Sensitometry and lens transmission formulae

Discussion in 'Exposure Discussion' started by polyglot, Oct 30, 2012.

  1. polyglot

    polyglot Member

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    I'm planning on doing a bit of sensitometry and am having trouble finding a formula that will tell me how much light a lens transmits, in compatible units at each side. The reason I want to know is that I want to calibrate my (flash) sensitometer using a flashmeter located where the film-under-test will be.

    Consider a simple imaging chain: light source, subject, lens and sensor. I take an incident meter reading at the subject (Es) and an incident meter reading at the film plane (Ef). What is the ratio of Ef/Es as a function of the lens' aperture?

    The only reference I could find is this one for CCTV sensitivity analysis, and it says:

    Ef/Es = 0.2 * t * R / (N^2)

    where t=lens transmission (any non-ideal filter-factor built-in to the lens due to imperfect coatings etc), R = scene reflectance and N = f-number. The presence of that 0.2 is at best highly unsatisfying. Given that their example has R=0.89, would I be right to suspect that the 0.2 is a poor approximation of the 18% reflectance standard therefore R=1 or R=0.9 for an 18% card? Or should I assume that if the subject is an 18% card and the lens is ideal, Ef/Es = 0.036/(N^2) ?

    Any better references welcome...
     
  2. ic-racer

    ic-racer Member

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    I think the question comes down to how do you meter at the film plane in a camera with a meter designed to measure light falling on the subject. That question is answered with the light meter calibration formula that has these variables:
    b = Constant with units of lx⋅cd–1⋅m2
    θ = Angle between subject and lens axis
    N = Relative aperture (f-number) of lens
    F = Lens flare correction factor
    f = Focal length of lens in m
    V = Lens vignetting factor
    Ls = Luminance of subject in cd⋅m–2
    T = Lens transmittance factor
    u = Subject distance in m

    However, if your flash meter can read directly in Lux-second then it takes out all the above variables and that is all you need.

    Personally, I 'calibrate' my sensitometer with some known film, like Ilford D100. There is a precedent in the literature on using film for this type of calibration.
    Also, did you see the lengthy discussion about Bill's EG&G sensitometer and how he was able to check it with a flashmeter?
     
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  3. polyglot

    polyglot Member

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    Yes I did see the big long discussion and I'll have to go back and have another look at it. However I recall concluding from that discussion that I needed a flashmeter reading of about f/16 at the film-plane to make a test and that doesn't match BTZS recommendations.

    My "sensitometer" is a flash bounced off the ceiling, exposing some film through a Stouffer wedge. I can put the flashmeter where the film will go and therefore measure the net flux though it will be in f-number units instead of lux-s. The aim of course is to adjust the flash power so that I get an exposure through my step wedge where I KNOW the H value for each step.

    For example, say the formula I posted above was correct (it or my interpretation of it is quite possibly not):
    Ef/Es = 0.18 / N^2
    For example with an f/1 lens, that gives Ef/Es = -2.47 stops. Or at f/8, Ef/Es = -8.47 stops, etc.

    Now I can hypothetically say that if I metered at an 18%-grey subject (filling entire field of view) and metered again at the film plane with an ideal f/1 lens, the reading at the film plane will be 2.5 stops lower. With my objective's aperture matching the Es reading, I can expect Zone V on a normally-developed film. Therefore I know that the flashmeter reading at the film will be 2.5 stops lower than f/1 to achieve Zone V when there is no step-wedge present.

    Now the step-wedge (Stouffer 21-step, 0 to 10 stops): I want the densest step (D=3.01; 10 stops) to be 5 stops below Zone V and the thinnest to be 5 stops above it, so I need an additional 5 stops of exposure. Therefore, if my flashmeter reads f/2+0.5 at the film-plane, I suspect that I should get an appropriate exposure spanning the useful part of the film's HD curve.

    In comparison, the BTZS book (4ed, page 71) says to expose for 0.4s in EV4 light metered at the film-under-test. It doesn't say what the expected wedge densities are or whether there's a correction factor in there for enlarger warmup/cooldown and/or spectral issues. Assuming perfect reciprocity though, I think that's equivalent to a flash meter reading of f/2+0.67, which means my guess is remarkably close to the exposure that Davis recommends for film-testing.

    As a backup, I have purchased a roll of Delta 100 for the explicit purpose of having a fairly-widely-agreed-to-really-be-ISO100 physical reference with good QC.


