Depth of field - effect of focal length

[Henryti]

Hi,

Grateful if any of you technical, engineering sorts out there can help with this. My question is - how is depth of field affected by the focal length of a lens if you keep the subject (eg. a person) the same size in the frame?

My understanding is that, all else being equal,

(i) the longer the focal length the shallower the depth of field and

(ii) the closer you are to the subject, the shallower the depth of field.

Using a 85mm and a 50mm lens as an example, in order to keep a subject the same size in the frame, you would have to be closer to the subject with the 50mm lens compared to the 85mm. So at the same aperture on the same camera, which would have the shallower depth of field? Is there any kind of formula or guide if you want to compare different focal lengths?

Thanks.

Henry

 

[ThrillaMozilla]

At close distances, pupil magnification (the exit pupil diameter) does matter.
In that case it's not just the entrance pupil.

The exit pupil diameter is f/N, where N is the f-stop number, and f is the focal length. That's the usual formula from every elementary text.

No. That's the entrance pupil.

Sorry, yes, that's what I intended to write. It's the entrance pupil.

Unfortunately, I can't go back and correct the original.

 

[bclaff]

Math is in this posting:

https://renesphotoblog.blogspot.com/2021/04/dof-bokeh-focal-length-f-stop-crop.html

If you want an explanation, you should look first to the writings of Tom Axford and others on DPR: https://www.dpreview.com/forums/thread/4559474 and https://www.dpreview.com/forums/thread/4495183 , for example. (Pay no attention to the dissent. It eventually dissolved, and it will probably not help you understand the subject.) There are also MANY other writings on the subject, going back over a century. Stripped of all the complications, depth of field is a very simple geometrical concept, and it is determined by a single angle defined by the diameter of the entrance pupil and the distance to the subject.

At close distances, pupil magnification (the exit pupil diameter) does matter.
In that case it's not just the entrance pupil.

For geometry and math: Optics Primer - Depth of Field

It is true that the pupil magnification is required if the subject distance is measured in the usual way (from the front nodal point).

However, if the subject distance is measured from the entrance pupil, this implicitly includes the pupil magnification and then the formula for the depth of field does not explicitly include the pupil magnification or the exit pupil.

There is a simple formula for the depth of field that depends on the size of the entrance pupil and the distance of the entrance pupil from the subject and nothing else apart from image magnification and circle of confusion.

More details - see the links above in bold.

Sort of but the distance from subject to entrance pupil includes pupil magnification !

Exactly. If you know the subject distance from the entrance pupil, it is not necessary to know the pupil magnification as well.

That subject distance is (1/m + 1/p) * f

Where f is focal length, m is magnification, and p is pupil magnification.

So yes, pupil magnification is involved.

 

[ThrillaMozilla]

Math is in this posting:

https://renesphotoblog.blogspot.com/2021/04/dof-bokeh-focal-length-f-stop-crop.html

If you want an explanation, you should look first to the writings of Tom Axford and others on DPR: https://www.dpreview.com/forums/thread/4559474 and https://www.dpreview.com/forums/thread/4495183 , for example. (Pay no attention to the dissent. It eventually dissolved, and it will probably not help you understand the subject.) There are also MANY other writings on the subject, going back over a century. Stripped of all the complications, depth of field is a very simple geometrical concept, and it is determined by a single angle defined by the diameter of the entrance pupil and the distance to the subject.

At close distances, pupil magnification (the exit pupil diameter) does matter.
In that case it's not just the entrance pupil.

For geometry and math: Optics Primer - Depth of Field

It is true that the pupil magnification is required if the subject distance is measured in the usual way (from the front nodal point).

However, if the subject distance is measured from the entrance pupil, this implicitly includes the pupil magnification and then the formula for the depth of field does not explicitly include the pupil magnification or the exit pupil.

There is a simple formula for the depth of field that depends on the size of the entrance pupil and the distance of the entrance pupil from the subject and nothing else apart from image magnification and circle of confusion.

More details - see the links above in bold.

Sort of but the distance from subject to entrance pupil includes pupil magnification !

