Let's say I have a micro-4/3rd camera and a full frame camera, both set to 1/60 at f/2.8, taking a picture of the same scene in the same lighting. Will the exposure be the same across both cameras despite the different sensor sizes?

The reason why I'm asking is because of the difference in depth of field between micro-4/3 and full frame sensors. I'm finding that, in order to take a picture of certain scenes with the full frame camera at the same depth of field as the micro-4/3rd camera, I have to increase the aperture, which in turn forces me to crank up the ISO.

  • 1
    \$\begingroup\$ You've good answers but I'd like to point out something you may find interesting. Even though you can get two pictures with the same exposure, they may not look the same due to different dynamic-ranges. You can have one camera with 9 stop DR and the other with 14 stops now. By squeezing 9 OR 14 stops of DR into a medium of fixed DR (such as an LCD display or print), the tonalities you see won't be the same. \$\endgroup\$
    – Itai
    Commented Jan 28, 2011 at 19:10

4 Answers 4


Yes. Exposure is based on the amount of light that hits any given point on the sensor (or film), not the total amount of light for the whole area. (The light hitting the corners doesn't have any effect on the light hitting the center, or anywhere else.) Or to put it the other way around, a full-frame sensor records more overall light, but for the same exposure, it's exactly as much more light as there is more sensor area.

Think of it this way: if you take a full-frame image and cropped out a small rectangle from the middle, the exposure there (ignoring vignetting and light falloff) is the same as the exposure for the whole thing.

Now instead of cropping, imagine replacing the full-frame sensor with a smaller one. Same exposure, just less of the image recorded.

Of course, a cropped image does have less light overall. The secret is that we "cheat" when enlarging. We keep the brightness the same, even though the actual number photons recorded per area is "stretched". That is, if on the sensor, 200 million photons collected in a square represents a medium gray, if we print so that square is 10"×10", we don't spread the brightness out making it much dimmer — we instead keep the brightness so it's the same gray.

Also, yeah, you have to increase the ISO (or shutter speed) to get the same final image brightness with a smaller aperture for higher depth of field on a larger sensor. But, assuming roughly equal technology, the larger sensor should give about the same amount of noise at that higher ISO as the smaller one did at lower sensitivities.

In concession to the long comments thread below, I will add: if you're literally comparing two camera combinations in the real world, the exact exposure may vary for several reasons. One of these is the actual transmission of light for a given lens at a certain f-stop — the lens elements themselves aren't perfect and block some light. This differs from lens to lens. Second, the lens makers round to the nearest stop when stating aperture, and may not be perfectly accurate. Third, the accuracy of ISO varies from manufacturer to manufacturer — ISO 800 on one camera may give the same exposure as ISO 640 on another. All of these factors should be (even cumulatively) less than a stop. And most importantly, these factors are all independent of and unrelated to the sensor size, which is why I left them out of the original answer.

  • 1
    \$\begingroup\$ Hang on... It seems like there would be additional variables to consider, right? I would have said that their exposures wouldn't necessarily be the 'same' unless both cameras were using the exactly the same lens. Is my logic flawed there? \$\endgroup\$ Commented Jan 28, 2011 at 3:40
  • 1
    \$\begingroup\$ As long as the f-numbers are the same between lenses, and ignoring things like manufacturer tolerances and actual-transmission factors, it will be exactly the same. At the same shutter speed and iso, f/2.8 on my iphone will give the same exposure as f/2.8 on a 4×5-format camera. Even though the latter has over 800× the surface area. :) \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 4:02
  • 1
    \$\begingroup\$ But that's an individual lens thing, not a format issue. It could well be that the micro 4/3rds lens being compared error on the side of brighter. For the purposes of the answer to the question, assume that all cows are spherical.... \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 4:34
  • 3
    \$\begingroup\$ :) But, again, the question is if the sensor format makes a difference, and the useful answer is that it doesn't. \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 4:51
  • 3
    \$\begingroup\$ I'm a sysadmin, so I'm hands-on in the real-world side of things, not a theorist. :) But I disagree with you here. The entire point of having standardized stops is so exposure can be compared regardless of particular equipment. You can buy a light meter that tells you that for a given shutter speed and ISO, set your aperture to f/X. That value is correct for any format, and that's important! The fact that individual equipment might vary from the standard is also useful, practical knowledge, but it's not helpful to just throw your hands in the air and say "it's all different so you can't tell!" \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 5:37

Let's say I have a micro-4/3rd camera and a full frame camera, both set to 1/60 at f/2.8, taking a picture of the same scene in the same lighting. Will the exposure be the same across both cameras despite the different sensor sizes?

