If APS-C and similar crop-sensor digital cameras have a focal length multiplying effect such that a 50mm lens has an apparent focal length closer to the field of view of an 80mm on a full frame camera, and yet at the same time the depth of field for the smaller sensor camera is more like the depth of field a 50mm lens would produce on a full frame camera (using the same aperture), then this would seem to suggest the concept of an "aperture dividing effect."

In other words, a 50mm f/1.8 lens on an APS-C camera would act more like a 80mm f/2.8 (approx. 1.8 * 1.6x) lens in 35mm equivalent — for depth of field, not considering exposure.

Can someone with a better understanding of the physics involved clarify this for me. I've never seen this concept explicitly mentioned anywhere, so I am a bit suspect of it.

  • I'm having trouble understanding the question. Do you want to compare DOF when using the 50mm on both cameras from the same shooting position? Or do you want to compare DOF when using the 50mm on the FF at say 10 feet subject distance and the 50mm on the 1.6x crop body at 16 feet to get the same subject size?
    – Michael C
    Feb 4, 2013 at 8:44

9 Answers 9


This answer to another question goes into detail on the math behind this. And there's a Wikipedia article with a section specifically about getting the "same picture" with different camera formats. In short, it is approximately true that adjusting both the focal length and aperture by the ratio of the format sizes (the crop factor) will give you the same picture. ¹

But this breaks down if the subject is within the macro range of the larger-format camera (focusing really close). In this case, magnification (and therefore actual sensor size) becomes crucial to the DoF equation, messing up the equivalence.

And, the Wikipedia article casually mentions but does not elaborate on another important point. The assumption is that for the same print size, the acceptable circle of confusion (roughly, the acceptable blur level still considered in focus) will scale exactly with format size. That might not actually hold true, and you might hope (for example) to get greater actual resolution from your full-frame sensor. In that case, the equivalence also isn't valid, but fortunately in a constant way. (You simply have to multiply in your pickiness factor.) ²

You mention "not considering exposure", and now you might be thinking (as I did): wait, hold on. If cropping+enlarging applies to "effective" aperture for depth of field, why doesn't it apply to exposure? It's well known that the basic exposure parameters are universal for all formats, from tiny point and shoots to DSLRs all the way up to large format. If ISO 100, f/5.6, ¹⁄₁₀₀th second gives correct exposure on one camera, it will on any other as well. ³ So, what's going on here?

The secret is: it's because we "cheat" when enlarging. Of course, in all cases the exposure for a given f-number on any area of a sensor is the same. It doesn't matter if you crop or just have a small sensor to start with. But when we enlarge (so that we have, for example, 8×10 prints from that point and shoot to match the large format), we keep the exposure the same, even though the actual photons recorded per area is "stretched". This also has the same correspondence: if you have a 2× crop factor, you have to enlarge 2× in each dimension, and that means each pixel takes 4× the area of the original — or, two stops less actual light recorded. But we don't render it two stops darker, of course.⁴


[1]: In fact, by changing the f/number, what you are doing is holding the absolute aperture of the lens constant, since the f/number is the focal length over absolute aperture diameter.

[2]: This factor breaks down too, as you approach the hyperfocal distance, because once the smaller format reaches infinity, infinity divided by anything is still infinity.

[3]: Assuming the exact same scene, and minor variations from real-world factors like lens transmission aside.

[4]: Basically, there ain't no such thing as a free lunch. This has the effect of making noise more obvious, and it's a reasonable approximation to say that this increase effectively means the crop factor also applies to noise apparent from ISO amplification.

  • This is exactly the direction in which my thinking was headed, but I had never come across a concise mathematical explanation. Thank you!
    – Sean
    Mar 25, 2011 at 3:20
  • @MichaelClark It's the direct ratio of 35mm format to the sensor size in question. So, yes, a smaller sensor has a larger crop factor.
    – mattdm
    Apr 22, 2013 at 3:51
  • Um. 36:24 = 3:2, or 1.5x. That's FF:APS-C. Which, sure, is the inverse of APS-C:FF.
    – mattdm
    Apr 22, 2013 at 4:13
  • In any case, crucially, whether you call it inverse or not, one adjusts the focal length (as a representation of field of view) and aperture (as a representation of depth of field) in the same direction in tandem for the purposes of this question.
    – mattdm
    Apr 22, 2013 at 4:15
  • 1
    @dmkonlinux No, that's different. Someone decided it'd be more clear to delete half the article about a year ago. Here's the last version of it from Wikipedia history. I'm not sure if it's more helpful to link to the old version of the article or just remove the reference.
    – mattdm
    Jan 9, 2020 at 18:01

Just as using a crop camera doesn't alter your focal length (that being a property of the lens, not camera) but does alter the field of view, there is no aperture dividing effect, a lens with aperture f/2.8 lens is still behaves like a lens with aperture f/2.8 lens for metering purposes, however when matching the field of view of a full frame sensor the depth of field will be the same as a lens with aperture ratio (f/ value) multiplied by the crop factor.


