Spectral sensitivity and its possible consequences

[loccdor]

This now make it hard to complete the calculation, the fall-off could be a bessel function (?) but we can't be sure of what blaze wavelength was used by Ilford.

Have you considered an email to Ilford? They seem to be a pretty responsive company compared to other film manufacturers.

 

[Mark J]

I'm ahead of you on that - I emailed their Tech guy yesterday from work.
It would be great if they could supply enough info to complete the task. I'll keep you posted.

 

[DREW WILEY]

Pretty interesting stuff. But for general photography practical purposes, the spec sheet graphs do give us enough to make approximate comparisons, or a reasonable valid starting point, along with recommended starting factors for contrast filter recommendations. For technical applications, however, I don't see any way around detailed specific testing or more serious instrumentation. Anyone who has had to make precision matched color separations from a film like FP4, for example, understands where things go off the rails, and how much up front work there is to correctly predict that, especially when spectral response divergences come into play in relation to reciprocity failure with long exposures.

 

[Mark J]

Yes, I think you're right on the whole but it would be fun to get it accurate enough to be able to make predictions with various types of film.
Recently I was looking at the Fomapan curves which look rather red-biased, but doing this work has made me think that a lot of that could be just the method of test.
Interesting, Fuji seem to show their graphs (spectrograms) for Daylight 5400 K , which would probably be the most relevant all-round method if everybody adopted it. It should be fairly easy to add another couple of columns to convert from 2850K tests to 5400K .

 

[DREW WILEY]

I've always wondered whether Fuji standardized on 5200K as their version of daylight or a little higher. It doesn't make a lot of difference with black and white film, but when doing precision duplicates or internegs from Fuji color film, I simply went the route of 5000K as my own shop standard, and it came out remarkably close. But that doesn't mean their lesser ventures into black and white, like Acros, were the same. That's an orthopan film anyway. I might shoot some more of it this afternoon.

 

[bernard_L]

Been away for 24hrs, and assumed (wrong) from no email notification that the topic had died off.

1. The discussion has remained factual and civilized; nice, and not a given on an internet forum.

2. Before connecting again to this thread, I sent a message to Ilford support; maybe with a similar message from @Mark J there will be some reaction. Too bad I did not keep a copy of my message.

3. @snusmumriken. Thank you for the copy of relevant pages of the Iford manual. There is an inaccuracy in the schematic: the wedge gradient should be perpendicular to dispersion, i.e. perpendicular to the plans of the figure, difficult to draw I agree. The text on the second page implies that the plot is uncorrected as of the manual writing (1960's).

4. @reddesert. I'm also an astronomer (retired). While I was not primarily involved in spectrophotometric calibration, I had to juggle a lot with SED's (spectral energy distribution) and convert between various units (janskys, magnitudes, Watt/m2/Hz, and several more).
"It is more technically correct when making a plot like the Kodak plot to put the y-axis in units of flux density"
Actually I like the Kodak representation. Conceptually it amounts to exposing the film with a tunable monochromatic source of constant (versus wavelength) power per unit area. (actually calibration steps deliver an equivalent result) This avoids scratching one's head whether the flux density is per unit wavelength or per unit frequency (lambda-squared factor between the two). Note that, more and more frequently, in astronomy, spectral densities are presented as nu*F_nu or lambda*F_lambda, because they are equal and this sidesteps the issue of freq versus wavelength units. But I digress.
"This means each unit of film area is seeing an equal wavelength range of incident light: this mm^2 of film sees light at 400-420 nm, the next mm^2 sees light at 420-440 nm, and so on. This is what we want to measure incident energy per spectral bandwidth, analogous to what I think the Kodak plot wants to show."
Good point. So wavelength is the proper variable to define the energy bins.
But that is not yet equivalent to the Kodak definition, there remains the non-uniform flux density (per wavelength bin) of the 2850K (or whatever is chosen) source. Then the correction by @Mark J should be valid in principle.

 

[bernard_L]

I gave some more thought to the topic under discussion, and I'm now pretty sure of what is the Ilford sensitivity and what can/cannot be done with the available graph.

Executive summary.