Yes - if it's the same lens or both lenses have the same transmission, and assuming that by saying "same exposure" you're using the same ISO rating (to even out differences in sensor efficiency).


  • Same ISO doesn't mean same noise level.

    Different sensors operating at the same ISO level will capture different amounts of light but turn them into the same exposure. However, even though the exposure is the same, the ability to resolve detail amongst the noise will be different. The ISO rating system is designed to factor out differences in sensor efficiency so you can set any sensor regardless of size or efficiency to ISO200 and get the same exposure. To achieve this, a full frame sensor working at ISO200 is gathering a lot more light than a 4/3 sensor at ISO200 for the same scene, and it is just internally applying a different amount of gain in order to translate the scene into the same brightness values.

    All will look equivalent in the end result in terms of exposure, except that the full frame will have lower noise levels since it started with more light information. Note that there can be differences in efficiency between sensors of the same size, too; hence it's not related solely to sensor size, though that is the major factor. In short, ISO 800 in FF is the same exposure as ISO 800 in 4/3, but you'll get different noise and dynamic range on them since it's not the same sensor efficiency.

  • Same f-stop doesn't necessarily mean same lens transmission.

    The common method of determining how much light comes through the lens is an f-stop. However, this measure is based on the diameter of the aperture, but does not take into account the transmissive properties of the lens elements (that is, how much light is absorbed by the glass in the lens). All lens glass absorbs some light. Modern lenses with multiple coatings absorb a good deal less, and it's not uncommon for a simple modern lens to transmit more than 99% of light.

    Without filters, the effect of transmission loss in a modern multi-coated lens is so small that in almost all cases it can be ignored, making this little more than an academic exercise with little practical value. Those cases in which it can't be ignored may include shooting for the cinema, where multiple consecutive shots should have the same exposure even though they may use a very different lens. That's why t-stops were invented; they're like f-stops by they take into account transmission properties of all your glass.

  • 2
    \$\begingroup\$ To add to all this t-stop discussion: there's no inherent reason that a full-frame lens would have a higher or lower t-stop relative to the f-stop than the equivalent micro-four-thirds (or other) lens would. That's literally a completely separate factor from sensor size. \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 5:43
  • \$\begingroup\$ Yes. Whether the same lens is used was not specified. It is indeed only relevant if you are also talking different lenses; it's not tied to sensor size. \$\endgroup\$ Commented Jan 28, 2011 at 6:10

Note: The following answer was originally written in answer to another question that, while very similar to this one, was concerned specifically with the differences between sensor sizes when shooting in low light situations.

Will 1 Inch sensor give same exposure at same aperture and ISO settings compared to APS-C sensor?

Exposure is a measure of the field density of light. The means it is an expression of how much light is captured per unit of area.

If you have the same ISO, f-number, and shutter time you will get the same exposure. There can be minor differences due to the inaccuracies of different cameras with regard to actual ISO, Shutter time, and aperture as well as the varying amount of light that is lost as it travels through various lenses. But for creative photography purposes anything within about 1/6 to 1/3 stop is viewed as close enough.

What you lose with a smaller sensor, especially when shooting in very low light conditions is the total amount of light collected. When the field density of light is the same, the amount of light falling on each square millimeter is the same, but the sensor that is four times as large in terms of area collects four times as many photons spread over four times the area. Assuming the angle of view is the same with both cameras due to different focal length lenses, the brightness of each mm² will be the same but the larger sensor produces a larger image. This is significant when we enlarge the image from the size it is on the sensor to the size with which we wish to display it.

If the images from both sensors are enlarged to the same display size, the image from the larger sensor requires less enlargement than the image from the smaller sensor. When images are enlarged from the size they are projected onto the sensor everything gets enlarged: the image from the light that was projected onto the sensor and recorded, the noise generated by the camera, the noise created by the random nature of light, blur due to motion and focusing/DOF issues, and any optical imperfections due to the lens.

