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If I take one photo with long exposure and dimmer lights (e.g. constant lighting) and then another photo with short exposure but lights brighter to the level enough to compensate for the shorter exposure (e.g. with a flash) - what differences between the two photos should I expect to see considering that in the both shots exposure time and light intensity are balanced so that the overall brightness of the photos is the same?
For example - would it affect noise level? Will long-exposure photo have richer details because the sensor was able to take more correct light measurements at a prolonged period of time? If I want to maximize quality of the picture in terms of noise - what approach should I pick?

Would be interesting to hear both theoretical and practical answers.

  • What kind of light? If it is well controled artificial light or natural light outside affects things like color temperature. – lijat Jun 11 at 19:19
  • Easy experiment to perform, assuming that you have a light source that can be dimmed without changing its shape or its color. – Solomon Slow Jun 12 at 11:08
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    This seems to be a round-a-bout way of asking about digital noise in long exposures. There are Many questions about that on this sight with good and correct answers. – Alaska Man Jun 12 at 17:31
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I do a lot of technical imaging and I know about image noise. You question is to the point, let me give you a to the point answer.

The difference in noise will probably be small and favorable in the short + bright exposure case.

If I simplify a bit, noise basically comes from:

  • noise related to the amount of photons of light captured during the acquisition (photon noise). This has to do with the physics of light. Because you suggest the same overall brightness this noise is equal in both bright + short and dim + long cases.
  • noise related to the acquisition of each image (eg. read noise,...). Basically this is noise that will be the same for every image, regardless of exposure.
  • dark noise caused by dark current in the photo diodes - pixels. This part is proportional to time as it is basically a value that creeps over time. The rate of creep is mostly constant but different for different pixels. It can be mostly compensated for by using "dark frame subtraction". This is basically taking a long exposure image in pure darkness and subtracting this from your actual photo. I say 'mostly' because dark noise is related to sensor temperature etc. This is why telescopes use super cooled sensors.

In practice I would say you don't have to bother yourself with dark noise for exposures under 1s or even much longer. So, if you are talking about a difference between 1/200 s v 1/20 s, do not bother.

You can experiment by covering your lens and taking "dark frame" images for different exposure lengths to see the effect of dark noise for your sensor.

Aside from noise, you will have motion blur at longer exposures.

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  • That answer actually answers my question, thanks! – Gill Bates Oct 30 at 20:52
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In properly exposed photographs, the only difference you (the viewer) should see is if something in the frame was in motion during the exposure. A sufficiently fast exposure will effectively freeze the object in motion. A longer exposure will blur the object.

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  • What about noise? – Gill Bates Jun 11 at 20:09
  • Digital or film camera? – BobT Jun 11 at 20:11
  • Digital, but an answer for film would also be interesting. – Gill Bates Jun 12 at 13:46
  • There is no such thing as noise in film, noise is digital only. In film there is grain and it is a completely different animal, not related to digital noise in any way. – Alaska Man Jun 12 at 21:52
  • @AlaskaMan, they both related in the sense that they are both result of compensation of imperfect light measurements. – Gill Bates Jun 14 at 9:25
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Film cameras make hardly a difference between long exposure and short exposure (after all, you can store a film for years in the dark). Digital cameras have an inherent noise level that makes longer exposures trickier. Many limit their maximum exposure times to the order of a minute. A frequent phenomenon are "hot pixels" that change state without light influence over time. This may well be a temperature-dependent effect: some cameras do "dark frame subtraction" for longer photographs by taking a photograph with closed shutter and the same duration after the main photograph and then subtracting the image, but once a hot pixel becomes saturated, this of course does not work any more.

So for really long exposures, you want film, or take a number of digital photographs and average them in post-processing.

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  • "Many limit their maximum exposure times to the order of a minute" I'm not too sure about this. Even my E-PL2 has bulb exposure mode that keeps the shutter open as long as you want. – wilkgr Oct 23 at 0:46
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    @wilkgr E-PL2 does not have bulb exposure "as long as you want" but limited to 30m. Admittedly more than the order of a minute. – user95069 Oct 23 at 14:15
  • @user95069 Fair. I forgot about the bulb timer. My bad, sorry. – wilkgr Oct 25 at 5:20
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Yes, film exposure is affected by unusually long or short exposures (but not for normal photo exposure ranges). This is called reciprocity failure, see https://en.wikipedia.org/wiki/Reciprocity_(photography)

Digital is not affected by reciprocity, but noise becomes a problem at long exposures.

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  • If exposure can be calculated to accommodate reciprocity failure, will there be any discernible difference between photos, assuming non-moving subject and appropriate light placement? – xiota Oct 26 at 9:21
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+100

@BobT made a good point about the effect of shutter speed (or exposure duration).

Light fall-off and the Inverse-Square Law

There is another aspect that hasn't been mentioned in the responses ... and that's the topic of light "fall-off".

The amount of light that lands on a subject change based on the distance between the light source and the subject. The change is based on the inverse square law.

Camera lighting is usually not the only source of light for a scene. There usually are some sources of ambient light. You can think of the ambient light as continuous lights. The longer the shutter remains open, the more continuous light will build up on your subject. Photo strobes, on the other hand, are not continuous lights. Your camera shutter opens, the strobe fires, and regardless of how much longer the shutter remains open it will not receive any additional light from the strobe.

Suppose you get an acceptable exposure without flash as long as the shutter is open for a long enough period. You decrease the duration of the exposure (which results in less ambient light) but you supplement with a photo strobe. If the power on the photo strobe was set to allow for a correct exposure of the intended subject, you may notice that anything in the scene nearer to the camera is especially bright and possibly over-exposed. Meanwhile, things in the scene located farther from the subject will appear much darker and possibly black.