The bigger the sensor, the smaller the depth of field for a specific aperture, assuming you are filling the frame with the subject. This is because you either need to use a longer focal length or get closer to fill the bigger frame.

In order to get the same depth of field with a full frame camera as you do with a cropped factor, you need to multiply both focal length and aperture by the crop factor. So to match a 35mm f/16 on a Nikon APS-C (crop 1.5), you'd need a focal length of 53mm and an aperture of f/24 on the full frame camera.


Yes, a sensor's crop factor can be used when calculating the change in depth of field (DoF) of a lens compared to that lens' use on a full frame (FF) camera. But it will not always lead to an increase in the DoF. If shot from the same distance and displayed to the same size, the DoF for the crop body camera will be reduced (because the image projected on the sensor, including the circles of confusion, will be enlarged to a greater degree). If, on the other hand, you adjust your shooting distance to frame the subject similarly the DoF will increase.

There are so many variables to deal with in this question and most of the answers assume several without specifying those assumptions. This leads to gross misunderstandings about the relationship of focal length, aperture, sensor size, shooting distance, display size, viewing distance, and even the visual acuity of the viewer to Depth of Field (DoF). All of these factors combined will determine the Depth of Field of an image. This is because DoF is a perception of what range of distances from the focal plane are in focus. Only one distance from the focal plane is actually in focus such that a point light source will theoretically produce a point of light on the focal plane. Point light sources at all other distances produce a blur circle that varies in size based on their proportional distance to the focal plane as compared to the focus distance. DoF is defined as the range between the near and far distance from the focal plane that the blur circle is still perceived as a point by the viewer of an image.

We ask questions such as, "How does depth of field change when using the same lens on a camera with a different sized sensor?" The correct answer is, "It depends." It depends on whether you shoot from the same distance (and thus change the framing of the subject) or shoot from a difference distance to approximate the same framing of the subject. It depends on whether the display size of the image is the same or the display size of the image is changed by the same proportion as the different sensor sizes. It depends on what changes and what stays the same in regard to all of the factors cited above.

If the same focal length is used at the same subject distance with the same aperture using the same sensor size with the same pixel density and displayed at the same resolution on the same size paper or monitor and viewed from the same distance by persons with the same visual acuity then the DoF of the two images will be the same. If any one of these variables change without a corresponding change to the others, the DoF will also be changed.

For the rest of this answer we will assume the image viewing distance and the visual acuity of the viewer are constant. We will also assume that apertures are large enough that diffraction does not come into play. And we will assume any printing is done on the same printer at the same number of dpi but not necessarily the same ppi and not necessarily on the same size paper.

For the sake of simplicity, let's consider a couple of theoretical cameras. One has a 36mm X 24mm sensor with a resolution of 3600 X 2400 pixels. This would be an 8.6MP full frame (FF) sensor. Our other camera has a 24mm X 16mm sensor with a resolution of 2400 X 1600 pixels. This would be a 3.8MP 1.5x crop body (CB). Both cameras have the same pixel size and pixel pitch. Both cameras have the same design and sensitivity at the pixel level. In other words the center 24mm X 16mm of the larger FF sensor is identical to the smaller CB sensor.

If you attach the same 50mm lens to both cameras and take a photo of the same subject from the same distance at f/2 (assuming all other settings are the same) and crop the FF sensor image to 2400 X 1600 pixels and print both images on 6" X 4" paper, the two images will be virtually identical, and the DoF will be the same in both photos.