So in the end what a larger sensor gives you is the ability to enlarge less to get to the same display size which means all of the imperfections in the photo are not as magnified as they would be with a smaller sensor.

For some situations, though, there are techniques that will allow the performance of both the smaller and larger sensors to be improved. Shooting at lower ISO for a longer exposure, for example, will reduce the influence of photon shot noise. Of course that might require a tripod or other means of stabilizing the camera. Using dark frame subtraction can reduce the influence of constant read noise produced by the camera. Stacking multiple images of the same scene will reduce the random noise in each frame. Stacking almost certainly requires a tripod. But any improvements you make using the smaller sensor can also be made using the larger sensor. Thus the larger sensor will always maintain its light collecting advantage when both are based on the same technology.


Shutter speed is an easy component of exposure to understand. Halve the shutter speed and you get half the amount of light striking the sensor. 1/50th on a small sensor yields the same amount of light per square meter as on a large sensor. The large sensor merely captures a larger area of it.

Field of view and aperture is an interesting component of exposure. This is why aperture is a relative size to focal length. If it wasn't, we'd need calculators in our pockets every time we changed it.

Imagine you have an aperture diameter of 5mm (78.5mm² area) and you increase your field of view by a factor of two (30º to 60º). This now increases the amount of light striking the same area by a factor of four (pi.R²), which would mean either your ISO would need to come down a factor of four, or your shutter speed shorten by a factor of four.

Now, if you keep the physical aperture size directly proportional to the field of view (determined by focal length & sensor size) you are cancelling out the field of view component. This is where the f-stop comes into play. All that matters now is the ratio. When your aperture is 1/2.8 the size of the focal length, for example, the same amount of light at a given shutter speed will strike the sensor regardless of focal length.

This means the aperture is getting physically smaller at wide angles (zooming out) and larger at smaller field's of view (zooming in).

How does this work on small and large sensors? Well on a large sensor the same field of view (cone of light) is restricted the same amount by the lens's aperture, but it is expanded to cover a larger are on the sensor.

ISO on the other hand is a standard. It determines a standard exposure at any given shutter speed and aperture.

Edited for clarification

The reason why a large sensor is able to produce a less noisy exposure is because the area of each pixel is larger (sometimes significantly larger). What this means is that the level of signal (light) compared to the level of noise hitting each pixel is greater. Think of it as a bucket of water with the same amount of soot at the bottom. A 5L bucket will have more water than soot versus a 2L bucket, increasing the usefulness of that bucket.

This is signal-to-noise ratio (SNR). On a point and shoot, the ratio of signal to noise is considerably less. Doubling the ISO for all intents and purposes halves the SNR. Because of these big bucket photosites on a digital SLR, ISO can be expanded considerably higher and still achieve less noise than a point and shoot, despite the same volume of light striking the sensor chip.

Phew. That's confusing stuff.

  • \$\begingroup\$ This is a fine answer as answers go, but I think it's the answer to a different question — the question is about sensor size, not focal length, which is a whole separate thing. \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 4:16
  • 1
    \$\begingroup\$ And actually, as I re-read, the part that is relevant at the end is, well, exactly wrong. The amount of light that the P&S sensor gets is just the same as what the equivalent area of a full-frame sensor gets, so amplification is exactly the same. Smaller sensors are noisier because 1) more electronics get packed in a much smaller area and 2) to make a print of the same size, you have to enlarge more (although one doesn't generally think of it that way when working with files) — not because they get less exposure. \$\endgroup\$
    – mattdm
    Commented Jan 28, 2011 at 4:22
  • \$\begingroup\$ That is true. I'll clarify that. \$\endgroup\$ Commented Jan 28, 2011 at 4:24
  • \$\begingroup\$ @Nick Bedford - In your edit part, "...hitting each pixel is greater" should be smaller. In "...ratio of noise to signal is considerably less" should be more. The SNR is higher in larger pixel sizes (larger sensors, same resolution). \$\endgroup\$
    – ysap
    Commented Jan 28, 2011 at 5:00
  • \$\begingroup\$ Thanks for that! Can't believe I got the words round the wrong way. \$\endgroup\$ Commented Jan 28, 2011 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.