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06-09-2012, 02:22 AM   #16
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QuoteOriginally posted by newarts Quote
The useful answer is yes. DOF is independent of focal length for macros
==================================================================

The precise answer is...

Focal length always affects DOF due to hyperfocal effects but sometimes not-so-much.

Total DoF = (stuff independent of focal length)/ (1- (distance/hyperfocal.distance)^2)

Hyperfocal distance = F^2/cN where c is circle of confusion & N is f-number

When F increases, Hyperfocal distance increases, therefore the denominator increases, therefore total DoF always decreases with increasing F.

But it is only important when distance is more than about one third of the Hyperfocal.distance.

Dave in Iowa
But not with close up!!!!

Wiki is correct for once.

Depth of field - Wikipedia, the free encyclopedia

This the formula


N = f number
c = circle of confusion
m = magnification
P = pupil magnification


It should be in the books of yours.

06-09-2012, 09:00 AM   #17
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QuoteOriginally posted by Anvh Quote
But not with close up!!!!

Wiki is correct for once.

Depth of field - Wikipedia, the free encyclopedia

This the formula


N = f number
c = circle of confusion
m = magnification
P = pupil magnification


It should be in the books of yours.
Can't see the formula. - perhaps you could type it in?

But believe me, there's always a hyperfocal term in the denominator. Its effect may approach zero, as in the close-up case, but it is always there (even if an author chooses to omit it because of its insignificance.)
06-09-2012, 09:43 AM   #18
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QuoteOriginally posted by newarts Quote
Can't see the formula. - perhaps you could type it in?
His graphic shows:

DOF ~ (2Nc (1 + m/P))/m^2

where m (magnification) is a function of distance and focal length
and p (pupil magnification) is the ratio of exit pupil diameter to the entrance pupil

In other words, we can't just observe the camera setup and plug-in numbers to solve for DOF. This formula also disregards frame size. It is not really usable.
06-09-2012, 12:42 PM   #19
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RioRicco, thanks for that.
If you mean with frame size mean the sensor size then that's covered with the circle of confusion, which in the formula.
And magnification is writing down on the focus indication of most macro lenses so it should be a problem to know the magnification of the lens, pupil magnification is the real problem here since it's not standard given by manufactures.
I'll write an email to pentax to see if they can give me some numbers.

c = circle of confusion


This as for as i know the only way to get accurate DOF calculation with close ups, the other one that Newarts use simply doesn't work in this case. Beside that it doesn't take into account the pupil magnification either which is important with these magnifications.
If Newart doesn't believe me he should look look his formula up and see what it says about this, they should mention it.



You can btw leave "P" out of the formula if you don't use it for close up.


Last edited by Anvh; 06-09-2012 at 12:58 PM.
06-09-2012, 01:56 PM   #20
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QuoteOriginally posted by RioRico Quote
His graphic shows:

DOF ~ (2Nc (1 + m/P))/m^2

where m (magnification) is a function of distance and focal length
and p (pupil magnification) is the ratio of exit pupil diameter to the entrance pupil

In other words, we can't just observe the camera setup and plug-in numbers to solve for DOF. This formula also disregards frame size. It is not really usable.
The posted equation simply omits the hyperfocal term in the denominator. Note that the "~" means approximately.

The exact equation is:

DOF = 2cN(1+m/p)/m^2/(1-(d/d*)^2) where d* is hyperfocal distance: d*= F^2/cN + F

Furthermore Wikipedia's sentence preceding the approximation says "When s<< H, (nb in my terms d << d*) the second term in the denominator becomes small in comparison with the first...." The preceding equation in Wikipedia is the exact one and reduces to the one I wrote here - including the influence of p.

Dave in Iowa

see http://www.largeformatphotography.info/articles/DoFinDepth.pdf eqn 104
06-09-2012, 02:29 PM   #21
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The 'confusion' is based in the 'approximately equals' sign in your equation, Anvh.
06-10-2012, 03:49 AM   #22
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QuoteOriginally posted by ElJamoquio Quote
The 'confusion' is based in the 'approximately equals' sign in your equation, Anvh.
They also do that with the formula of Newarts so what does that say?

And the PDF that he links with says the error gets larger with magnification...

On page 43 it says you can't use the formula for Macro lenses and since many things aren't known it's the best just to use the manufacture parameters then.

Last edited by Anvh; 06-10-2012 at 05:11 AM.
06-10-2012, 10:48 AM   #23
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QuoteOriginally posted by Anvh Quote
They also do that with the formula of Newarts so what does that say?

And the PDF that he links with says the error gets larger with magnification...

On page 43 it says you can't use the formula for Macro lenses and since many things aren't known it's the best just to use the manufacture parameters then.
From p.43
When such lenses [nb unsymmetrical lenses] are used at high magnification, the DoF may differ significantly from that given by formulae for symmetrical unit-focusing lenses. In most cases, the parameters required for DoF calculations with such lenses are not known


I agree. But what's that have to do with the existence of a DOF dependence on focal length?

This is a bizzare discussion: I'm demonstrating, based on optics and math, that although a DOF dependence on focal length exists, it becomes vanishingly small at high magnifications.