Not "sort of". The distance from subject to entrance pupil doesn't "include" anything. It's just a distance.

However, you are correct that if you need to know the exact position of the entrance pupil (and for many purposes you do not need to know that), the pupil magnification is one way to calculate it. Thank you for including that handy formula in a preceding message.

 

[bclaff]

Math is in this posting:

https://renesphotoblog.blogspot.com/2021/04/dof-bokeh-focal-length-f-stop-crop.html

If you want an explanation, you should look first to the writings of Tom Axford and others on DPR: https://www.dpreview.com/forums/thread/4559474 and https://www.dpreview.com/forums/thread/4495183 , for example. (Pay no attention to the dissent. It eventually dissolved, and it will probably not help you understand the subject.) There are also MANY other writings on the subject, going back over a century. Stripped of all the complications, depth of field is a very simple geometrical concept, and it is determined by a single angle defined by the diameter of the entrance pupil and the distance to the subject.

At close distances, pupil magnification (the exit pupil diameter) does matter.
In that case it's not just the entrance pupil.

For geometry and math: Optics Primer - Depth of Field

It is true that the pupil magnification is required if the subject distance is measured in the usual way (from the front nodal point).

However, if the subject distance is measured from the entrance pupil, this implicitly includes the pupil magnification and then the formula for the depth of field does not explicitly include the pupil magnification or the exit pupil.

There is a simple formula for the depth of field that depends on the size of the entrance pupil and the distance of the entrance pupil from the subject and nothing else apart from image magnification and circle of confusion.

More details - see the links above in bold.

Sort of but the distance from subject to entrance pupil includes pupil magnification !

Not "sort of". The distance from subject to entrance pupil doesn't "include" anything. It's just a distance.

However, you are correct that if you need to know the exact position of the entrance pupil (and for many purposes you do not need to know that), the pupil magnification is one way to calculate it. Thank you for including that handy formula in a preceding message.

OK, but if you're going to claim to accurately compute DOF from that distance then you have to have an accurate measurement.
I doubt that anyone reading this thread is using a "nodal" slide to locate their entrance pupil and then accurately measuring the distance from that point to their subject.
With a direct measurement you wouldn't need to "know" pupil magnification.
But, you also don't accurately know focal length either, so that doesn't help.
From a geometric math standpoint at close distance pupil magnification matters (that's my original statement).
When 1/m is much larger than 1/p then 1/p becomes "irrelevant".

 

[ThrillaMozilla]

ThrillaMozilla wrote:

If you just want to understand the phenomenon, then the amount of blur and the depth of field are determined by a single parameter: the angle of acceptance. Axford's approach leads easily to many useful conclusions about comparisons of lenses, sensor formats, apertures, and focal lengths -- such as the question asked by the OP.

Here's a diagram to show you how simple this is. This is a direct answer to the original question.

Here are two lenses, one with twice the focal length of the other and twice the aperture (both are set at the same f-stop). They are both focused on subject plane S, but out-of-focus point P appears as a circle with diameter c. If we take c as the maximum blur we will accept, then the depth of field is d. Notice that both lenses show the same depth of field d! Notice also that the depth of field is determined by the angle of the lenses viewed from point P.

Alternatively, both lenses could have the same focal length, but the closer of the two lenses has only half the aperture (diameter) because it is stopped down by two stops. They still have the same depth of field!


Both lenses are focused on subject plane S. Out-of-focus point P appears as a circle with diameter c. The depth of field is d. Both lenses have the same depth of field.

Here's a bonus. Light from distant points (at "infinity") is parallel. Distant points appear as circles, with diameters shown by the red and blue lines. Even though both lenses show the same depth of field, there is a big difference in the appears of very distant objects!


The same diagram as above, but also showing the diameters of blur circles from distant points (red and blue lines).

 

[bclaff]

ThrillaMozilla wrote:

If you just want to understand the phenomenon, then the amount of blur and the depth of field are determined by a single parameter: the angle of acceptance. Axford's approach leads easily to many useful conclusions about comparisons of lenses, sensor formats, apertures, and focal lengths -- such as the question asked by the OP.

