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I have often heard it said that to have a blurred background, it is beneficial to shoot full frame. However, is this really true?

The useful largest aperture is anyway limited by the amount of depth of field that can't be too shallow or else parts of the subject are not in focus, and therefore, I want to know if the relationship between background blur and depth of field favors full frame over crop sensors.

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  • "The useful largest aperture is anyway limited by the amount of depth of field that can't be too shallow or else parts of the subject are not in focus..." Sometimes this is the case, but many times it is not. – Michael C Apr 21 at 12:27
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    See also Can a smaller sensor's “crop factor” be used to calculate the exact increase in depth of field? – mattdm Apr 21 at 13:41
  • As is typically the case with questions such as this one, what is variable and what is held constant isn't specified. Same focal length at different subject distance? Same subject distance with different focal length? Same display size or same enlargement ratio? Voting to close as "Unclear what you are asking." – Michael C Apr 22 at 2:54
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Larger sensor size cameras yield shallower depth-of-field. The reason is: We fit lenses to cameras based on the corner-to-corner measurement of the format (film or digital). In other words, the “normal” lens for any camera is one with a focal length that is approximately equal to the diagonal measure of the frame size. As a rule-of-thumb, this assures that the circle of good definition projected by the lens accommodates the frame size.

Additionally, a camera sporting the classic rectangular format delivers an angle of view of 45°, when camera is held in the landscape (horizontal) position. A lens shorter than the diagonal measure will be classified as wide-angle; a lens longer than the diagonal measure is classified as telephoto. A portrait lens is traditionally 2X thru 2.5X of the diagonal measure.

Now the rest of the story: A larger sensor sports a longer lens, and this combination forces the photographer to work in closer in order to fill the viewfinder. It is this action that results in shallower depth-of-field.

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Short answer: If you want to increase the background blur, then yes... using a larger sensor format digital camera (or larger format film camera) will generally get that result.

The longer answer: The reason this happens isn't quite obvious. Equations to find depth of field mostly depend on three factors: (1) focal length, (2) aperture ratio, and (3) focus distance. Notice what it does not include is sensor size.

The reason sensor size makes a difference isn't because the rules of optics magically change... but rather because the photographer is likely going for a specific composition. The change in sensor size will change the true angle of view. The true angle of view will likely cause the photographer to change their subject distance in order to get the composition they were after (usually by moving the camera closer to the subject). Moving the camera closer to the subject will change the depth of field (it changes the focus distance -- one of the three factors that determines depth of field.)

This is based on an assumption that the photographer will move the camera position closer to achieve a similar composition. If the photographer does not change the focus distance then the depth of field won't change.

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    Depth of Field absolutely depends on sensor size. The acceptable CoC limit is computed as directly proportional to sensor size. A smaller sensor has to suffer greater enlargement to reach the standard 8x10 inch print for DOF reference, which reduces DOF, so DOF is absolutely less in smaller sensors if all else is equal. HOWEVER, all is not equal, smaller sensors normally must use a shorter lens to see the same Field of View, which is a larger factor (focal length squared), giving the notion that smaller sensors have more DOF. But which is only due to the shorter lens they necessarily use. – WayneF Apr 21 at 21:53
  • Common formulary circle of confusion size one thousandth of the focal length renders sensor size moot. True because larger sensor size mandate longer focal lengths. P = distance focused on D=diameter circle of confusion f= f-number F=focal length. Near point = P/1+PDf/F^2 Far point = P/1-PDf/F^2. – Alan Marcus Apr 22 at 2:07
  • #Alan Marcus No,sorry, far from moot, sensor size is very fundamental to DOF. CoC is NOT 1/1000 of focal length. CoC is sometimes 1/1500 of sensor diagonal, which takes greater enlargement of smaller sensors into effect. Actually, 1/1500 is rarely used, as coarse rounding means that actual results of normal 0.03 mm full frame or 0.02 mm APS crop frame has to be 1/1442 of sensor diagonal. Yes, smaller sensors do use shorter focal length, which is the larger offsetting factor. dofmaster.com/equations.html or Wikipedia has better DOF formula. Note H has a f-squared / CoC factor. – WayneF Apr 22 at 4:04
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    @WayneF, I hate to post on these things because the concept of "crop factor" is so mis-understood. My analogy: Point a video projector at a screen and focus the image. Next, cut away a section of the screen - but leave the rest. How has the image quality changed on the section of screen that you did not cut away? (it isn't - do the experiment yourself if you still have doubts) CoC calculations that include sensor size do so because they make assumptions that when producing a print, you will enlarge the image more if your sensor or film negative size was smaller. – Tim Campbell Apr 22 at 15:01
  • Yes, the remainder Not being cut away and Not changing is exactly correct (especially pertinent for macro work), but that is not the point about DOF. Yes, the lens image is exactly the same there, but nevertheless, a small sensor simply must be enlarged more (to compare at the same standard viewing size). That enlargement of course enlarges blur too (easier to see), so CoC must be computed smaller (to compare at the same view size). Focal length and f/stop and distance do determine blur, but CoC is all important to computing visible DOF limits. CoC is ( sensor diagonal / 1500) more or less. – WayneF Apr 22 at 16:15
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Starting from the depth of field and background blur equations, we can derive the following approximate equations for a background at infinity:

