Macro vs extension tube

Michael Fontana · Jun 11, 2009

Does anybody experienced the difference of these two methods of close up?

Well, I know that a macro has been calculated for close focus, and for rather closed f-stops, but how shows the diffrence up in a image?

Is it correct, that a lens with a extension tube will have a more shallow DOF than its counterparth - same focale lenghts - the macro?

Doug Kerr · Jun 11, 2009

Hi, Michael,

Michael Fontana said:
Does anybody experienced the difference of these two methods of close up?

Well, I know that a macro has been calculated for close focus, and for rather closed f-stops, but how shows the diffrence up in a image?

Is it correct, that a lens with a extension tube will have a more shallow DOF than its counterparth - same focale lenghts - the macro?

First, note that an extension tube does not alter the focal length, nor the actual f/number, of the lens with which it is used. It merely shifts the range of distances at which the lens is able to focus.

Then, note that for a given focal length, a given f/number, and a given magnification, (and assuming we keep a consistent acceptable circle of confusion criterion, the basis on which we consider that objects at certain distances are or are not in acceptable focus) the depth of field should be identical.

In fact, for a given f/number and a given (and relatively large) magnification (and again assuming a consistent acceptable circle of confusion value), the depth of field is very nearly the same for any focal length.

Generally speaking, we may find a more desirable image result (for any given needed magnification, and any desirable focal length) with a lens that can inherently focus at the needed distance (a "macro" lens) than with another lens and an extension tube, owing to the (hopefully) optimized design of the "macro" lens for such work. For example, the "macro" lens may exhibit better flatness of field, or less geometric distortion, than another lens we might use with an extension tube.

But not necessarily.

Jack_Flesher · Jun 11, 2009

Michael Fontana said:
Is it correct, that a lens with a extension tube will have a more shallow DOF than its counterparth - same focale lenghts - the macro?

Kind of, but due to a specific reason: there are two styles of "macro" lenses.

The first is a basic design where to focus closer, the lens is simply extended to 2x the focal length to achieve a 1:1 reproduction ratio. Whether this lens extension is due to the built-in focus helical on the lens body itself OR supplied by extension tubes or macro bellows, the result is the same. A sidebar point, is that when you extend the lens group to focus closer, you also lengthen the lens' focal*. Hence at 1:1 magnification ratio, a 100mm lens is in all actuality an effective 200mm focal and behaves as such -- and why the net aperture value as marked on the lens 'changes' and why the DoF is affected significantly.

(*Note to Doug: Sorry to disagree with what you wrote, "First, note that an extension tube does not alter the focal length," but what I wrote above is correct. As a fixed lens-group extends, it's effective focal-length changes correspondingly; the marked focal on any lens is only valid at infinity focus. This is precisely why a large format shooter -- 8x10 and above -- can often shoot portraits with a standard lens and not have the spatial distortion issues we see when doing the same with 35mm. A human portrait on 8x10 is around 2:3 magnification, so the "standard" 300mm lens becomes extended to almost 500mm, and generates a comparable effect to using an 85mm lens on a 35mm camera. It is also why we have to add an exposure factor for bellows extension in LF shooting; the lens focal, f, gets longer but the aperture diameter, D, remains constant, so the actual aperture number defined by N = f/D gets numerically larger.)

The second design is generally referred to as "internal focus." IF lenses are actually a type of zoom lens where instead of extending the lens group, it alters (reduces) the focal length of the lens. Since the flange-focal distance remains constant, the lens focusses closer as it is "zoomed" shorter. In normal lenses used at normal focussing distances, the shift of focal required is very tiny and for the most part goes unnoticed; it is effectively an amount equal to the conventional focus design, though in the opposite direction. The current Canon 100 Macro is of this design, or more accurately a hybrid (the old one was conventional), and at 1:1 reproduction is essentially a 70mm lens, which ironically, is *the same* focal length as their own 50mm macro with the special 1.4x macro converter tube attached(!)

Cheers,

Olaf Ulrich · Jun 11, 2009

Michael Fontana said:
Does anybody experienced the difference of these two methods of close up?

Huh? I don't understand the question in the first place ... what 'two methods'? Using extension tubes is macro. So what is 'the other method' supposed to be?

Doug_Kerr said:
Then, note that for a given focal length, a given f-number, and a given magnification, (and assuming we keep a consistent acceptable circle of confusion criterion [...]), the depth of field should be identical.

But it's not. At large magnifications, depth-of-field also depends on the pupil magnification. Smaller exit pupils give more depth-of-field.

Jack_Flesher said:
The first is a basic design where to focus closer, the lens is simply extended to 2× the focal length to achieve a 1:1 reproduction ratio.

As a matter of fact, not just a certain kind but all kinds of lenses extend to twice their focal length when being focused at 1:1 magnification.

Jack_Flesher said:
Whether this lens extension is due to the built-in focus helical on the lens body itself or supplied by extension tubes or macro bellows, the result is the same.

Exactly.

Jack_Flesher said:
A sidebar point, is that when you extend the lens group to focus closer, you also lengthen the lens' focal*. Hence at 1:1 magnification ratio, a 100 mm lens is in all actuality an effective 200 mm focal and behaves as such -- and why the net aperture value as marked on the lens 'changes' and why the DoF is affected significantly.

This is nonsense. Obviously you are confusing focal length and image distance.

Jack_Flesher said:
Note to Doug: Sorry to disagree with what you wrote, "First, note that an extension tube does not alter the focal length," but what I wrote above is correct.

No, it's not. Doug is right; neither extension tubes nor macro bellows will alter the lens' focal length.

Jack_Flesher said:
As a fixed lens-group extends, it's effective focal-length changes correspondingly; the marked focal on any lens is only valid at infinity focus.

Actually just the contrary is true: a lens with internal focusing (i. e. non-fixed elements) will change effective focal length on focusing closer, so their nominal focal length is valid only at infinity.

Jack_Flesher said:
This is precisely why a large format shooter -- 8×10 ins and above -- can often shoot portraits with a standard lens and not have the spatial distortion issues we see when doing the same with 35mm.

The reason why a large-format shooter can do that is because, contrary to common belief, the angle of view does not depend on focal length (for a given image format) but on the image distance. At or near infinity focus, focal length and image distance are the same---as a matter of fact, that's how focal length is defined: it's the image distance at infinity focus. So we are used not to distinguish too precisely between these two things. And that's where your confusion is coming from.

