Just an odd thought I had today. Does the weight (ie. mass) of a photographic emulsion change post exposure?

You guys need to get out more. Interesting question, though--I take it because the energy of the photon is converted to mass in interacting with the emulsion? Photons themselves are massless, are they not? Man, I shoulda read Feynman's book more carefully.

Here is part of an article I wrote some time ago: (Quote) It is one thing to say photography is a physical process but I emphasise that the assertion is meant completely literally without metaphoric content at all. I do not know why the following calculation has never turned anywhere up in my forty-odd years of photographic research. Maybe my doing it here is the first time ever. How much subject matter hits the film? An approximate calculation goes like this: From published characteristic curves for film a middling exposure in Lux.seconds is 10 times the reciprocal of the numerical ISO speed. This secret (?) formula is how light-meters are calibrated as exposure-meters. For an ISO 100 speed material this middling exposure is 0.1 Lux.seconds. How much energy is there in 0.1 Lux.seconds? Converting Lux, an illuminance unit that says "how bright looking" to an irradiance unit, Watts per square metre, that says "how much energy" relates to the photometric efficiency of the human eye. This varies with wavelength and is maximal at 555 nanometres for a photoptic (daylight) adapted eye. At this wavelength 0.1 Lux.seconds equals 0.00015 Watts per square metre.seconds. Of course Watts multiplied by seconds equals Joules so the previous quantity becomes 0.00015 Joules per square metre. An 8x10 inch sheet of film has an area of about 0.05 square metres so if it was ISO100 material getting middling exposure at 555nm wavelength it would absorb 0.0000075 Joules of energy. All this energy has come from the subject matter and it is important that the light-tightness of cameras and film holders ensures that the energy comes from nowhere else. So far so good; the next part is courtesy of Albert Einstein. E = M.C squared ! Those 7.5 x 10 to the -6 Joules have a mass equivalent that is easily calculated by dividing them by the square of the speed of light (3x10 to the 8 metres.sec-1 )2 . A few deft pokes of the computer calculator keys yields: 8.3x10 to the -23 kilograms. This is the minimum amount of subject matter that has to embed itself in the film. Because, in practice, wavelengths of lesser illuminance value than 555nm are involved more kilograms of subject matter has to hit. Going even further in this argument one can calculate the mass of a latent image. Assuming a quantum efficiency for film of (say) 1% (with 99% of absorbed energy just leaking away as heat) then a 8x10 sheet of ISO 100 film receiving middling exposure gains and keeps, at minimum, a mass of: 8.3x10 to the -25 kilograms. These kilograms are the mass equivalent of the up-changed chemical potential energy of exposed film compared to unexposed film. One may say the kilogram quantities are tiny but it is also true that they are incommeasurably greater than zero. Things are different in the digital world where pictures are made from information. Information is pretty certainly (short of absolute theoretical proof) massless. Research on this point continues and I suspect (don't quote me) that the final outcome will be that information has no energy (mass) but energy has to be expended to cancel a data point and write an alternative one in its place.(Unquote) The mathematical formatting in this extract is clumsy but those folks familiar with the concepts will know what the numbers mean. And, of course, the arithmetic is worth checking.

How would one ever prove this? Exposing film, then weighing it in the dark...nah, I doubt a 'scale' can weigh the tiny amount you're talking about. Surely the physical/chemical changes from processing has a much larger effect.

Thanks Maris, Now I can tell if my exposed film will need push, normal, or pull processing! Seriously though, Thanks for the answer. Doyle