Here's a diagram to show you how simple this is. This is a direct answer to the original question.

Here are two lenses, one with twice the focal length of the other and twice the aperture (both are set at the same f-stop--assume that the apertures are circular). They are both focused on subject plane S, but out-of-focus point P appears as a circle with diameter c. The depth of field is d. Notice that both lenses show the same depth of field d!

Alternatively, both lenses could have the same focal length, but the closer of the two lenses has only half the aperture (diameter) because it is stopped down by two stops. They still have the same depth of field!


Both lenses are focused on subject plane S. Out-of-focus point P appears as a circle with diameter c. The depth of field is d. Both lenses have the same depth of field.

Here's a bonus. Light from distant points (at "infinity") is parallel. Distant points appear as circles, with diameters shown by the red and blue lines. Even though both lenses show the same depth of field, there is a big difference in the appears of very distant objects!


The same diagram as above, but also showing the diameters of blur circles from distant points (red and blue lines).

d is not the depth of field, only the far portion of it, there is also a near portion.

You can't know d from c at close distances because using similar triangles does not work unless m is small.

--
Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

 

[ThrillaMozilla]

d is not the depth of field, only the far portion of it, there is also a near portion.

Yes, that's right, but the same principle applies. I encourage anyone who wants to remember this to draw their own diagram by dragging point P to the right and see what happens. There's nothing like doing it yourself if you want to understand it and remember it.

You can't know d from c at close distances because using similar triangles does not work unless m is small.

You can, actually. Similar triangles are always similar, no matter what shape they are or how close the subject is.

But you raise a valid point. I'm pretty sure you are referring to the fact that there's a little complication at close distances if one tries to do the calculation. I was contemplating adding a footnote when you replied.

What I glossed over is that the large triangle does not show the distance from the entrance pupil to the subject. That complicates the algebra slightly, but the principle still applies, and the diagram is still valid as drawn. The statement about the focal lengths and f stops (twice the focal length and same f stop), however, is not accurate at close distances. f and N are not part of the diagram. The statement of the relation between them was intended as an approximate explanation , but it should not be assumed when one tries to do the calculation. There are many ways to calculate the depth of field, and they all give the same result if done correctly.

 

[bclaff]

...

You can't know d from c at close distances because using similar triangles does not work unless m is small.

You can, actually. Similar triangles are always similar, no matter what shape they are or how close the subject is.

...

But for an exact answer similar triangle never work and are always an approximation.

Before arguing with me consider the image side and the difference between depth of focus (where similar triangles work) and depth of field (where they do not).
And perhaps, look again at my article (which is not only correct by observation but proves out with ray tracing of actual lenses).

FWIW, this (similar triangles not working) is also why depth of field isn't symmetrical.

--
Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

 

[ThrillaMozilla]

...

You can't know d from c at close distances because using similar triangles does not work unless m is small.

You can, actually. Similar triangles are always similar, no matter what shape they are or how close the subject is.

...

But for an exact answer similar triangle never work and are always an approximation.

Note to the reader: Similar triangles always "work". Euclid determined over 2000 years ago that The sides of similar triangles have constant ratios, and that isn't an approximation.

Aw, come on, Bill. The ratio c/d, as shown in my diagram, is always determined by the angle subtended by the entrance pupil from point P, and there's nothing approximate about it.

You're apparently complaining that you think I don't know the distance from P to the entrance pupil, but you and many others have told us many times how to calculate it from the pupil magnification if we need to. You're knocking down a straw man that I didn't erect.

And I've already written in my previous post -- as edited -- that one statement I made was an approximation intended as a simple explanation. Do you mind reading that carefully?

 

[Tom Axford]

...

You can't know d from c at close distances because using similar triangles does not work unless m is small.

You can, actually. Similar triangles are always similar, no matter what shape they are or how close the subject is.

...

But for an exact answer similar triangle never work and are always an approximation.