I'm not sure what Anvh's objecting to.

Perhaps the objection is that I'm unnecessarily nit-picking & showing off by giving a confusing answer to a simple question.

Q. Is DOF at macro scales independent of focal length?

A(1). Yes
A(2). Yes for all practical purposes.
A(3). No but the dependence is insignificant & here's why

I'll choose to explain why whenever possible.

Dave

@Anvh, you say..They also do that with the formula of Newarts - where?

06-10-2012, 11:11 AM   #24
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We are talking about DOF at macro here since that's what this topic is about if you want to discuss something else feel free to start a new topic about it.
Focal length is part of DOF if you want but you can also say that DOF depends on magnification and the aperture, like riorico say magnification is focallength and focus distance but with macro we often talk about magnification and so with 1:1 magnification there is no difference in DOF when pupil magnification and the aperture are the same.
That's why i say that with macro photography DOF is independent from focal length, if you chose to disagree then please do so.
Beside that at 1:1 the focal length difference should be gone right?

About the rest, your first formula simply didn't work for macro and so that's why i commented on it, not sure about your second one, never seen that one before.


Where they do that with your formula, well you commented they use ~ with the formula i quoted but they also use ~ with your formula at Wikipedia.
Anyway all DOF calculation are approximations when look at it, pupil magnification often is not used and non take into account aberrations.
06-10-2012, 12:38 PM   #25
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QuoteOriginally posted by Anvh Quote
.....
That's why i say that with macro photography DOF is independent from focal length, if you chose to disagree then please do so.
Beside that at 1:1 the focal length difference should be gone right?...
I don't think we disagree - I always have said the effect is insignificant - but that is not quite the same as saying it doesn't exist - an important philosophical/scientific detail.

Like you say, at 1:1 the effect is almost gone. The focal length term in the denominator is

(d/d*)^2 = [(m+1)/(m(f/cN+1))]^2

for a 28mm lens, f:2.8, 1:1 mag, it is about1/10000. A small number indeed! Certainly small enough to ignore - but not zero -

but close enough to zero to completely ignore (only about 1/25 a wavelength of light) - maybe I'll have to change my philosophy on reporting such meaningless detail.

It seems to me that the effect of focal length on DOF is insignificant for magnifications greater than about 1:50. That's a lower mag than I would have guessed.

Dave in Iowa
06-10-2012, 01:23 PM   #26
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QuoteOriginally posted by Anvh Quote
They also do that with the formula of Newarts so what does that say?
I'm guessing it's there (I haven't looked) because of the 'thin lens' approximation. But I haven't done any math substantiating or disproving that.... in fact I haven't even looked at any of the equations here, so take my comments with a big grain of salt.
06-10-2012, 01:46 PM   #27
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Actually how did you came with this formula.
DOF = 2cN(1+m/p)/m^2/(1-(d/d*)^2)
It seems to me you simply added the hyperfocal distance and the question is why?
06-10-2012, 01:48 PM   #28
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QuoteOriginally posted by ElJamoquio Quote
I'm guessing it's there (I haven't looked) because of the 'thin lens' approximation. But I haven't done any math substantiating or disproving that.... in fact I haven't even looked at any of the equations here, so take my comments with a big grain of salt.
If you want to have it absolutely precise then you need to take some many things into account that it isn't interesting any more if you ask me
06-10-2012, 04:17 PM   #29
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QuoteOriginally posted by Anvh Quote
Actually how did you came with this formula.
DOF = 2cN(1+m/p)/m^2/(1-(d/d*)^2)
It seems to me you simply added the hyperfocal distance and the question is why?
I added it because it should be there.. Here's the detail on how it gets there

The previous equation in the wikipedia article is: first factor fm/Nc out of the denominator, then factor 1/m from the numerator

DOF =2f(1/m+1/p)/(fm/Nc -Nc/fm) = 2f(1/m+1/p)/((fm/Nc)(1-(Nc/fm)^2) = 2cN(1+m/p)/m^2/(1-(Nc/fm)^2)

Using the fact that m= f/(d-f) =(f/d)/(1-f/d), the denominator becomes (1-(Nc/fm)^2) = (1-(Nc/(f(f/(d(1-f/d))))^2) =(1-(d/(f^2/(Nc(1-f/d))))^2)

(1-(d/(f^2/(Nc(1-f/d))))^2) = (1-(d/d*)^2) where d* = f^2/Nc + f - the "hyperfocal distance".

so DOF = 2cN(1+m/p)/(1-(d/d*)^2)

But it is just as useful for macros to leave the denominator in terms of magnification (so you don't have to calculate hyperfocal distance).

DOF = (2cN(1+m/p)/m^2)/(1-(Nc/fm)^2)

Say a focal length effect less than 1% DOF can be ignored, and choose some fairly extreme parameters for normal macro use, like: N < f:22, c~ .02mm, f>10mm, then 1/10 > Nc/fm for m ~ 1/2. That is, focal length effect on macro DOF is less than 1% for magnifications greater than about 1:2.

Dave in Iowa

Last edited by newarts; 06-10-2012 at 05:51 PM.
06-10-2012, 05:37 PM   #30
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Thanks got it, it actually makes sense... XD
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