Jack_Flesher said:
The current Canon EF Macro 100 mm is of this design, or more accurately a hybrid [...], and at 1:1 reproduction is essentially a 70 mm lens, which ironically, is *the same* focal length as their own 50 mm macro with the special 1.4× macro converter tube attached!

Yes, a modern 100 mm macro lens, focused at 1:1, really has an effective focal length of approx. 70 mm or thereabouts. But that is still longer than 50 mm which is the focal length of a 50 mm macro lens, no matter whether it's on a macro tube or not. A 50 mm lens will become 70 mm only on a 1.4× tele converter, not on a macro tube.

-- Olaf

Michael Fontana · Jun 11, 2009

thanks for your posts, seems more complex, than my questions expected ;-)

Dougs
"For example, the "macro" lens may exhibit better flatness of field, or less geometric distortion, than another lens we might use with an extension tube."

Yep - I noticed the very little distortion my two Canon macro's (50 and 100) have.

Thanks Jack - your LF-analogy helped a lot and grabbing in memory - thinking in terms of LF - helps understanding the initial question better - just moving the ground glass (or the lens - type 1) - but then, it's still different, as in LF the lens groups (type 2) doesn't move, but the entire lens only.
Now, the modern 100 mm macro, beeing in 1: 1 a 70 mm, does it has the DOF of a 100 or a 70 mm?

As even in LF, macro-lenses are in use, there must a difference between a macro lens and a °normal° lens with extension tube (longer bellow). - - wasn' the exposure factor as big as 4 (square of 2 x focal lenghts) for 1:1?

Conclusion:
The 100 mm macro will have a similar (70/85) DOF in the 1:1- region as a 85 mm - non macro - with a extension tube. Correct?

How about 1:2 or 1 :3?

Olaf Ulrich · Jun 11, 2009

Michael Fontana said:
Now, the modern 100 mm macro, being in 1:1 a 70 mm, does it has the DOF of a 100 or a 70 mm?

Focused at 1:1, it has the depth-of-field of 1:1.

-- Olaf

Jack_Flesher · Jun 11, 2009

deleted, double post

Jack_Flesher · Jun 11, 2009

Olaf Ulrich said:
This is nonsense. Obviously you are confusing focal length and image distance.

No, it's not. Doug is right; neither extension tubes nor macro bellows will alter the lens' focal length.

Yes, a modern 100 mm macro lens, focused at 1:1, really has an effective focal length of approx. 70 mm or thereabouts. But that is still longer than 50 mm which is the focal length of a 50 mm macro lens, no matter whether it's on a macro tube or not. A 50 mm lens will become 70 mm only on a 1.4× tele converter, not on a macro tube.

-- Olaf

No I'm not, and yes it will -- you need to go back to optical school Olaf. Explain why the aperture value changes as you add extension please? (Hint, the formula for aperture number is, N = f/D.) There is a very good primer on this subject in Stroebels "View Camera Technique."

And yes, as a matter of fact the Canon macro adapter for the 50mm macro lens DOES have a converter element in it and why I wrote my original comment including that little tidbit...

Jack_Flesher · Jun 12, 2009

Michael Fontana said:
Thanks Jack - your LF-analogy helped a lot and grabbing in memory - thinking in terms of LF - helps understanding the initial question better - just moving the ground glass (or the lens - type 1) - but then, it's still different, as in LF the lens groups (type 2) doesn't move, but the entire lens only.
Now, the modern 100 mm macro, beeing in 1: 1 a 70 mm, does it has the DOF of a 100 or a 70 mm?

As even in LF, macro-lenses are in use, there must a difference between a macro lens and a °normal° lens with extension tube (longer bellow). - - wasn' the exposure factor as big as 4 (square of 2 x focal lenghts) for 1:1?

Conclusion:
The 100 mm macro will have a similar (70/85) DOF in the 1:1- region as a 85 mm - non macro - with a extension tube. Correct?

How about 1:2 or 1 :3?

1) The modern Canon macro at 1:1 will have DoF that matches a 70mm lens at focus distance for the 1:1 magnification. The problem is the marked aperture is NOT accurate any more, so you'll have to recalculate it for a now 70mm lens. Since the stated aperture is for a 100mm lens, this is easy, just use 70% of the set aperture --- so f22 becomes ~ f15.4, or close enough to f16 that you can just use one stop lower than set. Unfortunately, the lens is physically longer than 70mm at lower image magnifications like 1:2 or 1:4 and since we don't the exact focal length other than it's somewhere between 70 and 100, the calculation is not so easy...

2) The difference between LF macro lenses and regular LF lenses is the optimal working magnification they are designed for. Most LF lenses are optimized for 1:10 and optimum usable range from 1:7 through infinity. Some of the more critical makers use 1:7 which they say is at its best in the 1:4 through 1:20 range, claiming that is a more realistic 'normal' working range. Macro lenses are usually optimized for 1:1 and usable from about 2:1 through 1:2, though some are specially marked "2x" or "4x" meaning they are optimized for 2:1 or 4:1 magnifications accordingly. Copy lenses have a further distinction of being "flat field" as well. The issue here is correcting curvature of field and fall-off is held to be more important design criteria than maximum resolution, so they typically do not have the raw resolving power of a macro design but will perform better corner-to-corner on flat, evenly lit subjects. Note that the science of lens design is an exercise in optical tradeoffs and why there are so many different options when you get into specialized lenses.

3) No, the new Canon 100 macro at 1:1 will not have similar DoF characteristics to the 85 with tubes. The reason goes back to the fact the 85 with tubes has a longer net focal length and its DoF will in fact behave like a longer lens, say a 135.

4) 1:2, 1:3... Again, with the Canon macro, we don't really know what the actual focal length is, so it's difficult to calculate. Your best bet is to use image magnification, and there are simple image calculation ruler tools you insert into your image and take direct readings from that help calculate all required factors.

Important note on DoF formulae: Most DoF formulas and tables are for "normal" shooting distances and good ones will warn you that they don't work for macro distances! The reason comes back to the fact that focal length changes due to extension are variable and significant enough to create errors in the basic formula. So you need to use special tables or formulas for macro shooting that rely on image magnification ratios instead of focal length, or one of the direct reading rulers I mentioned above.