    PS I'm not seeing a formula in your post, just a list of variables... and u should cancel out entirely unless you want to include bellows factor for being focused at u. Which I don't.
     
  4. ic-racer

    ic-racer Member

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    You may be making that too complicated. "Fitting" the step wedge exposure on the film is really by trial-and-error. For example the Wejex sensitometer has an adjustment pot. Instructions indicate to adjust it so the important part of the curve is evident. One would really need to adjust a sensitometer for each film speed, so you can get the maximum number of steps to plot, but I leave mine the same so I can do some relative film speed comparisons without touching the adjustment.
    Also, the EG&G is 'adjusted' for each film type with ND filters. But again, I keep it set up for 400 film and let the slower films fall off on the end and don't get as many steps to plot out on the shoulder.

    If you really want to try to figure it out with as much math and as little testing as possible you can also work backward from the ISO equation. In this case using iso 100 film. So, you want the 0.1 log d (speed point) to fall on , say step 17 which we will say to be 2.5 log d on your step wedge. You can add the densities if they are both represented as log (2.5log + 0.9log) = 3.4 log. By taking the anit-log of that you need 2512 millilux-seconds exposure on the step wedge.

    Where did the 0.9 come from? That is the exposure needed to give 0.1 log density on the film and comes from the following relationship:
    ISO = 800/(millilux-seconds) (by definition)
    100 = 800/(millilux-seconds)
    millilux-seconds = 800/100
    millilux-seconds = 8
    millilux-seconds = 0.9 log


    The short of it is that I think Phil's number of EV4 is just a suggested starting point. Not part of the whole BTZS calibration parade.
     
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  5. ic-racer

    ic-racer Member

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    Those variables were from the calibration equation which is too complicated to type out. Steve sent me the paper. I can e-mail it to you if you like. Calibration Levels of Film and Exposure Devices D. Connelly, The Journal of Photographic Science, Vol. 16, 1968.
     
  6. Rafal Lukawiecki

    Rafal Lukawiecki Subscriber

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    I went through this only a few weeks ago. Have you seen my thread, including the flash meter calculations, on the last page? In the end, film-plane needed to read less than f/0.7 measured at ISO 100 for a 400 film. Since my meter could not measure flash less than f/1.0 I had to get it to that level, then reduce further using flash power adjustment.

    Watch out for uniformity of flash output at the film surface, and check the repeatability of your flash. Finally, bear in mind the comments about suitability of a flash sensitometry exposure vs. your expected target lighting scenario.
     
  7. Stephen Benskin

    Stephen Benskin Member

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    I'm not exactly sure what you are looking for. There's the way to determine what the log-H is with a step tablet, and there's the classic camera exposure equation. There is also "P" which is the midpoint exposure that the meter aims to make. P = 8 / ISO

    Finding log-H
    log((1 / antilog Step tablet density) * Illuminance) = log of transmitted light or log-H

    Finding Transmitted Light
    Take the speed you have assigned to the film and plug it into:
    0.80 / film speed = transmitted light necessary to create the 0.10 over Fb+f density
    example 0.80 / 125 = 0.0064 mcs

    Finding the Incident Light
    Incident light = transmitted light / transmittance
    Find the step tablet density that created a film density of 0.10 over film base plus fog
    Take that step tablet density along with the transmitted light value and plug it into the equation:

    Incident light = transmitted light / transmittance
    example: 0.0064 / (1/10^2.75) = 3.6 mcs (lxs)

    Calculating Exposure


    From "Is the K factor relevant to me or should I cancel it out?" thread.

    Defining K, part 1.jpg
     
  8. Chan Tran

    Chan Tran Member

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    although a flash meter does not display in lux.sec but if it displays f/1.0 @ ISO it's 2.5 lux.sec with the flat diffuser. So one can convert the f/stop to lux.sec.
     
  9. polyglot

    polyglot Member

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    The BTZS book does say to adjust exposure by trial and error, but then of course the result is entirely uncalibrated wrt true speed. I'm trying to avoid that. I had another poke through the previous thread, which confirms that AV0 = f/1 = 2.5 lux-s.

    I had a read of Wikipedia and it also gives the same S=0.8/Hm formula as you, which is probably (in conjunction with AV0 = 2.5 lux-s) what I need.

