bclaff
Forum Pro
If I understand you then yes; R relates to the image size and applying it to the object side implicitly is assuming pupil magnification of unity.
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Entrance pupil determines depth of field
- Thread starter Tom Axford
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Tom Axford
Forum Pro
Oh dear, you have lost me again!
As I see it, the image size = m x object size, where m is the image magnification.
So a circle of confusion of size c in the image corresponds to a circle of confusion of size c/m in the object being imaged.
Is that not the case?
It seems very much a matter of use cases, if I am not confused.
Thin lens, long subject distances: entrance pupil is fine, the simple geometry solution works.
For macro use and a thick lens, things become different.
I don't think they are different if the lens is thick compared to the focal length but the subject is still at infinity. If it is different it would be a good criterion for lens purchase (how much depth of field do you want for this FL/aperture combination).
Thin lens, long subject distances: entrance pupil is fine, the simple geometry solution works.
For macro use and a thick lens, things become different.
I don't think they are different if the lens is thick compared to the focal length but the subject is still at infinity. If it is different it would be a good criterion for lens purchase (how much depth of field do you want for this FL/aperture combination).
bclaff
Forum Pro
Yes, my bad (this time) ;-)
bclaff
Forum Pro
You will find that even at long distances if pupil magnification is significantly away from unity there are real differences. Naturally YMMV.
Pixel Pooper
Senior Member
Bill is correct about the exit pupil. Depth of field depends on the angle θ in the diagrams below. With a symmetrical lens at infinity focus, the entrance and exit pupils are the same so you can use the entrance pupil to calculate the angle, but if you look at the second diagram you can see that the angle will be determined by whichever pupil accepts a narrower cone of light.
If the exit pupil is smaller than the entrance pupil (pupil magnification < 1), then this angle will be determined by the exit pupil, not the entrance pupil, and as the exit pupil moves farther from the sensor to focus closer than infinity, the effect of the exit pupil on the cone of light increases. This is why in macro photography we use the effective f-number which is accurate at close focus distances, rather than the f-number which is only accurate at infinity focus.
If the exit pupil is smaller than the entrance pupil (pupil magnification < 1), then this angle will be determined by the exit pupil, not the entrance pupil, and as the exit pupil moves farther from the sensor to focus closer than infinity, the effect of the exit pupil on the cone of light increases. This is why in macro photography we use the effective f-number which is accurate at close focus distances, rather than the f-number which is only accurate at infinity focus.
alanr0
Senior Member
It makes no difference if you work with effective or working f-number.
The thickness of the lens makes no difference, provided you keep the image magnification the same. If you have a thicker lens of the same focal length, then to achieve the same magnification, the image plane will be further from the subject by the difference in principal plane separations.
Pupil magnification influences DoF via the relationship between image distance (and therefore magnification) and the angle subtended by the exit pupil at the lens. If you adjust the aperture to maintain the same effective f-number, then depth of field is independent of pupil magnification. In object-space, any difference in entrance pupil position is then exactly compensated by the change in aperture diameter.
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Alan Robinson
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ThrillaMozilla
Veteran Member
Yes, I think that's correct. The geometry is quite simple, and calculations from either image or object space must be consistent.
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I don't understand what the whole fuss is about except being poorly explained by both sides.
The object side geometry is similar to to the usual geometry for DOF calculation on the image side (see the link to the doc by Jeff Conrad, which I think was posted by DM elsewhere). And, at a casual glance the equations for near and far side DOF presented by Tom Oxford in the OP are structurally similar to the equations 9 and 10 in Conrad's document. Just throw in the parameters and things seem to check out. And, that argument about R can be resolved as it is just a ratio so it transfers from image side to object side easily. Except that it appears to me that Tom's equation for D should be the reciprocal of what he wrote - but again, this is just a glance and I haven't bothered to check the math.
NOTE: The underlying fact here is that there are points on the object side that correspond to the blur dimension as exhibited by CoC on the image side. The same initial reasoning was also used to figure out the DoF using the diffraction properties as explained briefly here . The idea is that free space propagation can be identified as an act of a quadratic phase filter. And, a lens can be modeled as an FM generator - FM as in frequency modulation. Therefore the kernel of free space propagation can be used to argue about CoC dimensions and on to DoF.