Cheers,

Doug Kerr · Jun 12, 2009

Hi, Jack,

Jack_Flesher said:
(*Note to Doug: Sorry to disagree with what you wrote, "First, note that an extension tube does not alter the focal length," but what I wrote above is correct. As a fixed lens-group extends, it's effective focal-length changes correspondingly; the marked focal on any lens is only valid at infinity focus. This is precisely why a large format shooter -- 8x10 and above -- can often shoot portraits with a standard lens and not have the spatial distortion issues we see when doing the same with 35mm. A human portrait on 8x10 is around 2:3 magnification, so the "standard" 300mm lens becomes extended to almost 500mm, and generates a comparable effect to using an 85mm lens on a 35mm camera. It is also why we have to add an exposure factor for bellows extension in LF shooting; the lens focal, f, gets longer but the aperture diameter, D, remains constant, so the actual aperture number defined by N = f/D gets numerically larger.)

I understand, and this is indeed an important point.

Nevertheless, putting an extension tube on a lens (if its focusing mechanism is at a certain point, so it has a certain actual focal length), does not change that focal length. It will cause you to put the focus mechanism in a different spot than without the tube, and that (as you aptly point out) may change the focal length.

Best regards,

Doug

Doug Kerr · Jun 12, 2009

Note that any lens operated at a magnification of 1.0 (1:1), whether within the range of its normal focusing system, or with the assistance of an extension tube, will have an effective f/number of twice the actual f/number of the lens (I am ignoring here transmission loss). (It may not work exactly that way in the case of a lens with a pupil magnification of other than unity.)

Best regards,

Doug

Jack_Flesher · Jun 12, 2009

Doug_Kerr said:
Hi, Jack,

I understand, and this is indeed an important point.

Nevertheless, putting an extension tube on a lens (if its focusing mechanism is at a certain point, so it has a certain actual focal length), does not change that focal length. It will cause you to put the focus mechanism in a different spot than without the tube, and that (as you aptly point out) may change the focal length.

Best regards,

Doug

Hi Doug:

This is correct, but only because the definition of lens' focal length is standardized for the cell group focused at infinity ;-) My point here is that the lens' true or effective focal length does increase with focus extension, and that is precisely why the effective aperture AND DoF changes in the fashion it does with extension (and why associated DoF formulae and tables don't work for macro).

Again, aperture number N = f/D where f is the lens' *actual* focal as calculated by measuring the optical center of the cell group's distance from the sensor plane, and D is the aperture iris diameter. So a 10mm aperture in a 56mm lens focused at infinity is f5.6. If we extend that lens to 1:1 magnification, we will have added another 56mm of extension for a total new focal length of 112mm, and the same 10mm aperture is now f11(.2), and precisely why we get the doubling or 2-stop change of aperture at 1:1...

(Trivia note. Convention allowed that all decimal points get rounded off after f10, so f11 is in fact f11.2 but it's just never written that way ;-). Note that f22 is really f22.4 and voilla, this is why instead of f44 we get f45, because it is really f44.8 and rounded up!)

Olaf Ulrich · Jun 12, 2009

Jack_Flesher said:
No I'm not, and yes it will -- you need to go back to optical school Olaf.

I'm afraid you need to go back to elementary school because your reading comprehension is severely lacking. Adding extension does not change focal length, period. Modern macro lenses do change (reduce!) their focal length upon focusing closer---however that doesn't happen due to added extension but due to moving lens elements. They also add extension but that doesn't affect effective focal length.

Jack_Flesher said:
Explain why the aperture value changes as you add extension please? (Hint, the formula for aperture number is, N = f/D.)

The nominal aperture doesn't change because f doesn't change on adding extension. The effective aperture does change because the formula for effective aperture isn't f/D but b/D (with b being the lens-to-image distance, as in the basic thin-lens formula 1/f = 1/g + 1/b; g being the subject-to-lens distance).

By the way, the thin-lens formula is the key to understanding this topic; do not try to teach others about all this unless you haven't fully digested it.

Jack_Flesher said:
And yes, as a matter of fact the Canon macro adapter for the 50 mm macro lens DOES have a converter element in it ...

Okay, then forget my remark about that one.

Jack_Flesher said:
1) The modern Canon macro at 1:1 will have DoF that matches a 70 mm lens at focus distance for the 1:1 magnification.

That's right. And the point is: a 70 mm lens at 1:1 has the same depth-of-field as any other lens at 1:1 on the same image format and at the same effective aperture (which may call for different nominal apertures, depending on the design of the lenses' focusing mechanisms). The effective aperture is the same when the shutter speed required for proper exposure is the same ... assuming everything else is constant, namely brightness and ISO setting.

Jack_Flesher said:
The problem is the marked aperture is NOT accurate any more, so you'll have to recalculate it for a now 70 mm lens.

Umm ... in fact, you have to re-calculate it for a now 70 mm lens focused at 1:1. So compared to the nominal value, the effective aperture value is one stop narrower, not wider. So if we are talking about a modern 100 mm f/2.8 macro lens, at 1:1 the full aperture has an effective value of approx. f/4. In an old-style 100 mm f/2.8 macro lens which focuses purely by extension, the effective aperture wide open at 1:1 would be f/5.6.

Jack_Flesher said:
No, the new Canon 100 macro at 1:1 will not have similar DoF characteristics to the 85 with tubes.

As a matter of fact, it will. At large magnifications where depth-of-field is small in relation to the size of the field of view, it depends solely on magnification and not on focal length. Of course, the images still won't be equal because perspective and background rendition will differ. But depth-of-field will be just the same.

Jack_Flesher said:
The reason goes back to the fact the 85 with tubes has a longer net focal length and its DoF will in fact behave like a longer lens, say a 135.

Nonsense.

Jack_Flesher said:
1:2, 1:3 ... Again, with the Canon macro, we don't really know what the actual focal length is, so it's difficult to calculate.

That's right, but a simple linear interpolation between the nominal focal length and the effective focal length at 1:1 gives a fairly good estimate for the effective focal lengths at magnifications between 0× (infinity) and 1× (at 1:1).