If you attach the same 50mm lens to both cameras and take a photo of the same subject from the same distance at f/2 (assuming all other settings are the same) and print all of both images on 6" X 4" paper there will be some noticeable differences. The image from the FF camera will have a wider field of view (FoV), the subject will be smaller and the DoF will be greater than the image from the CB camera. This is because the FF image was printed at 600 ppi and the CB image was printed at 400 ppi. By enlarging each pixel from the CB camera by 50%, we also enlarged the size of each blur circle by the same amount. This means that the largest blur circle projected on the CB sensor that will be perceived as a point is 33% smaller (the reciprocal of 3/2 is 2/3) than on the FF sensor. If we had printed the FF image on 9" X 6" paper and the CB image on 6" X 4" paper the DoF would have been the same (both printed at 400 ppi), as would the subject sizes in both prints. If we then trimmed the center of the 9" X 6" print to a 6" X 4" print we would again have near identical prints.

If we attach the same 50mm lens to both cameras and take a photo at f/2 of the same subject from different distances so that the subject size is the same and print both images on 6" X 4" paper there will be some noticeable differences. The perspective will have changed because the CB image was taken at a greater distance from the subject. The subject will appear compressed in the CB image compared to the FF image. If background details are visible the background will also appear closer to the subject than in the image from the FF sensor. Because the 50mm lens was focused at a 50% greater distance, the DoF also increased by 50%. If the subject was at 10' using the FF camera and 15' using the CB camera here are the resulting DoF calculations:

  • 50mm @ f/2 from 10' on FF: 9.33' to 10.8'. DoF of 1.45' (17.4"). The DoF ranges from 8" in front of to 9.6" behind the 10' point of focus (PoF).
  • 50mm @ f/2 from 15' on CB: 14.0' to 16.2'. DoF of 2.18' (26.16"). The DoF ranges from 12" in front of to 14.4" behind the 15' PoF.

These calculations are based on a circle of confusion (CoC) of .03mm for the FF camera and .02mm for the CB camera. This is because we are printing at 600 ppi for the FF and 400 ppi for the CB (and the pixels are the same size for both- 0.01mm or 10µm).

In reality, we all know the pixels on most FF sensors are larger than the pixels on most newer CB sensors. They range from 6.92µm on the 18MP FF Canon 1D X to 7.21µm on the 16MP D4 to 4.7µm on the 36MP FF Nikon D800. The crop bodies go from 4.16µm for the 18MP Canon 7D to 3.89µm for the 24MP Nikon D7100 (the D7200 will be around 3.0µm) to 5.08µm for the 14MP Sony SLT Alpha 33. In all cases the pixel size is considerably smaller than the generally accepted CoC of .03mm (30µm) for FF cameras and .02mm (20µm) for 1.5x CB cameras. For 1.6x CB Canon cameras 0.019 (19µm) is generally used. The largest size pixels Canon has used in the last decade or so was 8.2µm for the 12.8MP FF 5D and the 8.2MP APS-H 1D mkII. What all this means is that at the pixel peeping level, focus blur will be visible even for objects within the accepted DoF because the accepted blur circle is anywhere from 4 to 7 times larger than the pixels on current DSLRs. To calculate DoF at the pixel level you would need to use a CoC the size of your camera's pixels which would be much narrower than most DoF calculators use.

  • Why are you so concerned about printing?
    – BBking
    Mar 18, 2015 at 12:56
  • 1
    Print is just a quick way of saying "display at a certain size/magnification". It all applies to display on a monitor as well.
    – Michael C
    Mar 19, 2015 at 12:17
  • 2
    printf("Oh right. I see now");
    – BBking
    Mar 22, 2015 at 10:23

The smaller sensor doesn't change the focal length or the aperture, it just captures only the center part of the image - it's almost the same as taking the full frame image and cropping it to leave only the center.

When you take only the center of the image it looks like you zoomed in - so the field of view of a 50mm lens on a 1.6 crop sensor look like a 80mm on a full frame sensor - but it just looks like that because you only see the center of the 50mm image, the focal length is still 50mm and the image you get is equivalent to the center of a 50mm image not to a true 80mm lens.

The same goes for aperture, a 50mm image taken at f/8 on a crop sensor is the same as the center of a 50mm f/8 image on a 35mm sensor, it is not the same as a 80mm image taken at f/12 (also not the same as 80mm f/8 obviously)

  • Actually, assuming enough resolution, if you blow up that cropped center to the same print size, the resulting DoF is quite close to proportional to the crop amount.
    – mattdm
    Mar 25, 2011 at 1:48
  • @Nir, your answer is correct if the shooting and focus distance is the same.
    – Michael C
    Apr 22, 2013 at 2:49
  • @mattdm, the resulting DoF will be inversely proportional. When you plan to enlarge the size of the print a smaller blur circle projected on the sensor is needed to allow it to be perceived as a point in the print (assuming the same viewing distance). Everything, including the size of the blur circle, is enlarged.
    – Michael C
    Apr 22, 2013 at 3:04
  • Yes, I phrased that poorly.
    – mattdm
    Apr 22, 2013 at 3:21

There is no "focal length multiplier effect", period. The focal length of the lens does NOT change magically because you use a smaller or larger sensor, it stays exactly the same.

