com/Materials/AgX/ ... 271688.PDF
Relative to what? Could you define 'relative sensitivity' , as opposed to 'absolute sensitivity'. Generally, plots such as these are normalised to unity, so that the maximum ordinate is, in effect, 1. Thus, if you're measuring intensity from some source, and I₀ is the maximum intensity, the ordinate is in units of I/I₀, making the maximum =1 (when I = I₀). In these cases, the ordinate is relative (normalised) to some constant. However, these curves don't specify any normalisation.
Ed Wesly wrote: Fri Apr 08, 2022 9:53 am
The absolute sensitivity never falls to zero,
Not entirely so. The sensitivity of silver halide is mostly in the uv/blue region. Silver halide film for photography/holography is infused with dye(s) to extend the sensitivity into the vis. The basic mechanism is a two-electron excitation process, whereby the dye molecule is excited by the radiation and the excited dye molecule passes an electron into the conduction band of the silver halide. However, because the actinic process whereby the dye gets excited occurs over a limited range, any individual dye has a limited range, so different dyes are used for different areas, such as X-Rays for dental photography. For holography, any given emulsion, with it's concommitant dye has a sensitivity over a limited wavelength range.

Ed Wesly wrote: Fri Apr 08, 2022 9:53 am
The -2 on the ordinate (vertical axis) could be translated as 100 microJoules/cm^2, and the -3 as 1000 microJoules/cm^2 or 1 full milliJoule/cm^2, since the axis is logarithmic. These figures are again borne out in practice, as 100- 200 microJoules/cm^2 is usually the bracketing point when I make test exposures at red wavelengths.
The scale actually states "Log S(S = m²/mJ). " Log(-2) = 1/100. So. S = 1/100 m²/mJ, or 100 mJ/m² (not cm²). Translated to μJ/cm², this comes to 10 μJ/cm², since 10⁴ cm² = 1 m²

From "Topics in Applied Physics, vol 20: Holographic recording materials"
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