Now to estimate the effective focal length at 1:1, set your lens to 1:1 and measure the distance from the focused-to subject to the camera's image plane. The effective focal length approximately is one fourth of that distance---approximately because we cannot include the distance between the lens' principal planes into the calculation, as that usually is unknown to the user ... still it's close enough for practical intents and purposes. If you happen to know your lens' principal plane distance then subtract it from the subject-to-image distance before dividing by four. Note that the principal plane distance can be positive or negative.

Jack_Flesher said:
DoF formulae [...] don't work for macro distances! The reason comes back to the fact that focal length changes due to extension are variable and significant enough to create errors in the basic formula.

It's right that typical normal-range depth-of-field formulas don't work well at high magnifications ... but the reason is not what you think. It's way more complex than that. Of course it's possible to give a depth-of-field formula that is accurate from infinity down to the highest possible magnification---but that formula would be pretty long and tedious to compute. At "normal" range, simplified formulas are accurate enough and much easier to memorize and to compute, so in textbooks you'll always see the simplified versions.

-- Olaf

Olaf Ulrich · Jun 12, 2009

Jack_Flesher said:
My point here is that the lens' true or effective focal length does increase with focus extension ...

Your point is that you don't understand what focal length is. You keep confusing it with image distance.

Jack_Flesher said:
(Trivia note. Convention allowed that all decimal points get rounded off after f/10, so f/11 is in fact f/11.2 ...)

Actually, it's f/11.3---or 8 * SQRT(2), to be exact. I'd say you should stop suggesting 'going back to optical school' to others until you learned your own lessons.

-- Olaf

Bart_van_der_Wolf · Jun 12, 2009

Doug_Kerr said:
Nevertheless, putting an extension tube on a lens (if its focusing mechanism is at a certain point, so it has a certain actual focal length), does not change that focal length. It will cause you to put the focus mechanism in a different spot than without the tube, and that (as you aptly point out) may change the focal length.

Exactly, the focal length doesn't change by using an extention tube/bellows. The focal length is determined by the optical construction, which can be altered with internal focussing or by adding a "close-up" lens element (e.g. screw-in filter).

Doug_Kerr said:
Note that any lens operated at a magnification of 1.0 (1:1), whether within the range of its normal focusing system, or with the assistance of an extension tube, will have an effective f/number of twice the actual f/number of the lens (I am ignoring here transmission loss). (It may not work exactly that way in the case of a lens with a pupil magnification of other than unity.)

Absolutely correct again. The apparent (or effective) aperture value changes because of the magnification (the F-number or physical aperture remains the same). The image circle of the in-focus image is magnified due to the longer image focus distance, thus spreading a given photo flux over a larger surface area which reduces the intensity per unit area. The easiest approximate calculation is (M+1)^2, with M being the magnification factor.
The pupil magnification is automatically incorporated in that calculation when we can measure the actual projected subject size, although transmission losses are not.

Jack_Flesher said:
Hi Doug:

This is correct, but only because the definition of lens' focal length is standardized for the cell group focused at infinity ;-) My point here is that the lens' true or effective focal length does increase with focus extension, ...

True focal length doesn't change. 'Effective' focal length is a very misguiding concept, and it doesn't help other readers in understanding the real issues, IMHO. The focus distance of the projected image does increase as one focuses closer, but that not the same as the focal distance. We need to make a clear distinction between Focus versus Focal distance, to avoid confusion.

Jack_Flesher said:
and that is precisely why the effective aperture AND DoF changes in the fashion it does with extension (and why associated DoF formulae and tables don't work for macro).

If one uses the correct formulae, they work perfectly fine. When the subject magnification factor becomes significantly larger than at near infinity focus, we need to incorporate the a-symmetry of the lens design if we want accurate calculation results. That includes the so-called "pupil magnification" factor, which often is not specified and therefore left out of the formula. Especially for Macro photography the exact subject magnification becomes more important, so we should include the pupil magnification for accuracy.

Fortunately, there are DOF tools that do allow to specify the pupil magnification if we want to calculate DOF that way, or we can use a reformulated calculation based on the observed magnification factor itself (regardless of what theoretical focal length and shooting distance is used). When a subject has the same size on the sensor as it has in real life, then the magnification factor is 1.0 (1:1), and the exposure needs to be (1+1)^2 = 4x as much as it would be at infinity focus.

Before sending others back to the school benches, I'd suggest all interested to read up on the matters on the very informative pages by Paul van Walree at http://toothwalker.org/optics.html. Relevant to the OP's question are the articles on DOF, and a free DOF calculator for Windows computer platforms, and the DOF calculation examples when using the pupil magnification and asymmetrical focal designs at a constant magnification.

I'll conclude by quoting from Paul van Walree's summary:

Paul van Walree said:
Hence, at close focus the depth of field depends only on the image magnification, the F-number, and the pupil magnification, regardless of the focal length.

Cheers,
Bart

Olaf Ulrich · Jun 12, 2009

Bart_van_der_Wolf said:
Before sending others back to the school benches, I'd suggest all interested to read up on the matters on the very informative pages by Paul van Walree ...

That's an excellent suggestion. I've read many texts, articles, and essays about the concept of depth-of-field, and Paul's articles are the only ones I found on the world's whole wide Internet that are completely free from errors, simplifications, and misconceptions.

-- Olaf

Doug Kerr · Jun 12, 2009

Let me clarify two wholly separate considerations at work in the topics in this thread. They are often entangled.

1. f/number

The f/number of a lens, N, is defined as f/D, where f is the focal length and D is the diameter of the entrance pupil.

In many lens designs, f changes as the position of the lens' own focusing mechanism is changed. It is also possible that in some designs, D changes as the position of the lens' own focusing mechanism is changed.

Thus N, the f/number of the lens, may well change as the position of the lens' own focusing mechanism is changed.

It does not change with the distance at which the camera is focused if that change is made without moving the lens' focusing mechanism (as is possible when we use an extension tube or closeup bellows accessory).

[Editorial note: Note that I make a distinction between "the position of the lens' own focusing mechanism" and "the distance at which the camera is focused", and particularly note in the latter my use of "camera" rather than "lens".

