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I was experimenting a lot with multiple exposures on the same frame on film and noticed an unexpected behaviour. While this might very well have to do with my cameras shutter not working properly, it's been so consistent that I thought I'd ask the community if I am missing something.

I am using a Minolta camera from the 80s and I'm shooting on Fuji color reversal film. When I took two picture on one frame I noticed I could get away with taking two pictures at "normal settings" (as in: what I would set the camera to if I just took a single picture) on the same frame and it didn't really look awfully overexposed.

Now, when I took several pictures on one frame I figured I'd just double the shutter speed if I took two pictures or quadruple it to take four pictures and so on (I'd actually just multiply the film's iso by the number of exposures and set the camera to that value). Using that system, I noticed the more exposures I did on one frame the seemingly darker it became.

Am I missing something here? I thought evenly spreading exposure time over all exposures would be the same as having one exposure with the full time but maybe the film has some properties to it that defy that logic?

I'd be thankful for clarification.

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  • \$\begingroup\$ A key point in running doubles (or triples) on the same frame is in the subject matter itself. If I take a low-key portrait using Rembrandt lighting, for example, then I've only properly exposed a very minimal amount of film...allowing the next shot to expose those unexposed bits of the first shot...thus it doesn't appear badly overexposed at all. So, my question is...what was your subject matter and did you plan for this type of situation? \$\endgroup\$
    – OnBreak.
    Nov 11, 2019 at 19:01
  • \$\begingroup\$ Of course you're right. I accounted for that. I am talking about the case when the subjects of each exposure overlap. \$\endgroup\$
    – Jeffrey
    Nov 11, 2019 at 19:48

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Film has an exposure curve. I'm guessing that when you are combining many multiples the amount of light is insufficient to generate any real density (you are on the heel/bottom of the curve). For your simple division method to work you need to be w/in the linear response region.

enter image description here

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  • \$\begingroup\$ Ok but how is making one continuous 4 second exposure different from making 4 separate 1 second exposures. For the sake of the example all of them with the same aperture setting and of the exact same subject \$\endgroup\$
    – Jeffrey
    Nov 11, 2019 at 21:54
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    \$\begingroup\$ @Jeffrey I would guess (no idea if I am right) it is similar to how you can boil kettle by putting it on fire for 4 minutes, but not by turning gas on and off in 4 one minute tries. \$\endgroup\$
    – Gnudiff
    Nov 11, 2019 at 22:46
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    \$\begingroup\$ @Jeffrey Viewing the total exposure as a superposition of the separate exposures would only work if the relation between time/negative density were totally linear. It is to some extent but not at the extremes (very fast shutter speed or longer than, say, 1 second). For a mathematical example take the inverse tangent function (arctan). 4*arctan(0.1) is very close to arctan(4*0.1), but 4*arctan(1) evaluates to almost 2.4 times the result of arctan(4 * 1). Basically put, your exposures are probably on the "flat" start or end of the curves Steven posted and the linear relation breaks down. \$\endgroup\$
    – G_H
    Nov 12, 2019 at 13:09
  • \$\begingroup\$ As you use reversal film I expect this to be even more pronounced. Slide film is known to be more susceptible to under- or overexposure than negative film. In the graph this comes down to a steeper slope between the flatter parts at the low and high ends of exposure. Maybe two 1/500 second exposures will produce roughly the same density as one 1/250 second exposure (all else equal), but four 1/2000 second exposures could be much darker than one 1/500th exposure. At the other end, four 1/2 second snaps could be fine while one 2 second snap is overexposed. Look up "reciprocity failure" too. \$\endgroup\$
    – G_H
    Nov 12, 2019 at 13:28
  • \$\begingroup\$ @ Jeffrey, On the response curves shown plot 1 second at the 0 LE vertical line (where the curves begin to rise). Plot 2 seconds at the -1 LE line, and 4 seconds at the -2 LE line. It should be apparent how two 2 second exposures can equal a 4 second exposure along the linear response (straight line) segment; whereas 4 one second exposures in the flat response section will not add up to anything near the 4 second exposure. (none of that is technically/specifically correct, just representative/corollary) \$\endgroup\$
    – Steven Kersting
    Nov 12, 2019 at 14:52
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Anything past one second, whether in a single exposure or over several multiple exposures of the same piece of film, are subject to what is known as reciprocity failure. It's also called the Schwarzschild effect after the man who extensively studied it in the 1890s, Karl Schwarzschild.

The sensitivity of films at longer exposure times is not linear. This must usually be taken into account when exposing film for longer than one second. This can very significantly impact exposure times, and it varies by the specific film in question. The manufacturer of your film should be able to provide information regarding how much compensation is needed for longer exposures. If you are doing multiple exposures, then you need to add the total time of each exposure together when consulting the manufacturer's data sheets for your film.

Most film manufacturers publish data sheets for each of their films that outline development times for shooting the film at different speeds as well as for developing the film when it is shot at the advertised sensitivity. They also include data regarding exposures longer than about one second (for most films) that are affected by the Schwarzschild effect. Each film has different characteristics, and how much compensation must be made for long exposures can vary significantly from one film to the next.

From the Wikipedia article for reciprocity:

At very low light levels, film is less responsive. Light can be considered to be a stream of discrete photons, and a light-sensitive emulsion is composed of discrete light-sensitive grains, usually silver halide crystals. Each grain must absorb a certain number of photons in order for the light-driven reaction to occur and the latent image to form. In particular, if the surface of the silver halide crystal has a cluster of approximately four or more reduced silver atoms, resulting from absorption of a sufficient number of photons (usually a few dozen photons are required), it is rendered developable. At low light levels, i.e. few photons per unit time, photons impinge upon each grain relatively infrequently; if the four photons required arrive over a long enough interval, the partial change due to the first one or two is not stable enough to survive before enough photons arrive to make a permanent latent image center.

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