All you get is an image that's cropped from the one you would have gotten had you used the same lens to record an image on a larger size sensor. The DOF will thus be the same it would have been had you used that larger sensor as well.

  • 1
    Not if you make the same sized print from two different sized sensor it won't! You're magnifying the image from the smaller sensor by a larger factor, which means a smaller blur circle is needed to allow it to be perceived as a point on the print.
    – Michael C
    Mar 18, 2015 at 0:22

In other words, a 50mm f/1.8 lens on an APS-C camera would act more like a 80mm f/2.8 (approx. 1.8 * 1.6x) lens in 35mm equivalent — for depth of field, not considering exposure.

Yes, a 50mm f/1.8 lens on an APS-C camera would act more like a 80mm f/2.8 (approx. 1.8 * 1.6x for Cannon) lens in 35mm equivalent, as far as DOF and to some extent image noise levels are concerned, presuming same shutter speed and reframing to compensate etc..


Yes, the span of the depth of field is exactly and inversely proportional to the crop factor (assuming all else is equal (focal length and focus distance and f/stop equal), and assuming CoC is computed from sensor diagonal.

This is easy to see in the calculator at http://www.scantips.com/lights/dof.html

This is because DOF is based on the final enlargement of the image, and smaller sensors require greater enlargement (to compare at the same size).


I did some comparisons using an on-line Depth-Of-Field Calculator. You have hit on something I did not know; good for you! As you have discovered, multiply the f/number by 1.6 works out to equivalent Depth-Of-Field. I am fascinated by this and I need to investigate the whys and wherefores.

To compare apples and oranges as to depth-of-field for two different formats, you must use a different criteria for the size of the circle of confusion. We are talking about the fact that a lens handles each point on the subject separately and then projects that point on film or digital chip. This tiny circle of light is the smallest fraction of an optical image that contains intelligence.

In order for us to pronounce some portion of the image as “sharp”, that image must consists of circles that are so tiny we cannot make them out as a disk, we see a point with no dimension. Newspaper pictures are made with too large points of ink, we say , newspaper images are not sharp. How big is the max size of the circles of confusion? They must be 0.5mm in diameter or smaller as viewed from normal reading distance. That means that a full frame (FX) must have a lens that projects circles that are small enough to tolerate enlarging. Kodak used a circle size of 1/1750 of the focal length and Leica used 1/1500 of the focal length, for crucial work. Using a fraction of the focal length is the industry standard way to do the computing as it mostly takes into account the degree of enlargement needed to make an 8X10 print or computer display. In other words, the circle size remain below 0.5mm after the miniature image is enlarged and viewed from normal reading distance.

Now the Kodak and Leica standards are too stringent so the industry normally uses 1/1000 of the focal length for everyday work. That works out to a circle size of 0.05mm for the 50mm lens and a circle size of 0.08mm for the 80mm.

Derived from on line Depth-Of- Field computer using these two circle sizes:

50mm @ f/1.8 focused 10 feet DOF 9.05 thru 11.2 feet circle of confusion 0.05mm 80mm @ f/2.8 focused 10 feet DOF 9.05 thru 11.2 feet circle of confusion 0.08mm

50mm @ f11 focused 10 feet DOF 5.96 thru 31.1 feet circle of confusion 0.05mm 80mm @ f/18 focused 10 feet DOF 6 thru 30feet circle of confusion 0.08mm

50mm @ f/4 focused 10 feet DOF 8.07 thru 13.2 feet circle of confusion 0.05 80mm @ f/6.4 focused 10 feet DOF 8.09 thru 13.1 feet circle of confusion 0.08

The 1.6 crop factor is actually a multiplying or magnification factor. The FX frame measures 24mm by 36mm with a diagonal measure of 43.3mm. The your APS-C measures 15mm by 22.5mm with a diagonal of 27.0 The ratio is 43.3 ÷ 27.0 = 1.6 (crop or magnification factor). By the way that’s 1/1.6 X 100 = 62.5%. The APS-C is 625% of the size of an FX.

Lots of math, I call it gobbledygook ! I can say this -- turned 79 today!

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