The reason is that a lens, not mounted at a camera, is not focused at a certain distance. (There must be an established "film plane" for that notion to have meaning.) Only a complete camera is focused at a certain distance. And, if we vary the mounting of the lens on the body (as when an extension tube or closeup bellows is used), the focus distance of the camera, for a given position of the lens' focusing mechanism, will vary.]

Thus the actual f/number, for a certain setting of the lens' focusing mechanism, may well be different from the "stated" f/number, which is by convention given for the lens at its "infinity" focus setting (the setting that would produce focus of the camera at infinity if the lens is mounted in the "normal" way, without the intercession of an extension tube of closeup bellows).

Typical, with the lens set to a closer focusing distance, the focal length will be less than the "stated" focal length, and probably the diameter of the entrance pupil is not much different. Thus, the actual f/number for such a "closer" focus setting will be smaller than for the infinity focus setting.

The notion of "effective f/number" is not involved in the above. We are speaking of a variation in the actual f/number (defined as at the outset).

2. Bellows factor

Totally separate from the matter in (1) above, but "piling on" to it, is that the parameter that tells us the "exposure impact" of a lens, which we may call the "effective f/number", is affected by the distance at which camera is focused as well as by the f/number itself (for the applicable setting of the lens' own focusing mechanism).

[Note again the distinction with "distance at which the camera is focused".]

If the lens has "unity pupil magnification", which means that the entrance and exit pupils are located at the first and second principal points, then the effective f/number, N', is given by:

N'=N (1+m)

where N is the (actual) f/number (yes, whatever that is for the position of the lens' focusing mechanism) and m is the image magnification at which we are operating.

(If the pupils are not located at the principal points - which we describe by saying that the pupil magnification is not unity - then the expression is a bit more complicated.)

The factor 1+m, which is the factor by which the effective f/number differs from the (actual) f/number, is often called the bellows factor for historical reasons.

At an image magnification of 1.0 (1:1), N' is twice N (a decline in exposure of 2 stops from this consideration).

Summary

Thus, when we focus our camera at a close distance with the aid of an extension tube of closeup bellows, the effective f/number (which is a factor in exposure) will differ from the "stated" f/number of the lens, often because of two quite separate considerations.

Ordinarily the decline in exposure owing to "bellows factor" is greater than the possible increase owing to change in focal length with focus setting, so we have a net decline in exposure when operating at high magnifications with the benefit of an extension tube or closeup bellows.

Now to breakfast.

By the way, today is Carla's and my tenth wedding anniversary.

Best regards,

Doug

Doug Kerr · Jun 12, 2009

Bart has urged caution in the use of the term "effective f/number", and perhaps even of the parameter that it designates. His point is that it can be misleading.

I fully agree.

But let me speak in defense of the term, and the parameter it designates (subject, as always, to the caveat that we must know what terms mean and how they can be properly applied).

The property photographic exposure results (always) from the interaction of effective f/number and exposure time. We usually simplify this, thinking of it as resulting from the interaction of f/number and exposure time. And, for focus at substantial distances, that approximation (which is what it is) is certainly workable.

But when we consider focus at smaller distances, our exposure equations (including those practiced by free-standing exposure meters), if fed the f/number, give an incorrect result. If fed the effective f/number, they give a correct result.

That, then, is the attraction of the parameter often labeled "effective f/number" (and it is hard for me to think of a more foolproof term). It can be directly used, in place of the "actual" f/number, in almost any of the exposure equations we use (again, as I said, in those practiced by free standing exposure meters), rendering them applicable for operation at whatever focus distance (that is, at whatever magnification).

So I suggest we not shy away from the term "effective f/number", or the parameter it designates. But, as in the case of terms like "full-frame 35-mm equivalent focal length", we just need to know what it is, and what it isn't.

Best regards,

Doug

Olaf Ulrich · Jun 12, 2009

Doug_Kerr said:
Bart has urged caution in the use of the term "effective f-number", and perhaps even of the parameter that it designates. His point is that it can be misleading. I fully agree.

I don't see why it would be misleading. The nominal aperture number is

(1) Nnom = f/D

and the effective aperture number is

(2) Neff = b/D

with f = focal length; b = image distance; D = diameter of entrance pupil.

In textbooks you'll often find the formula

(3) Neff = Nnom * (m + 1)

with m = magnification and (m + 1) being the "bellows factor" you mentioned earlier. As according to the thin-lens formula m = (b/f) -1, you can easily see that (2) and (3) in fact are equivalent. I don't see anything misleading here; everything is crystal clear ... provided you understand the basic thin-lens formula and its implications.

Doug_Kerr said:
But, as in the case of terms like "full-frame 35-mm-equivalent focal length", we just need to know what it is, and what it isn't.

Exactly. In particular, f is not always equal to the number engraved on the lens' barrel. Neither it is always equal to the image distance (as Jack keeps suggesting).

-- Olaf

Bart_van_der_Wolf · Jun 12, 2009

Hi Doug,

Congrats to you and Carla on the anniversary.

Doug_Kerr said:
The property photographic exposure results (always) from the interaction of effective f/number and exposure time. We usually simplify this, thinking of it as resulting from the interaction of f/number and exposure time. And, for focus at substantial distances, that approximation (which is what it is) is certainly workable.

But when we consider focus at smaller distances, our exposure equations (including those practiced by free-standing exposure meters), if fed the f/number, give an incorrect result. If fed the effective f/number, they give a correct result.

I prefer to look at the exposure effect from close focusing as an exposure (usually time or intensity) correction factor.

The real focal length and the real aperture or F-number can change due to opical conditions. The f/number is a ratio between focal length and F-number, both of which are subject to apparent or real changes.

When you mention "effective f/number", to me it seems like a confusing concept, because it is not clear which of the two caused the 'effect'. Paul van Walree mentions the "effective aperture" and the "effective F-number", but not the "effective f/number". Also David Jacobson mentions the f-number or f-stop and uses the symbol N in his formulas. He also uses the "effective f-number, symbol Ne. He doesn't use the effective ratio f/number.

Kind regards,
Bart

Doug Kerr · Jun 12, 2009

Hi, Bart,

Bart_van_der_Wolf said:
The real focal length and the real aperture or F-number can change due to opical conditions. The f/number is a ratio between focal length and F-number, both of which are subject to apparent or real changes.