Before arguing with me consider the image side and the difference between depth of focus (where similar triangles work) and depth of field (where they do not).
And perhaps, look again at my article (which is not only correct by observation but proves out with ray tracing of actual lenses).

FWIW, this (similar triangles not working) is also why depth of field isn't symmetrical.

No, similar triangles do work and they give an exact result.

Consider this diagram in which I have labelled some of the important points:





Triangles PQR and PTU are similar, so

y/a = distance from P to the subject plane / distance from P to the entrance pupil

Also, triangles SQR and STU are similar, so

y/a = distance from S to the subject plane / distance from S to the entrance pupil

It is then just a matter of simple algebra to work out formulae for the far side and near side depth of field in terms of a, y and the distance between the subject plane and the entrance pupil.

The formulae obtained are exact and apply for all distances, large or small.

Further details here.

 

[ThrillaMozilla]

OK, I need to edit this and add one word: "approximately". The word is added three times, and it is shown in bold.

Here's a diagram to show you how simple this is. This is a direct answer to the original question.

Here are two lenses, one with twice the focal length of the other and approximately twice the aperture diameter (both are set at the same f-stop). They are both focused on subject plane S, but out-of-focus point P appears as a circle with diameter c. If we take c as the maximum blur we will accept, then the depth of field is d. Notice that both lenses show the same depth of field d! Notice also that the depth of field is determined by the angle of the lenses viewed from point P.

Alternatively, both lenses could have the same focal length, but the closer of the two lenses has approximately half the aperture (diameter) because it is stopped down by approximately two stops. They still have the same depth of field!

Both lenses are focused on subject plane S. Out-of-focus point P appears as a circle with diameter c. The depth of field is d. Both lenses have the same depth of field.

The reader can verify for himself that the result is very similar if point P is on the near side of the subject.

Here's a bonus. Light from distant points (at "infinity") is parallel. Distant points appear as circles, with diameters shown by the red and blue lines. Even though both lenses show the same depth of field, there is a big difference in the appears of very distant objects!


The same diagram as above, but also showing the diameters of blur circles from distant points (red and blue lines).

 

[bclaff]

...

You can't know d from c at close distances because using similar triangles does not work unless m is small.

You can, actually. Similar triangles are always similar, no matter what shape they are or how close the subject is.

...

But for an exact answer similar triangle never work and are always an approximation.

Note to the reader: Similar triangles always "work". Euclid determined over 2000 years ago that The sides of similar triangles have constant ratios, and that isn't an approximation.

Aw, come on, Bill. The ratio c/d, as shown in my diagram, is always determined by the angle subtended by the entrance pupil from point P, and there's nothing approximate about it.

You're apparently complaining that you think I don't know the distance from P to the entrance pupil, but you and many others have told us many times how to calculate it from the pupil magnification if we need to. You're knocking down a straw man that I didn't erect.

And I've already written in my previous post -- as edited -- that one statement I made was an approximation intended as a simple explanation. Do you mind reading that carefully?

I have no objection to approximations when they are stated as such

You and Tom have both reacted to a statement I made but I think you both read in some intent that wasn't there.

Every exact formula for depth of field includes pupil magnification one way or another.
That is my sole point.

Any blanket statement (eg. one without stated assumptions) about depth of field that omits pupil magnification is wrong. One should at least state the assumption that magnification is significantly less than 1.

 

[ThrillaMozilla]

Every exact formula for depth of field includes pupil magnification one way or another.
That is my sole point.

Any blanket statement (eg. one without stated assumptions) about depth of field that omits pupil magnification is wrong. One should at least state the assumption that magnification is significantly less than 1.

Here is Tom Axford's result (equations 1 and 2). Pupil magnification is not mentioned, and the result is valid for any magnification.

Axford's results were well vetted by experts Alan Robinson and others in the DPR Photographic Science and Technology Forum. Robinson wrote a very scholarly, careful comparison compared formulations by Jeff Conway, you, Tom Axford, and himself, and concluded that they "differ in form but are mathematically equivalent" . In other words, there is more than one way to do it, and Axford's equations are correct.

