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# Crop factor and DOF

Canon 30D (1.6 crop) + 20mm lens

Canon 1Ds MkII (Full Frame) + 32mm lens

Obviously these two setups would give a very similar angle of view/composition. But depth of field would be completely different (at a given aperture)?

That's really confusing? Is this important to anyone?

Scroll down to the para entitled DOF and focal length

Note: "It is well known that short focal length lenses have large apparent depths of field and long telephoto lenses have small apparent depths of field. There are some very practical reasons for this conception, but it isn't quite true. DOF is much more closely related to magnification and f-stop; DOF expressed in distance is nearly independent of focal length. It appears smaller with telephoto lenses because it is smaller when expressed as a fraction of the lens-to-subject distance, s"

Quote:No, depth of field will be the same. See the last link I gave!

Mr Croftsphoto Sir, did you read the article you linked to?

The distance to subject was changed to keep the subject the same size, hence the similar DOF. He didn't

**crop**the images.

The reason the OP is confused, is because your conflicting arguement is flawed, you are changing two different variables. All three variables will change the outcome in the end, that's all I was saying.

EDIT - in your second link, he is also changing two variables, distance and focal length. I give up!

Strawman - I understand what you're saying, but my posts are an attempt to help answer the OP's question, correctly.

Quote:EDIT - in your second link, he is also changing two variables, distance and focal length. I give up!

The second link is just giving a mathematical formula - he isn't changing the distance to subject as it isn't been given a value in the formula, it's represented by a letter.

But he does then go on to quote the well known means of calculating acceptable depth of field for a print based on what appears acceptable in a print from a normal viewing distance. It's definitely the case that with a wider angle lens image that hasn't been cropped the DoF is less apparent (even though it's the same at a given aperture for different focal lengths). To put it another way, you can get away with less depth of field in a wide angle lens image. I'm not trying to dispute the fact that you can get away with less depth of field in a wide angle image without it being so noticeable!

Let me quote his full explanation, note especially the bits I've highlighted in bold:

Quote:It is well known that short focal length lenses have large apparent depths of field and long telephoto lenses have small apparent depths of field. There are some very practical reasons for this conception,

**but it isn't quite true.**DOF is much more closely related to magnification and f-stop;

**DOF expressed in distance is nearly independent of focal length**. It appears smaller with telephoto lenses because it is smaller when expressed as a fraction of the lens-to-subject distance, s.

Equations for Total Depth of Field

Combining the equations for Df and Dr from the first box of equations, we can obtain the total depth of field.

Total DOF = Df + Dr = sC(s-f )/( fa-C(s- f )) + sC(s- f )/( fa+C(s- f ))

= 2 fasC(s- f )/(( fa)2-C2(s-f )2)

Now, substitute magnification M into the equation using M = d/s = f / (s-f ); s-f = f / M.

Total DOF = 2 fasC( f /M )/(( fa)2-C2 f 2 / M2) = 2asCM / (M2a2 - C2)

Eliminate s using s = f + f / M = f (1 + 1/M) = Na (1 + 1/M), where N = f-stop = f / a.

Total DOF = 2Na2C (M+1)/ (M2a2 - C2) = 2NC (M+1)/ (M2 - (CN / f )2)

**No approximations yet, but we haven't entirely eliminated the focal length f. Fortunately, the (CN / f )2 term is usually much smaller than M 2, except for very distant images (with very small magnification).**As we point out below, c/f is a constant, independent of format, equal to about 1/1600 for a "normal" lens. For example, for the 35mm format with a standard 50 mm lens at f/8, cN / f = 0.03*8/50 = 0.0048 ~= 1/200. So the (cN / f )2 term can be eliminated from the equation (the error will be less than 1%) for magnifications M larger than 1/20 (a 20x30 inch or smaller field for 35mm format), which covers most portraits and still lives.

Total DOF ~= 2NC (M+1)/ M2

This approximation holds for large magnifications: portraits, still lives, etc. (M > 1/20 in the above example).

Now let's look at Depth of Field for M ~= 1/20 at f/8 for several focal lengths, using Jonathan Sachs' Depth of Field Calculator set for 30 lp/mm resolution (the default).

DOF for f/8, M ~= 1/20, 35mm format Focal length

f mm Distance

S mm Near DOF

limit mm Far DOF

limit mm Total

DOF mm Total

DOF/S %

20 mm 400 319 536 217 54.2

50 mm 1000 908 1113 205 20.5

100 mm 2000 1904 2107 203 10.1

200 mm 4000 3901 4104 203 5.1

1000 mm 20000 19899 20102 203 1.0

**For a specific format, depth of field, expressed as distance, is independent of focal length.**But depth of field, expressed as a percentage of the distance to the subject (Total DOF/s %), is inversely proportional to focal length. It can be very small for long telephoto lenses.

Using a long telephoto lens is an effective way of isolating a subject from busy, uncontrolled backgrounds without sacrificing actual depth of field.

Quote:But depth of field, expressed as a percentage of the distance to the subject (Total DOF/s %), is inversely proportional to focal length. It can be very small for long telephoto lenses.

Indeed, the distance used in the calculation for a 20mm lens is 400mm, the distance used for a 1000mm lens is 20,000mm. That's why the total DOF is roughly the same.

If both distances were the same, the DOF would be wildly different.

In science, if you want to test a theory, you change one variable only, to prove it. As a result of changing focal length, with your subject being at the same distance, the DOF will be shallower as a result of the greater magnification, which agrees with his theory. So, in practical terms, changing focal length does alter the DOF.

If however, you take a step back so that your subject is magnified exactly the same, the DOF will be very similar.

Anyway, I'm off to the pub to numb this wasp sting! It's actually been an interesting discussion. You had me doubting myself for a moment then. :-P

Quote:Visual acuity and viewing distance are other factors which come into it as well.

Just to complicate things!

Agreed, but that's when viewing the print, which is a slightly different matter!

Quote:Indeed, the distance used in the calculation for a 20mm lens is 400mm, the distance used for a 1000mm lens is 20,000mm

The 20mm lens/400mm and 1000mm lens/20,000mm is only brought in where he's quoting Jonathan Sachs' Depth of Field Calculator - it's not in the author's formula at all. The Sach's depth of field calculator is used for the purpose of determining what's acceptable resolution of lines per millimetre taking into account the final viewing distance of print etc. That's different from the issue of what the optical depth of field of the lens is. For that - see the author's own formula set out above that!

To put it another way, with a wide angle lens you would need a much higher resolution of lines per millimetre on the sensor/film to be able to detect the fact that the DoF in that part of the subject is the same as in the telephoto lens, which has 'magnified' that area.

Have a good drink

Aperture, focal length, distance-to-subject, and subject to background, all have an effect on "apparent" DOF, which I suppose is what really matters, how the DOF appears?

Since crop factor effectively increases a given focal length, and f/stop is the ratio of aperture diameter, which directly effects DOF, to focal length, then crop factor also effectively increases DOF.

Is that fair in layman's terms? Greatly oversimplified I realize after reading all the responses.