• The plot is derived from a wedge spectrogram (it's written in the datasheet, sono big news)
• The vertical scale is sensitivity on a log scale
• But the vertical axis has been rescaled to 0-1
• So the amplitude of the log scale is lost
• To apply a correction we need the actual sensitivity on a linear scale (give or take a global factor)
• But we cannot invert the log
• So we are stuck with a qualitative plot

More details

As I stated in my previous post, to obtain a wedge spectrogram, the gradient of attenuation of the wedge must ab at right angle to the direction of spectral dispersion. Contrary to what the diagram in the Ilford diagram might suggest; see post 37 by @snusmumriken.
At each wavelength, starting from the "clear" side of the wedge, the film is strongly illuminated; going along the wedge gradient, the light intensity decreases, and so does the exposure of the film, up to a point where a set density of the developed film (say B+F+1.0) is reached. The farther one can go (decreasing illumination) into the wedge gradient to reach that condition, the higher the sensitivity. Once developed, the wedge spectrogram looks like this, borrowing from a post by Ron Mowrey in a past thread with similar concerns as the present one https://www.photrio.com/forum/threads/equal-energy-vs-wedge-spectrogram.64615/

Next is a little sketch I made to illustrate the various relevant quantities. Those familiar with these notions need not feel offended by the level of explanation; I'm just trying to keep everybody on board.

Assume a wedge with total thickness 6mm, where the absorbing "gray" glass is such that 2mm transmits 1/10 of incident light. Then 2x2mm transmits 1/100, and 3x2mm transmits 1/1000. Density is minus log of transmission, so ranges (in this example) from 0 to 3. and sensitivity is the reciprocal (give or take a scale factor) of the amount of light needed to achieve a pre-set darkening of the film (avoid "density" just to avoid confusion with the wedge density).
Just to make things simpler, one might assume that the film is a graphic arts film. Then the dotted curve is just the boundary between the opaque and clear regions in the negative after development.

So the "y" coordinate of the curve is proportional to the logarithm of the sensitivity. But to apply a correction for the spectral distribution of the light source, one first needs the natural value of the sensitivity.
Say we have u=log10(s). We can undo the log with s=10^(u). But if the scale for u is unknown, or we use a wrong scaling for u, weird things result; e.g. s'=10^(2*u)=s^2. Clearly a non-linear operation. Applying the spectral energy correction to the log of the sensitivity (the published curve) is also incorrect. And taking the log (a second time!) does not improve the situation.

If I can be allowed to fantasize, imagine the following scenario. Ilford engineer records the wedge spectrogram; looks like the one from Ron Mowrey; not pretty enough for the published datasheet. So he/she tasks a summer intern with the rest of the work. Develop film. Enlarge on grade 5 paper. Trace the black/white boundary, and transfer to graph paper. Prettify the curve using a French curve. And, because the intern is fresh from University and has been taught how a proper graph should look, he/she makes up a vertical scale 0-1 normalized to max value.

 

[Mark J]

I have read your comments briefly but need to go through it in more detail when I'm back in spreadsheet mode ( the sun is shining this weekend ! ).

However I will say that I don't think either the OP or myself is hoping to get an accurate absolute sensitivity out of the conversion. Clearly we have no information from Ilford in their plot at the start. This is a minor issue, since we have the ISO sensitivity for the films.
I think we can get a relative (spectral ) comparison at the end of this, if that's what you mean by 'qualitative' ?

From your previous posting, I do think that there is a function of varying efficiency from the diffraction grating that must be allowed for ( thanks Reddesert ) and I'm interested to see what Ilford say on this subject.

More soon...

 

[koraks]

not pretty enough for the published datasheet

Or, also possible: far too informative for a published datasheet. Surely, Kodak will have a go with the Ilford film, but absolutely no need to make them any wiser than strictly necessary. Let them come to their own conclusions. A little smoke and mirrors - and if they look pretty, anyway...

 

[Mark J]

If I can be allowed to fantasize, imagine the following scenario. Ilford engineer records the wedge spectrogram; looks like the one from Ron Mowrey; not pretty enough for the published datasheet. So he/she tasks a summer intern with the rest of the work. Develop film. Enlarge on grade 5 paper. Trace the black/white boundary, and transfer to graph paper. Prettify the curve using a French curve. And, because the intern is fresh from University and has been taught how a proper graph should look, he/she makes up a vertical scale 0-1 normalized to max value.

Maybe this was how it was done; but alternatively, I'm sure a company like Ilford could have added a simple X,Y stage for the film to a densitometer head, and looked for a specific density ( eg. 1.0 ) along the curve for a set of wavelengths.

 

[bernard_L]

I have read your comments briefly but need to go through it in more detail when I'm back in spreadsheet mode ( the sun is shining this weekend ! ).

Enjoy! Definitely more gratifying in the end than splitting hairs as we do (I include myself).

However I will say that I don't think either the OP or myself is hoping to get an accurate absolute sensitivity out of the conversion. Clearly we have no information from Ilford in their plot at the start. This is a minor issue, since we have the ISO sensitivity for the films.
I think we can get a relative (spectral ) comparison at the end of this, if that's what you mean by 'qualitative' ?