Tom Axford and I fully acknowledge that the way you calculate DOF is one valid way to do it. We have never disputed the validity of that approach. He and I are both trying to show another way of doing it that we think is particularly instructive, and which has been well vetted by experts. I think it helps to answer the OP's question in a simple way in a general forum.

 

[KristinSh]

Thank you for this question and the answers to it - I had problems...

 

[bclaff]

Every exact formula for depth of field includes pupil magnification one way or another.
That is my sole point.

Any blanket statement (eg. one without stated assumptions) about depth of field that omits pupil magnification is wrong. One should at least state the assumption that magnification is significantly less than 1.

Here is Tom Axford's result (equations 1 and 2). Pupil magnification is not mentioned, and the result is valid for any magnification.

Axford's results were well vetted by experts Alan Robinson and others in the DPR Photographic Science and Technology Forum. Robinson wrote a very scholarly, careful comparison compared formulations by Jeff Conway, you, Tom Axford, and himself, and concluded that they "differ in form but are mathematically equivalent" . In other words, there is more than one way to do it, and Axford's equations are correct.

Tom Axford and I fully acknowledge that the way you calculate DOF is one valid way to do it. We have never disputed the validity of that approach. He and I are both trying to show another way of doing it that we think is particularly instructive, and which has been well vetted by experts. I think it helps to answer the OP's question in a simple way in a general forum.

Since you agree the approaches are mathematically identical (as Alan proved) then you acknowledge that pupil magnification is involved (since that's what's in my approach). You simply deny it when it is not explicitly shown.

 

[ThrillaMozilla]

Since you agree the approaches are mathematically identical (as Alan proved) then you acknowledge that pupil magnification is involved (since that's what's in my approach). You simply deny it when it is not explicitly shown.

You should know the answer. To prove that they are mathematically identical, it is necessary to rewrite the equations so they're both in the same form. He could just as easily have rewritten your equations without p.

If you want to litigate this further, please do it in the Science and Technology Forum, where people understand the math. This isn't suitable for the general public.

Tom Axford and have presented another way of viewing the problem that we think is particularly instructive. We both acknowledged the method you use, years ago. You've made your point. Do you mind if we also make ours without undue interference?

 

[bclaff]

Since you agree the approaches are mathematically identical (as Alan proved) then you acknowledge that pupil magnification is involved (since that's what's in my approach). You simply deny it when it is not explicitly shown.

You should know the answer. To prove that they are mathematically identical, it is necessary to rewrite the equations so they're both in the same form. He could just as easily have rewritten your equations without p.

If you want to litigate this further, please do it in the Science and Technology Forum, where people understand the math. This isn't suitable for the general public.

That's already been done and you seem to have forgotten the result.

Tom Axford and have presented another way of viewing the problem that we think is particularly instructive. We both acknowledged the method you use, years ago. You've made your point. Do you mind if we also make ours without undue interference?

Good bye!

(And good luck to anyone who is mislead by your intuitive but wrong approach.)

--
Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

 

[Brett8883]

So my job at work can be described in its simplest form as translating between technical and non-technical people. This is necessary because god love them non-technical people never ask the question they really want answered and god love them the technical people almost never answer just the question that was asked. I’m the buffer.

So how id answer this question (based on the responses from the technical people) is that to compare the background blur of different focal length lenses and apertures from different distances while keeping constant framing you would take the focal length divided by the f-number. Higher the number will give you more background blur. Any objections?

No objection, but your answer can be refined. It is correct only for very distant backgrounds. In practice, this means that the background must be several times as far away as the subject or more.

If the background is very close to the subject (e.g. a head and shoulders portrait and the background is a couple of feet behind the subject), then the blurring changes very little with focal length (if the f-number and framing are constant).

In between situations can be worked out using the formula:

size of blur = (focal length / f-number) x distance from subject to background / distance from camera to background

Yea that’s a much more through and comprehensive answer to the question. Technical people are very good at giving comprehensive answers (that’s what makes them good at what they do) but where I come in is helping them condense their message into the lowest simplest form possible. It can be painful for technical people since they just aren’t wired to think that way.