These two photo examples (the "with" and "without" flash) wont actually resemble each other at all. The supplemental light changes the ratio of light from ambient (continuous light) sources vs. strobes (momentary sources).

In general, if you place a light source and set its intensity to correctly illuminate your subject at some distance, then you can calculate the amount of light that will illuminate other elements of your scene based on their distances from the same light source using the relationship described in the formula below.

Inverse Square relationship

For example, suppose you have a photo of a person who is located 10 feet away from the light source. There is a person a bit nearer to the camera and they are located 7 feet away... and another person located 14 feet away. Here is what would happen:

In the first instance where a 2nd subject is located 4 feet farther away (14 feet from the light) the equation looks like this (I'll use the value of 1 for the correct amount of light intensity because it makes it easier to see what the intensity is of the other subjects):

14 vs 10 equation

This is solved as 14 vs 10 solved which works out to .51 ... so the person located 14 feet away would get only about half as much light as your main subject.

Meanwhile another person located 3 feet nearer than your intended subject looks like this:

10 vs 7

This is solved as 10 vs 7 solved which works out to 2.04 ... so the person located 7 feet away would get twice as much light as your intended subject. Also note that means that the person 14 feet away is only 1/4 as bright.

This is what is meant by the light fall-off problem.

You can decrease the severity by moving the flash farther away. So imagine the subjects are all still 7, 10, and 14 feet away from the camera location ... but the strobe is located 20' away from the main subject. So the distance from the light source to your subjects is now 17, 20, and 24' away.

If we assume we have the light adjusted to properly illuminate the subject at the 20' distance then

... the person at the 17' distance receives 1.38 times more light (a little more than a third brighter)

... the person at the 24' distance receives .69 times as much light (about a third dimmer)

And those differences aren't too bad. This assumes the light source and any modifiers (reflectors, soft-boxes, etc.) can adequately light the subject(s) at those distances.

(On a side-note, Photography StackExchange doesn't support LaTeX (some other StackExchange communities do) ... so the equations had to be generated as .png files and imported as images.)

Noise Differences

There are many kinds of noise. But generally the type that dominates most images tends to be read noise. The amount of "read" noise in an image is fixed. Extremely long exposures can result in a build-up of heat and that introduces a new type of noise ... but for purposes of this answer I'll assume we are discussing exposures limited to a few seconds ... rather than several minutes.

Noise becomes noticeable when signal is insufficient (e.g. under-exposure). When the signal is boosted to compensate for under-exposure, the noise is boosted as well. Essentially if the image had a poor "signal to noise ratio" (SNR) then boosting the signal will boost the noise and now the noise is noticeable. If the signal was already sufficient (not an under-exposure) then it doesn't require boosting and this means noise isn't boosted ... so it isn't noticeable (it is there ... but you probably wont see it).

This means discussions of whether or not there are differences in noise are really discussions of whether the two exposures had a different SNR.

If exposure 1 was adequately exposed but did not use flash (or other supplemental lighting ... only already present ambient lights) and if exposure 2 was half as long (not a sufficient exposure based on ambient light ... but supplemental light was added to compensate) then those two photos could have the same SNR ... which means there would not be a different in noise.

Noise is mostly the result of insufficient exposure. Keep in mind that ISO in digital photography is a gain applied to the image but... this gain is not applied under after the shutter closes and the exposure is complete. It is convenient to discuss ISO as if it were part of exposure, but it technically a post-exposure process and not part of true exposure. In other words, boosting ISO to compensate for lack light isn't really an increase in true exposure. When I mention that noise is generally the result of insufficient exposure ... it means I'm not counting the ISO boost as a true part of the exposure (and it also explains why photos taken at high ISO seem to have more noise. They have the same amount of noise ... it's just that the noise was amplified to make it more noticeable.)

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  • The question states "brighter to the level enough to compensate for the shorter exposure". Inverse square law is about light lost to distance, but the question implies that this is already taken into account. – xiota Oct 26 at 9:16
  • @xiota, the OP used the example of a flash ... which involves light fall-off. The most significant differences will be motion blur and light fall-off. – Tim Campbell Oct 26 at 15:21
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Yes.

Let's start with the basics: shutter speed has two potential uses. It can be used to control exposure, and it can be used to depict the passage of time for artistic purposes. Short shutter duration freezes action. Long durations show moving items as a blur or ghostly effect. Carefully chosen shutter speed can also be used with panning to keep the subject sharp while blurring the background to show movement. If you use brighter light instead of varying the shutter speed, you are choosing not to exercise the artistic control it can give you.

If we look solely at exposure, shutter speed controls how long you let light into the camera. But the digital and film results are different given the ways the two media record light.

Digital sensors also capture noise that is inherent to the photon sampling process they use. Longer exposures = more time for noise to accumulate. Also, noise affects shadows much more than highlights, because a 1-bit error on a shadow luminosity reading of (say) 3 is far more significant than a 1-bit error on a highlight at 254. So using brighter light can result in a lower noise image on a digital sensor. An alternative is to take several sequential images of shorter duration, and then average them. Because thermal noise is random, it should cancel itself out over many images of the same scene -- provided nothing is moving.

Film on the other hand captures light using a chemical change. Noise does not accumulate with length of exposure -- it's inherent to the medium itself: chemical grain. So brighter light and longer exposure can be interchangeable, except at the extremes.

Film has a "roll off" response to light intensity at very low and very high ends: reciprocity failure. It takes longer to blow highlights on film, and it also requires more time to capture very dark shadows. Thus there can be a visible difference between equivalent exposures that trade brighter light for shutter speed.

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