I'm not sure what you mean by "F-number". Usually, that is just an editorial alternative to "f/number". But it almost seems from your discussion as if by"F-number" you mean the actual diameter of the entrance pupil (although that seems to be a peculiar name for it).

Can you clear that up for me?

When you mention "effective f/number", to me it seems like a confusing concept, because it is not clear which of the two caused the 'effect'.

Here, the root "effect" does not mean "what causes this number to be different from another number". So we need not be concerned that the term doesn't clarify this.

Rather the term "effective" means that this number describes the overall, bottom-line effect of the lens, it its current situation in the camera, on exposure (in the same form for which the f/number is an approximation).

As such, it takes into account:

• The diameter of the entrance pupil (with the lens focus ring set where it is)

• The focal length of the lens (with the lens focus ring set where it is)

• The magnification, which is a creature of:
-- Where the focus ring of the lens is set, and
-- Any gratuitous extension of the lens from external mechanical accessories, such as an extension tube of closeup bellows

and

• (If we are being really scrupulous) the effect of the lens transmission not being 1.0.

That number, plugged into any of the exposure equations in the place we usually, for convenience, plug the f/number, will make the equation give us the right result for the overall situation in existence. (I know you know that - I'm just restating it to complete my presentation!)

The numerical parameter is meaningful and useful. Perhaps you would suggest a name for it that you feel is better than "effective f/number".

Best regards,

Doug

Bart_van_der_Wolf · Jun 12, 2009

Doug_Kerr said:
Hi, Bart,

I'm not sure what you mean by "F-number". Usually, that is just an editorial alternative to "f/number". But it almost seems from your discussion as if by"F-number" you mean the actual diameter of the entrance pupil (although that seems to be a peculiar name for it).

Can you clear that up for me?

I've grown accustomed to using "F-number" just like David Jacobson defines it in his Lens tutorial, namely f/N:

David Jacobson said:
Part II - Apertures, f-stop, bellows correction factor, pupil magnification

We define more symbols

D diameter of the entrance pupil, i.e. diameter of the aperture as
seen from the front of the lens
N f-number (or f-stop) D = f/N, as in f/5.6
Ne effective f-number, based on geometric factors, but not absorption

Doug_Kerr said:
The numerical parameter is meaningful and useful. Perhaps you would suggest a name for it that you feel is better than "effective f/number".

"Effective aperture" is the term that Paul van Walree uses most. I think that makes it clear as to what the actual variable is, when one assumes that the focal length of an optical design is a fixed parameter after focusing (sometimes the only parameter, when focusing doesn't involve separate internal element/group shifts). It also ties in better to the concept of pupil magnification, which has more to do with asymmetrical lens design than focal length per se.

Bart

Olaf Ulrich · Jun 12, 2009

Bart_van_der_Wolf said:
Paul van Walree mentions the "effective aperture" and the "effective F-number", but not the "effective f/number".

That's because the 'effective f/number' doesn't exist in the first place. Aperture is someting like 1:2.8. The respective f-number (formula symbol 'N') is 2.8---i. e. the reciprocal of 1:2.8. In contrast to these, 'effective f/number' is a term you and Doug just made up yourselves.

-- Olaf

Doug Kerr · Jun 12, 2009

Hi, Olaf,

Olaf Ulrich said:
That's because the 'effective f/number' doesn't exist in the first place. Aperture is someting like 1:2.8. The respective f-number (formula symbol 'N') is 2.8---i. e. the reciprocal of 1:2.8.

Indeed. And that number, as you mention, has a name - which I like to render as "f/number" (although we see it in many typographical variants). "f-number" is certainly as good, or even "f number".

Just for thoroughness, the physical property that is being discussed is formally called the "relative aperture". As you pointed out just above, in different contexts, we express it as a ratio ("1:3.5") or as the "f/number" (N), which is normally presented as the denominator of a symbol like this: f/3.5.

Now, when we have a number that has the same dimensionality, and concept, but reflects the the actual, not idealized, effect of the lens on exposure (even focused at infinity, the effect of a lens on exposure, in these terms, is not exactly f/D, because of transmission considerations), we need to have a formula symbol for it (I often use just " N' "; "Neff", which I have read somewhere today, would be just as good), and a name.

The name "effective f-number" (to present it in one of its typographical variations) has been used for that parameter for some while, and I find it very descriptive.

In contrast to these, 'effective f/number' is a term you and Doug just made up yourselves.

No, I can't take the credit/blame for "making up 'effective f/number'". The term has been around for a long time (perhaps rendered as "effective f number", or "effective F number", or "effective f-number", or "effective F-number" - my point here does not involve the diversity of typographical presentations).

Best regards,

Doug

Doug Kerr · Jun 12, 2009

Hi, Bart,

Bart_van_der_Wolf said:
I've grown accustomed to using "F-number" just like David Jacobson defines it in his Lens tutorial, namely f/N.

Well, that's not really a definition of the quantity - just an description of the notation for "f/number" (or F-number, or f-number). The definition comes when we say:

N=f/D

But you used it in a sense inconsistent with that:

"The f/number is a ratio between focal length and F-number".

The f/number is the F-number (f number, f-number, F number). It's numerical portion (N) is the ratio between focal length and the diameter of the entrance pupil.

Perhaps that phrase was just an editorial misstep.

"Effective aperture" is the term that Paul van Walree uses most.

Yes, and I use the same term when describing the concept.

But there's a problem when we begin to use numerical values. If we speak of just the nominal "aperture" (meaning, to be fussy, the relative aperture) of a lens, and don't want to use the f/number notation, then for an "f/3.5" lens, we write "1:3.5" (as is done in most lens markings, and often in European photographic practice generally). That is numerically 1/3.5 (per the definition of a ratio), or about 0.286).

That numerical value expresses the relative aperture. It's the aperture - the diameter of the entrance pupil - expressed in relation to the focal length; that's why it actually has that name (even though we rarely use it). And not surprisingly it is numerically D/f (not f/D, which is the f/number).

Now suppose that the effective exposure performance of that lens in a certain close-focus situation is 2 stops worse than its nominal relative aperture would suggest. Thus, its effective aperture, stated in the "ratio" way, is 1:7.0 (which can properly be thought of numerically as 1/7.0).

We can write that as an "f/number", in this form: "f/7.0". If we want the "numerical value" of that f/number, it is 7.0 (f/7.0 is not a number - just a special symbol).