The way I understood the question wasn’t how to find the exact amount of background blur, it’s how do you compare two systems. All non-technical people really want to know is which system provides more subject isolation. System A or system B. The exact amount is unnecessary/unhelpful to a large degree honestly.

Isn’t it true that focal length/f-number will always give the correct answer to this question when the subject framing and distance from subject to the background is the same? No matter the focal length, f-number, or sensor size combination? Is there any situation that wouldn’t be true? Given of course the background is far enough away from the subject that depth of field doesn’t extend into the background.


Again I’m not adding anything to the technical discussion here, just taking the information you guys have laid out and trying to help condense down your message. You seem to be interested in getting the message out so people can understand it or I wouldn’t bring this up.

I think “size of blur = (focal length / f-number) x distance from subject to background / distance from camera to background” is still too complicated for non-technical people. In fact by the time you know distance from camera to subject and subject to background it’s actually too late to make practical use of this equation because you can’t change the camera system and lenses you brought with you at that point.

 

[Tom Axford]

So my job at work can be described in its simplest form as translating between technical and non-technical people. This is necessary because god love them non-technical people never ask the question they really want answered and god love them the technical people almost never answer just the question that was asked. I’m the buffer.

So how id answer this question (based on the responses from the technical people) is that to compare the background blur of different focal length lenses and apertures from different distances while keeping constant framing you would take the focal length divided by the f-number. Higher the number will give you more background blur. Any objections?

No objection, but your answer can be refined. It is correct only for very distant backgrounds. In practice, this means that the background must be several times as far away as the subject or more.

If the background is very close to the subject (e.g. a head and shoulders portrait and the background is a couple of feet behind the subject), then the blurring changes very little with focal length (if the f-number and framing are constant).

In between situations can be worked out using the formula:

size of blur = (focal length / f-number) x distance from subject to background / distance from camera to background

Yea that’s a much more through and comprehensive answer to the question. Technical people are very good at giving comprehensive answers (that’s what makes them good at what they do) but where I come in is helping them condense their message into the lowest simplest form possible. It can be painful for technical people since they just aren’t wired to think that way.

The way I understood the question wasn’t how to find the exact amount of background blur, it’s how do you compare two systems. All non-technical people really want to know is which system provides more subject isolation. System A or system B. The exact amount is unnecessary/unhelpful to a large degree honestly.

Isn’t it true that focal length/f-number will always give the correct answer to this question when the subject framing and distance from subject to the background is the same? No matter the focal length, f-number, or sensor size combination? Is there any situation that wouldn’t be true? Given of course the background is far enough away from the subject that depth of field doesn’t extend into the background.

Yes, that is perfectly true. My post, that you have quoted above (in green), was simply refining your answer by explaining exactly the meaning of your last sentence above. For anyone interested, I explained exactly what happens if the background is not far enough away for your statement to be true.

Again I’m not adding anything to the technical discussion here, just taking the information you guys have laid out and trying to help condense down your message. You seem to be interested in getting the message out so people can understand it or I wouldn’t bring this up.

I think “size of blur = (focal length / f-number) x distance from subject to background / distance from camera to background” is still too complicated for non-technical people.

I am sure that is true. However, some readers here are quite technical and it was included for their benefit. I find that it is always difficult to decide how much of the mathematical details to include. We will probably always disagree on that one, but one of the merits of the dpreview forums is that they bring together people with widely differing points of view. You can have your approach and I can have mine. ;-)

 

[ThrillaMozilla]

Ooh, I found it! This may be the first incarnation of the diagram. For the other part: Also here.

Other people appreciate the simplicity of viewing depth of field from the subject side. I particularly like the subsequent remarks by BobN, John Sheehy, Iliah Borg ("Should be on every depth of field site") , and of course Tom Axford. John Sheehy wrote that he considers DOF from the point of view of the subject. He also wrote that concerning his usual subjects: "Once you start thinking this way, there is no point in looking at things the conventional way, and even in situations where frame quality is more relevant, the model still works...."

.