    For ISO100 we want Hm=0.008 lux-s for the speed point, plus 9 stops so that it occurs at step 18 => 4.096 lux-s = AV0.71. So my meter would in theory read f/1+0.7 at the film plane, which is about 2 stops less light than BTZS is suggesting that I use. Any idea why the discrepancy?

    PS I'm sure you're meant to take logH from lux not millilux, so the speed point is logHm = -2.1 not 0.9. Doing the computation like this in D instead of AV, we have (step 18) D=2.71, logHm = -2.1, logH = 0.61 => 4.096 lux-s as previously. See also the diagram in that Wiki page that has the speed point at about -2.4 therefore being ISO200.

    Second question is "what is the offset between speed point and metered reading?". In other words, where on that H-D graph is the "average reflected reading" going to fall? Is it at n, i.e. Hn = Hm+1.3? This is basically a re-statement of my original question for this thread, trying to directly relate light readings in lux-s at subject and at film.


    Edit: woah, delayed my posting and there's lots of replies...

    The P=8/ISO is probably what I was looking for and closes the loop for me. That means that Hp=Hm+1.0 and therefore (ignoring flare) the speed point should be about 3.3 stops below the metered reading in a scene.

    As for f/0.7 for ISO400, that agrees well with these calculations rather than the BTZS suggestion. Hm = 0.002 lux-s plus 9 stops = 1.024 lux-s = -1.29AV, which is 1/3 stop less than f/0.7; or f/0.5+0.7 if your meter will read f/0.5 :wink: Did you successfully make a test-strip at that exposure and have the speed-point land where you expected it on the test strip? What if you set the meter to ISO400, that should get it to read a larger aperture: f/1+0.7 at the correct light level (assuming the limitation is the meter's display and not its sensitivity).


    thanks guys. I'll ignore the BTZS suggestion for now and do my first exposure test (for ISO100) at 4 lux-s (f/1+0.7) instead of 16 lux-s.
     
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  10. Bill Burk

    Bill Burk Subscriber

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    I think I wrote that... That's what the EG&G does bare bulb at the film plane. Then you are supposed to add lots of Neutral Density because that's too much light for a 400 speed film. Rafal reduced the light 8 stops to f/1.0 and then a bit lower.

    Calculating Neutral Density doesn't involve difficult math. Because the numbers are logarithms, you add or subtract them. Every 0.3 density is a whole stop. Step wedges with 0.1 density and 0.15 density intervals have steps that are 1/3 stop and 1/2 stop respectively.

    You don't necessarily have to take the antilog to find the actual light in meter candle seconds - you are welcome to do that ... it helps to mentally see the whole picture ... but for practical purposes just count steps and work in stops.

    All that heavy math of meter calibration and optics comes when you try to relate the film tests to light meters and cameras calibrations... If you just want an exposure index and curves, you are going to get that without any of those calculations.

    For the film tests, the only math you need is addition and subtraction, multiplication and division. Make a wild guess and expose a sheet of film through a step wedge and then develop it.

    If you did not give enough light, the high numbered steps will all be clear film. Count how many wasted steps you have (how many clear steps). Multiply by 0.1 or 0.15 (depending on your step wedge steps) and divide by 0.3 - That's about how many more stops of light you need to hit the film with. Same goes for overexposure, try to guess how many steps down you would have to go to hit clear film if every step is too dark. You could graph it or compare it with other tests, and get pretty close. Ideal is when you have one, two or three clear steps in the high numbers 21, 20, 19 - and then you hit 0.1 density on your film test with a step near 19. You don't aim for exact - you aim to get it on there.
     
  11. Bill Burk

    Bill Burk Subscriber

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    Sounds right... just don't do all 5 sheets before you see where it lands.
     
  12. Stephen Benskin

    Stephen Benskin Member

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    You've already answered this. The ratio is usually noted as n sub 1. I did a thread on this subject "What is the relationship between film speed and camera exposure?" I've also attached a paper that you might find interesting, Connelly's "Calibration Levels of Film and Exposure Devices."

    View attachment Calibration Levels of Films and Exposure Devices, Connelly.pdf

    My sensitometer is (or at least was) calibrated. I did use those Log-H equations to determine which step tablet density I wanted the speed point to fall. Worked perfectly.
     