Suppose I have an equation that expects the relative aperture as an f/number, and I want it to work based on the "effective aperture" in this situation.I will need to feed it, in this case, 7.0.

Now should I call 7.0 (the number part of the f/number) the "effective aperture"? Well, I said just above that the effective aperture here is 1/7.0. So we would have one term attached to two numbers. Is the effective aperture here 7.0 or is it 1/7.0 (0.1428...)

And that's why the term "effective f/number" is needed as well as the term "effective aperture". They refer to two different, and inverse, numeric expressions of the same thing (rather like "frequency" and "period"). It is the same reason we have the terms "relative aperture" and "f/number".

Best regards,

Doug

Olaf Ulrich · Jun 13, 2009

Doug_Kerr said:
And that number, as you mention, has a name - which I like to render as "f/number" ...

You may prefer it that way---but it's wrong. You may write it with or without a hyphen but not with a slash. And Bart even tried to suggest that "f-number" and "f/number" were two different things---which is utterly wrong.

Doug_Kerr said:
"f-number" is certainly as good, or even "f number".

That's not as good; that's better.

Doug_Kerr said:
Just for thoroughness, the physical property that is being discussed is formally called the "relative aperture".

Yes, sure. In the context of photography, "aperture" virtually always means "relative aperture" ... or "relative geometric aperture," to be even more precise. If we're talking about absolute aperture then we'd usually say, "diameter of the entrance pupil" (formula symbol 'D'). By the way, when astronomers are talking about their telescopes and say "aperture," they usually refer to the absolute aperture.

Doug_Kerr said:
Now, when we have a number that has the same dimensionality, and concept, but reflects the the actual, not idealized, effect of the lens on exposure (even focused at infinity, the effect of a lens on exposure, in these terms, is not exactly f/D, because of transmission considerations), we need to have a formula symbol for it (I often use just " N' "; "Neff", which I have read somewhere today, would be just as good), and a name.

You are confusing two different concepts again. Transmission loss usually is not taken into consideration because (1) it's small in most cases, and (2) it would mess up things even more, and (3) most photographic effects---namely depth-of-field---really depend on the pure geometric aperture, not on transmission. If we want to take transmission loss into consideration then we'd say, "T-number" which is the equivalent f-number at a hypothetical 100 % transmission (equivalent in terms of exposure, not depth-of-field).

The effective f-number (formula Symbol 'Neff') is not the same as the T-number. Instead, it still is a purely geometric concept. It takes the actual focal length as well as extension (the "bellows factor") into consideration---and also, if we're scrupulous, the effects of a non-unity pupil magnification---but not transmission loss.

Doug_Kerr said:
No, I can't take the credit/blame for "making up 'effective f/number'".

Well ... it's your article (#2 further up this thread, first reply to the original question) which introduced "f/number"---which I never saw before in 30 years. Do not invent or adopt new typographical variations! They only mess up things and confuse people and don't add anything useful to the discussion.

Doug_Kerr said:
... for an "f/3.5" lens, we write "1:3.5" (as is done in most lens markings, and often in European photographic practice generally). That is numerically 1/3.5 (per the definition of a ratio), or about 0.286).

Don't get carried away with "1:3.5" and "f/3.5" ... these two ways of expressing (relative geometric) aperture really are perfectly equivalent. By convention, it's common to use "1:3.5" for the lens speed, i. e. the maximum relative geometric aperture, and "f/3.5" for an actual aperture setting which may or may not be equal to the lens' maximum aperture, as in "I took this photograph at f/8" (by the way, the f-number here is not f/8; it's 8).

Doug_Kerr said:
We can write that as an "f-number", in this form: "f/7.0". If we want the "numerical value" of that f-number, it is 7.0 (f/7.0 is not a number - just a special symbol).

Umm ... but "f/7.0" sure is a number---or more precisely, it represents a number. The number is focal length divided by 7.

The word aperture, in the context of photography, has several meanings. First of all, it's the hole formed by the iris. It also refers to the effective diameter of that hole. Effective here means that we're not talking about the actual diameter of the hole formed by the aperture blades but the apparent diameter thereof, as seen through the elements placed before it---in other words, the diameter of the entrance pupil. This is also referred to as the absolute aperture. Then there's the relative aperture which is the diameter of the entrance pupil in relation to the focal length (nominal relative aperture) or to the image distance (effective relative aperture). Then there's the aperture number which is nothing but the reciprocal of the (nominal or effective) relative aperture. The nominal aperture number is also referred to as f-number; this is what's written on aperture rings (do you still remember aperture rings?), and it also is what we use in most formulas dealing with exposure or depth-of-field. And finally, mentioned here for completeness, there's a thing called the aperture value which is the logarithm of the aperture number to the base of the square root of two; it's part of the APEX system of photographic values.

In other contexts besides photography, there are even more meanings of the word aperture. For an example, see the definition of aperture in the context of microscope lenses which is completely different from what we're accustomed to.

-- Olaf

Doug Kerr · Jun 13, 2009

Hi, Olaf,

Olaf Ulrich said:
You may prefer it that way---but it's wrong. You may write it with or without a hyphen but not with a slash.

Well, you're right. I will revert to the accepted typographical convention.

Please remind me what that is.

You are confusing two different concepts again. Transmission loss usually is not taken into consideration because (1) it's small in most cases, and (2) it would mess up things even more, and (3) most photographic effects---namely depth-of-field---really depend on the pure geometric aperture, not on transmission.

Indeed. But effective f-number (the parameter that takes into consideration the "bellows factor") is not used in connection with DoF formulas, so that argument is not pertinent.

If we want to take transmission loss into consideration then we'd say, "T-number" which is the equivalent f-number at a hypothetical 100 % transmission (equivalent in terms of exposure, not depth-of-field).

But the T-number does not take bellows effect into account, so that is not an appropriate notation for a parameter that is the "bottom line" of the exposure impact of a lens - the parameter I am seeking to fully characterize, and which is indeed only useful in exposure calculations.

Note that I use a separate symbol for the "bottom line" parameter (including the matter of transmission) - N'' rather than N' (my symbol for the effective f-number on a geometric basis).