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Dj Joofa
http://www.djjoofa.com
The object side geometry is similar to to the usual geometry for DOF calculation on the image side (see the link to the doc by Jeff Conrad, which I think was posted by DM elsewhere). And, at a casual glance the equations for near and far side DOF presented by Tom Oxford in the OP are structurally similar to the equations 9 and 10 in Conrad's document. Just throw in the parameters and things seem to check out. And, that argument about R can be resolved as it is just a ratio so it transfers from image side to object side easily. Except that it appears to me that Tom's equation for D should be the reciprocal of what he wrote - but again, this is just a glance and I haven't bothered to check the math.
NOTE: The underlying fact here is that there are points on the object side that correspond to the blur dimension as exhibited by CoC on the image side. The same initial reasoning was also used to figure out the DoF using the diffraction properties as explained briefly here . The idea is that free space propagation can be identified as an act of a quadratic phase filter. And, a lens can be modeled as an FM generator - FM as in frequency modulation. Therefore the kernel of free space propagation can be used to argue about CoC dimensions and on to DoF.
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Dj Joofa
http://www.djjoofa.com
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Tom Axford
Forum Pro
That is essentially my position. I am not claiming that image space calculations are wrong. All I am saying is that there is a simple derivation of the DoF in object space that depends only on the size and position of the entrance pupil in relation to the object whose image is being captured. Of course there will be an equivalent derivation of the DoF in image space which will depend on the relationship between the exit pupil and the image plane. I would expect the image space analysis and the object space analysis to produce the same answers in the same situation.
Right. When I look at Conrad's equations 9 and 10 it leads me directly to your equations except D seems to be the reciprocal of what was in your OP.
Which if you look further in Conrad's document are given as eqs 89 and 90 for an asymmetrical lens' case. With P =1 the eqs 89 and 90 are the same as eqs. 6 and 7, which he further reduces to eqs. 9 and 10. And, eqs 9 and 10 are structurally similar to your equations in the OP (ignoring the reciprocal and a sign issue in the denominator of one eq.). And, I guess that is Bill Claff's position that situation only applies when P = 1.
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Dj Joofa
http://www.djjoofa.com
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A
AiryDiscus
Guest
Here's a math problem for you - when does that quadratic phase filter equal a Fourier transform?
A bit hard to go to the focus without a Fourier transform, which is "the Fourier transforming property of a lens" which you can find widely in The Good Book, B&W, etc.
A bit hard to go to the focus without a Fourier transform, which is "the Fourier transforming property of a lens" which you can find widely in The Good Book, B&W, etc.
Tom Axford
Forum Pro
If that is so, then by reasoning back from the argument I gave, it is not possible for a lens to produce a perfect image unless P=1.
Well, since you have a ‘degree on the topic’, notwithstanding just a Bachelor’s, I hope that you are more qualified to answer that question
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Dj Joofa
http://www.djjoofa.com
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AiryDiscus
Guest
Ooh, a combo - casting doubt on my education and a deflection! Do I have to pay extra for that?
I find the "degree of degree" gatekeeping repulsive. There isn't an education watermark on competence. Education just increases the probability you're competent in a topic.
I think my employment and publication history demonstrate that competence. Perhaps you may disagree. My advisor, or my academic grandfather (Goodman himself) may disagree with your disagreement.
bclaff
Forum Pro
They won't be "equivalent" unless you take pupil magnification into account when you to the object side which means you're not just doing the object side.
bclaff
Forum Pro
Yes !
alanr0
Senior Member
< snip >
Tom's D = d / (R a) = b / a = c / (m a)
where m is the image magnification and c is the image-space CoC.
Conrad's equation (9) is un = u / (1 + Nc / fm) = u / (1 + c / ma)
So Conrad's near side DoF = u - un = (u c / ma) / (1 + c/ma)
Which corresponds to Tom's near side DoF = s D / (1 + D)
How so, or am I missing something?
Tom's D = d / (R a) = b / a = c / (m a)
where m is the image magnification and c is the image-space CoC.
Conrad's equation (9) is un = u / (1 + Nc / fm) = u / (1 + c / ma)
So Conrad's near side DoF = u - un = (u c / ma) / (1 + c/ma)
Which corresponds to Tom's near side DoF = s D / (1 + D)
bclaff
Forum Pro
Remember it's not only the size of the disc that matters but the angle of the marginal rays at the edge.
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