  13. polyglot

    polyglot Member

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  14. Rafal Lukawiecki

    Rafal Lukawiecki Subscriber

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    f/0.7 proved to be too high exposure, and I had not enough of a toe in my tests. I estimated a new exposure, about 1.5 f lower, and that worked well for HP5+. I fudged it with 320TXP, which needed a little less exposure, about 0.5f, but I dialled down too much, and ended up having pretty toe but no DMax for the 4, 5.5, and 8 minute dev times.

    I second Bill's suggestion to shoot just one sheet to test exposure. You can develop it quickly in a tray, even if the main test uses a tank etc. That's what is holding me back from doing it once more (3rd time!) for 320TXP: needing to do 5 tank-loads of several sheets, as I am lead to believe I would get incorrect results if doing less than the full load each time (I usually do 5 or 6 sheets at a time). Also, all of this testing is getting film-cost pricey. :smile:
     
  15. ic-racer

    ic-racer Member

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    Also, note you can convert transmittance, density, illuminance etc to logs then you can add and subtract rather than divide and multiply. So the exposure to your film at the speed point is the sum of the log of step density, log of additional ND over the light and the log of the lamp's meter-candle-second (or lux-seconds)
     
  16. Stephen Benskin

    Stephen Benskin Member

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    Here's a graph showing the standard model of how the scene luminance fits on the characteristic curve. The log-H values are for a 125 speed film.

    Speed Point - Standard Model.jpg
     
  17. ic-racer

    ic-racer Member

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    Metered exposure to ISO speed point = 1.0 log H? Not 1.2?
     
  18. polyglot

    polyglot Member

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    That's what you'd get from the ratio of Hm=0.8/ISO and P=8/ISO. Not that I have any citation/justification for the latter formula other than posts in this thread.
     
  19. Stephen Benskin

    Stephen Benskin Member

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    I'm surprised you don't remember either thread. The 1.0 ratio is the reason why there's a consistent discrepancy between ISO speeds and Zone System EIs.

    The proof for the ratio is with exposure meter calibration and how it relates to the metered camera exposure point.
     
  20. Bill Burk

    Bill Burk Subscriber

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    I see how this discrepancy (between 1.2 Zone System and 1.0 ISO) can explain 2/3 stop (of the 1 stop typical) discrepancy between ISO and Zone System speeds...

    Zone System: Meter Zone V, stop down once Zone IV (0.3), twice Zone III (0.6), three times Zone II (0.9), four times Zone I (1.2)... where you expect 0.1 density speed point to exist on Zone I (four stops below metered point)...

    versus

    ISO: Meter, stop down 1 stop (0.3), 2 stops (0.6), 3 stops (0.9), 1/3 more stop (1.0)... where you expect 0.1 density speed point to exist at 3 1/3 stops below metered point.

    Tell me... That's why I shoot TMY2 at 250 instead of 400... It's that simple?
     
  21. ic-racer

    ic-racer Member

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    I sometimes skip or ignore threads and posts that say "zone system" so I may have missed that. :D I do remember a thread where I was posting about uncertainties in zone exposure being at the beginning, middle or end of the zone, so it was probably in that thread.

    But is the "1.0 difference" a definition? If so, wouldn't all meters be required to have the same "k" constant?
     
  22. Stephen Benskin

    Stephen Benskin Member

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    Dale, the only way a single meter can be used for exposure with both negative and positive materials is through the speed constant. The exposure meter only wants to make one exposure at 8 lux * 1/ISO. The speed constant is necessary to determine where the exposure range falls. For black and white negative film, it's 0.8. With color transparency film, it's 10.

    There are a lot of variables that combine to make K. The K factor thread really goes into detail on this. In a nut shell, many of the factors have to do with the physicality of the meter and the camera optical system. But no matter what the value of K is, the exposure is intended to be 8 / ISO.

    Here's a page that introduces the equation for K and if you're wondering about where the value of 8 in 8 / ISO comes from, check out the value of K1.

    Defining K, part 2.jpg
     
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  23. Stephen Benskin

    Stephen Benskin Member

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    I forgot to mention that the equation Eg = q*Lg / A^2 equals approximately 8. This is the same thing as Sunny 16. Eg * t (shutter speed) = Hg or Hg = 8 / ISO.

    Exposure for Sunny 16.jpg
     
  24. ic-racer

    ic-racer Member

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    Steve, that is great that you can understand that discrepancy (1.0 vs 1.2) and show us where it comes from. All these years I have used 1.2 and did not pickup on it in the papers I read.