No. By qualitative I mean it differs from the "true" one by more than just a scale factor. And is better than plain wrong because nevertheless it goes up and down at the right places. But for that the graph in the Ilford datasheet is already good enough.

From your previous posting, I do think that there is a function of varying efficiency from the diffraction grating that must be allowed for ( thanks Reddesert ) and I'm interested to see what Ilford say on this subject.

More soon...

Probably. Impact at the 10% level? 50%?
But that was not my main point. My main point is that you want to apply this correction and other (like the 2850K blackbody distribution) to the real sensitivity function, not to its logarithm. And we are missing the scale for the log as displayed, which is not equivalent to a scale factor for the sensitivity itself.
I spent maybe 2 hours trying to prepare a clear presentation of my arguments; please spend 10min reading through.

 

[bernard_L]

Or, also possible: far too informative for a published datasheet. Surely, Kodak will have a go with the Ilford film, but absolutely no need to make them any wiser than strictly necessary. Let them come to their own conclusions. A little smoke and mirrors - and if they look pretty, anyway...

Yes. I have a persistent suspicion that the logE-D curves for enlarging paper as they appear in datasheets are "idealized". When going to higher grades, the slope increases equally over the whole curve when actually the toe (highlights) lags and the contrast builds up first in the shadows. Ditto for the humps from imperfect overlap of the blue/green emulsions.

So there is (IMO) reasonable suspicion that the film sensitivity curve has been massaged.

 

[bernard_L]

Maybe this was how it was done; but alternatively, I'm sure a company like Ilford could have added a simple X,Y stage for the film to a densitometer head, and looked for a specific density ( eg. 1.0 ) along the curve for a set of wavelengths.

In 2024, I'd scan (2-D image) together the wedge spectrogram and a Stouffer transmission wedge, find out from the Stouffer wedge the calibration of the image in density units, and feed the wedge spectrogram (now in density units) to an appropriate contouring software, with appropriate noise filtering, etc...

 

[snusmumriken]

@bernard_L, @Mark J, @reddesert: thank you all so much for the thought you have been putting into this, and your thoughtfulness (@bernard_L, post #57) in endeavouring to keep us all on board. Self-evidently, I am not a physicist, and - worse - have been retired for 3 years, so all this exercises my little grey cells quite severely. Please understand if my questions below are naïve.

1. I don't think there is any suggestion, is there, that Ilford have logged the y-axis after reading the density? From your explanation of how the step-wedge works, the light inputs are already a logarithmic series, and one could read the Ilford manual text as simply pointing this out. It's different for the Kodak graphs, because they have used the same energy input at each wavelength, so they've had to log the resulting data on exposure-to-reach-the-referent-density.

2. The function relating density to light intensity is essentially the characteristic curve for each wavelength bin, isn't it? We know that curve to be non-linear w.r.t. log(light input), and it may conceivably differ across wavelengths. (Presumably this is what @DREW WILEY was saying back in post#11.) However, if you pick the density you trace (from the spectrogram image) to be well up in the straight-line portion of the curve, the assumption of linearity could be a reasonable approximation, perhaps? The Kodak graphs show curves for two densities, presumably to show whether or not there are differences between them. As they seem nicely parallel in the Tri-X and Double-X examples, this is perhaps not an issue after all.

3. In the end, we may all have to accept that the Ilford graphs can't be converted into a form comparable with the Kodak ones. But can we do the reverse? Can the Kodak graphs be converted into something comparable with the Ilford ones?

 

[bernard_L]

Please understand if my questions below are naïve.

So if they are questions (rather than remarks) I'll oblige and try to respond without looking argumentative...
And, don't worry, naïve questions are sometimes the most pertinent.

1. I don't think there is any suggestion, is there, that Ilford have logged the y-axis after reading the density?

No

From your explanation of how the step-wedge works, the light inputs are already a logarithmic series, and one could read the Ilford manual text as simply pointing this out.

Yes. Yes. The log scale is how the data come out first from the analysis of the wedge spectrogram.

It's different for the Kodak graphs, because they have used the same energy input at each wavelength, so they've had to log the resulting data on exposure-to-reach-the-referent-density.

No. Kodak having an equal-energy source is a bonus (more on this below §) that allows the sensitivity to be expressed in proper physical units (see Kodak datasheet or my earlier post; no need to repeat once more). Now to log or not to log that equal-energy sensitivty graph is a pure cosmetic issue. As I wrote in my post #40:

then log for the usual reason (display a large dynamic range on a graph).