b   = f^2 / (x_d * N) 
DoF = 2 * x_d^2 * N * C / f^2

where b is background blur disc size, DoF is depth of field, f is the focal length, x_d is the subject distance, N is the aperture F-number and C is the circle of confusion.

We immediately notice that b can be represented in terms of DoF and vice versa:

b   = 2 * x_d * C / DoF
DoF = 2 * x_d * C / b

Now, since we are after large background blur b, we should consider how it varies as a function of sensor size. x_d, the subject distance, is the same for equivalent image. DoF is also the same for an equivalent image, provided that you can achieve the DoF you're after on both full frame and crop body. What is different is the circle of confusion C: it's divided by the crop factor on a crop sensor. So, background blur is also divided by the crop factor on a crop sensor. However, the sensor size is too divided by the crop factor on a crop sensor.

So, as conclusion, the amount of background blur as a percentage of the sensor size stays the same (with sensor size and background blur disc size being divided by the crop factor). Thus, on full frame and crop sensor bodies, you achieve equivalent amount of background blur, provided that you can achieve the DoF you're looking for.

So, we must analyze whether it's possible to achieve the desired DoF on a crop sensor body. By mounting a f/1.8 lens (such as the Canon Nifty Fifty, EF 50mm f/1.8 STM), it's possible to have a depth of field of merely centimeters as this example shows:

shallow DoF

Clearly, we can see from the image the depth of field is too shallow (image info: 50mm focal length, 1000mm subject distance, f/1.8, crop sensor body, giving DoF of 27mm). Thus, it is usually possible to achieve a shallow enough DoF on a crop sensor body.

Now, in some cases the photographer may intentionally want an extremely shallow depth of field. Can such a shallow depth of field be achieved on crop sensor body? The answer depends on whether an equivalent lens is available, which depends on the focal length. An equivalent lens has both the aperture F-number and the focal length divided by the crop factor. Then both the numerator and denominator of the fraction formula for b is divided by the crop factor squared, giving equivalent depth of field.

For example, for Canon telephoto primes, an equivalent lens is available:

FF 135mm f/2   -> crop 84mm f/1.25  (it's 85mm f/1.2, close enough)
FF 200mm f/2.8 -> crop 125mm f/1.75 (it's 135mm f/2, close enough)
FF 300mm f/4   -> crop 188mm f/2.5  (it's 200mm f/2.8, close enough)

I won't include the 500mm f/4 because of its high cost, and because a 300mm lens with equivalent cost is not available.

But for Canon short telephoto and standard lenses, the situation is different:

FF 85mm f/1.2 -> crop 53mm f/0.75 (it's 50mm f/1.2, not available)
FF 50mm f/1.2 -> crop 31mm f/0.75 (it's 28mm f/1.8, not available)
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    "Thus, it is usually possible to achieve a shallow enough DoF on a crop sensor body." Well, if you're always shooting at 1m subject distance it is more likely to be the case. Not so much when one is shooting at longer distances. You're taking a specific set of (fairly extreme) parameters and trying to argue a general rule from them. – Michael C Apr 21 at 12:26
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    ”Now, I will admit that it's easier to achieve extremely shallow DoF on a full-frame camera, but that is typically not what the photographer wants” - citation needed. Evidence to the contrary: I’ve many pet portraits that rely on throwing everything but the eyes to isolate the emotion in them. I want everything else thrown - and this is very typical of my style of pet portrait. – Hueco Apr 21 at 15:48
  • Now I consider extremely shallow depth of field as well, and it seems to depend on the focal length: for telephoto primes, an equivalent lens is available, but for short telephoto and standard lenses, an equivalent lens is not available. – juhist Apr 22 at 8:04
  • While I applaud playing with math, this is wrong from the start. Hint: in your DoF equation, what happens if x_d is set to the hyperfocal distance, H = (f^2/Nc) + f, or even the simpler approximation H = f^2/Nc? In your DoF equation, nothing happens, it's just math. But the hyperfocal distance is precisely when DoF is infinity. – scottbb Apr 22 at 14:31

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