(§) Equal energy source. I wrote in my post #56:

Conceptually it amounts to exposing the film with a tunable monochromatic source of constant (versus wavelength) power per unit area.

Conceptually only. It's difficult (impossible) in the optical domain to deliver constant power output versus wavelength. It is possible however to have a so-called bolometric detector with constant efficiency, e.g. a Golay cell. So: measure, then calibrate.

2. The function relating density to light intensity is essentially the characteristic curve for each wavelength bin, isn't it? We know that curve to be non-linear w.r.t. log(light input), and it may conceivably differ across wavelengths.

Let's not get side-tracked, the discussion is complex enough already.
(a) I was careful in my previous post to use different words for the "blackening" of the (once-developed) film and the density of the wedge. The film linearity is not an issue at this stage of the discussion.
(b) The relative sensitivity of the film under test versus wavelength is measured at an arbitrary density, no matter where it is on the film's characteristic curve. Kodak chose to measure it at B+F+0.3 and B+F+1.0. To measure the ISO/ASA sensitivity, one would need to measure at the density of the speed point, but pleeez let's not open that box in this discussion.
(c) Yes, conceivably the shape of the characteristic curve might change with wavelength. Maybe that is what the Kodak engineers had at the back of their mind when they decided to measure at two densities. And indeed there is a hint of the two curves not tracking perfectly below 400nm. That is a second-order issue. Again, pleeez let's stick to first-order issues.

However, if you pick the density you trace (from the spectrogram image) to be well up in the straight-line portion of the curve, the assumption of linearity could be a reasonable approximation, perhaps?

It is not that linearity which is in question. Please read again the Executive Summary of my post #57. I don't think there is a benefit in copy-pasting it here.

3. In the end, we may all have to accept that the Ilford graphs can't be converted into a form comparable with the Kodak ones.

That is what I'm arguing.

But can we do the reverse? Can the Kodak graphs be converted into something comparable with the Ilford ones?

An interesting suggestion. Three steps are required:

1. Multiply the equal-energy sensitivity from Kodak by the unequal-energy distribution of the 2850K blackbody. Since the Kodak data is in already log form (like the Ilford data), we just add the logarithm of the blackbody distribution. Plus, need to be careful about energy binning; I'd have to write it down carefully before doing it. But definitely do-able.
2. Correct for non-uniform (versus wavelength) efficiency of grating, optics, and geometry (if angles are not small). Only Ilford knows (if they ever cared). Not do-able from first principles. Needs access to the apparatus, calibrated measurements, and supervision by a good and experienced physicist.
3. Putting the Kodak and Ilford curves, both with a log scale, on the same scale. meaning a factor 10 change in sensitivity to be represented by the same number of cm on paper. Or, equivalently, associated with a proper vertical scale (0.1, 1.0, 10, 100). Missing that information, you could make one film look like it is terribly unbalanced between different wavelengths by expanding the vertical scale (without a labeled vertical axis). Such information is missing for the Ilford graph, therefore that step is not do-able.

I hope I have answered your questions.

 

[Mark J]

I spent maybe 2 hours trying to prepare a clear presentation of my arguments; please spend 10min reading through.

I will do, trust me. There's a lot I want to go back and read in detail. It's just the best weather that we've had for over 6 months !

 

[Mark J]

Conceptually only. It's difficult (impossible) in the optical domain to deliver constant power output versus wavelength. It is possible however to have a so-called bolometric detector with constant efficiency, e.g. a Golay cell. So: measure, then calibrate.

I think we have at least two instruments of this type at work. My best guess of how it works for the visible is a halogen or xenon source, then a mirror-based monochromator ( eg. Czerny–Turner ) , combined with a rotating ND gradient filter that is electrically coupled to the grating angle to take out the non-equal energy of the incandescent source and the variable grating efficiency. I can ask my friend Dave, in test engineering, tomorrow - he looks after this sort of kit. However, whatever the exact details, we do test optics across a spectral range and require the data to be accurate to 1 or 2%

The Golay cell is a clever concept but looks to be confined to the IR, given the use of absorption in glass for its operation.

 

[Bill Burk]

You should compare TMAX films too.

A simple solution would be to use a light yellow filter with Double-X.

Decreased blue sensitivity means you don’t need a blue filter to see wispy clouds in blue sky.

So increased blue sensitivity means you need a yellow filter.

 

[DREW WILEY]

In the real world, not so simple, Bill. You might get a little distinction between sky and clouds that way, without any filter, but how well will it reproduce in printing paper itself without jumping through a lot of inconvenient hoops? Current TMax films have reduced blue sensitivity compared to most pan films; but I sure can't get them to do that job without any filters.

It's amazing just how much our normal color vision factors into this when viewing such scenes and appraising them. It's also a bit surprising just how much panchromatic films differ from one another in handling that issue, especially if we factor in some of the older choices no longer around. Seemingly minor differences on a graph aren't always so small after all in sensitometric results. Of course, most skies aren't as blue as they once were, either.

A lot of that has to do with the other end of the spectrum. A film with a little bit of extended red sensitivity, capable of handling a deep red filter well, like a 29, will in turn, using such a filter, deeper blue significantly more than a standard medium red 25 filter can. With old Bergger 200 film, for example, that combination would work like a sledgehammer on blue. Now I never even carry a 29 unless TMax is involved, and only optionally then.

 

[bernard_L]

I think we have at least two instruments of this type at work. My best guess of how it works for the visible is a halogen or xenon source, then a mirror-based monochromator ( eg. Czerny–Turner ) , combined with a rotating ND gradient filter that is electrically coupled to the grating angle to take out the non-equal energy of the incandescent source and the variable grating efficiency. I can ask my friend Dave, in test engineering, tomorrow - he looks after this sort of kit. However, whatever the exact details, we do test optics across a spectral range and require the data to be accurate to 1 or 2%

The Golay cell is a clever concept but looks to be confined to the IR, given the use of absorption in glass for its operation.

I stand corrected. Most of my professionnal life in microwaves and 10yr in near infrared.

 

[Bill Burk]

In the real world, not so simple, Bill. You might get a little distinction between sky and clouds that way, without any filter, but how well will it reproduce in printing paper itself without jumping through a lot of inconvenient hoops? Current TMax films have reduced blue sensitivity compared to most pan films; but I sure can't get them to do that job without any filters.

I used to feel guilty about not always using a yellow filter until I printed this. No filter TMY2. Pretty much a straight print on Galerie.

 

[Bill Burk]

It’s interesting that the Ilford document clearly says if you filtered for daylight the curve would show higher blue.

They almost make it sound simple.

Haven’t had a chance to try out my glass wedge but I am hoping to use it to better see the toe shape. A lot happens in that last 0.15 step and I am curious to see how sharp TMY2 toe really is.

 

[Mark J]

It’s interesting that the Ilford document clearly says if you filtered for daylight the curve would show higher blue.
They almost make it sound simple.

Well yes, the curve they show is produced from Tungsten light which has a lot less blue.
Which document is that, by the way ?

 

[DREW WILEY]

Bill - you really need more steps at smaller increments to determine that. TMY not only has a relatively steep toe, but an exceptionally low fbf in some developers. The nearly straight line launches off the toe pretty fast, but not as fast as a classic "straight line" films like Super-XX or Bergger 200 in optimal developers.

But just like those two old classic films, a blue separation using FP4 yields a significantly lower gamma than red and green separations. That's obviously related to color reproduction work; but still, it tells you something about the film's spectral response per se. Extra development of the blue separation can bring things closer, but still not ideally.

The only film I have ever found that can handle all three colors identically - producing overlapping curves from identical development times is TMax100. And in that case, you need to identify the "sweet spot" in exposure time for each. TMY400 oddly matches blue and red, but with the green separation being out of synch and need more development.

All this informs us that at long exposure times with strong filters, as often needed for that kind of work, the normal published spectral expectations can get seriously skewed.

So it isn't just an issue of light source color temperature potentially at play. In other words, what Ilford is telling you is generally correct for ordinary photography, since people who use this film for specialty applications instead probably already know they're in for a headache calibrating them.

I'm not the right person to try to quantify this other than by plotting actual separate curves. I have a lot of densitometer and color spectrophotometer experience, but nothing directly related to the immediate question. Those particular spectrophotometers did read a step past visible light both directions, into near-UV as well as near-infrared. The old IBM one, as big as an upright grand piano, was halogen with a rotating drum, and gave a true continuous spectral plot, while the later more convenient pulsed xenon units interpolated and extrapolated it. All of them were capable of being fooled by certain things.

In her Biotech days, my wife operated what was perhaps the most expensive spectrophotometer ever made, costing six million dollars. But it was such a trade secret that it was kept in a room with four foot thick reinforced concrete walls behind a timed bank vault door. The software was an even better kept secret. She was merely trained to operate it. Only the two inventor-owners really knew what made it tick.

 

[Bill Burk]

Bill - you really need more steps at smaller increments to determine that.

It’s literally a wedge