Well ... it's your article (#2 further up this thread, first reply to the original question) which introduced "f/number"---which I never saw before in 30 years.

Oh, I introduced it long before that - probably in about 2004.

Do not invent or adopt new typographical variations! They only mess up things and confuse people and don't add anything useful to the discussion.

Yes, you're right. And again, I will adopt the accepted convention, if somebody will remind me what it is (I don't find it satisfactory that there are maybe four that are OK, but not five).

Now, a final semantic conundrum you may be able to assist me with.

Is "f/3.5" an illustrative f-number ? (Note that "f/3.5" is not actually a number but a string, unless we consider it an expression and evaluate it with the pertinent value of f), in which case it turns into D). In any case, "f/3.5" is not the number 3.5.

Or, do we take the view that, in the example above, the "f-number" is in fact the numerical value 3.5, and "f/3.5" is just a special way we present it?

If the latter is our outlook, then we can quite reasonably say, in connection with equations, that "N represents the f-number".

If the former is our outlook, then we cannot reasonably say that (since N does not represent "f/3.5", either as a string or an expression.

Note that a reasonable quantitative meaning of "relative aperture" is D/f (of which N is the inverse).

That is, we consider "aperture" here to mean the diameter of the aperture (precisely, the diameter of the entrance pupil), and the quantity "relative aperture" compares it to the focal length.

Thanks for your inputs.

Best regards,

Doug

Olaf Ulrich · Jun 13, 2009

Doug_Kerr said:
But effective f-number (the parameter that takes into consideration the "bellows factor") is not used in connection with DoF formulas

Huh!? But it is! Well, except maybe in simplified formulas not intended for use in the context of macro photography.

Doug_Kerr said:
But the T-number does not take bellows effect into account, so that is not an appropriate notation for a parameter that is the "bottom line" of the exposure impact of a lens ...

That's right. The T-number doesn't behave the same way as the f-number in the presence of a bellows factor. That's why it mustn't be included in those formulas.

Doug_Kerr said:
... the parameter I am seeking to fully characterize, and which is indeed only useful in exposure calculations.

Note that I use a separate symbol for the "bottom line" parameter (including the matter of transmission) - N'' rather than N' (my symbol for the effective f-number on a geometric basis).

Yeah, I see ... N is the nominal aperture number, N' is the effective aperture number which takes geometric issues like bellows factor and pupil magnification into account, and N" finally is N' plus the effect of a non-perfect transmission, right?

Simply think of the transmission loss as a neutral-density filter attached to the lens. It takes away a certain percentage of the light, no matter what's happening behind it.

Doug_Kerr said:
Oh, I introduced it long before that -- probably in about 2004. [...] I will adopt the accepted convention, if somebody will remind me what it is ...

It's "f-number."

Doug_Kerr said:
Is "f/3.5" an illustrative f-number?

No, it's an aperture (a relative aperture, to be specific). The f-number therein is 3.5.

Doug_Kerr said:
(Note that "f/3.5" is not actually a number but a string ...

Well ... if you start discussing on that level then please note that 3.5 is not a number but just a string, too. A number is an abstract concept, an idea as ancient Greek philosophers would say. "3.5" is just a symbol we use to denote it. Another notation would be "three point five" ... another would be the string of bits in the binary system, and then there a many more. Still the number behind all these is always the same. But don't let us get carried away with the difference between names and named things.

Doug_Kerr said:
... unless we consider it an expression and evaluate it with the pertinent value of f), in which case it turns into D).

Exactly. It is an expression actually.

Doug_Kerr said:
In any case, "f/3.5" is not the number 3.5.

That's right. The former is an aperture; the latter is an aperture number, also called an f-number. Keep separate things separate, even if they are closely correlated to each other. In an ideal world, different things would have different names, and no thing would have more than one name. Unfortunately, our world is far from ideal ... umm, or maybe it is fortunately so ;-)

-- Olaf

Doug Kerr · Jun 13, 2009

Hi, Olaf,

Olaf Ulrich said:
Huh!? But it is!

But not correctly.

The rigorous basic forms of the general equations (applicable to any focus distance) for calculating depth of field involve (among other parameters) focal length, f, and the diameter of the entrance pupil, D.

For convenience in use, these equations are often recast to ask instead for focal length, f, and (actual) f-number, N. (In effect, the equation gets D from f/N.)

We need not, and should not, feed such an equation the effective f-number, even for focus at a short distance.

It's "f-number."

I'm good with that.

Well ... if you start discussing on that level then please note that 3.5 is not a number but just a string, too. A number is an abstract concept, an idea as ancient Greek philosophers would say. "3.5" is just a symbol we use to denote it. Another notation would be "three point five" ... another would be the string of bits in the binary system, and then there a many more. Still the number behind all these is always the same. But don't let us get carried away with the difference between names and named things.

Agreed.

It is an expression actually.

Which identically equals D.

Best regards,

Doug

Olaf Ulrich · Jun 13, 2009

Doug_Kerr said:
The rigorous basic forms of the general equations (applicable to any focus distance) for calculating depth of field involve (among other parameters) focal length, f, and the diameter of the entrance pupil, D.

You can design formulas for any set of input parameters.

Doug_Kerr said:
We need not, and should not, feed such an equation the effective f-number, even for focus at a short distance.

Indeed. As a matter of fact, such a formula would compute the effective aperture number internally from the input parameters.

Doug_Kerr said:
Which identically equals D.

Sure. After all, D is the aperture ... the absolute aperture, that is.

In case you're interested, here is the most accurate formula for depth-of-field that I am aware of. It is valid at any distance. Still it's not perfect as it takes neither principal plane distance nor diffraction effects into account. The input parameters are focal length f, diameter of entrance pupil D, focus distance d, pupil magnification p, and of course the diameter of the acceptable circle of confusion z. If you prefer it to use the f-number N rather than D as one of the input parameters, simply replace all occurrences of D by f/N.

DOF = 2 * f * D * z * (G(f, d) - f) * ((p - 1) * f - p * G(f, d)) / (p * z² * (G(f, d) - f)² - p * D² * f²)

It uses a sub-function G which is defined as follows:

g = G(f, d) = d/2 + sqrt(d * (d/4 - f)) for magnification < 1,
g = G(f, d) = d/2 for magnification = 1, and
g = G(f, d) = d/2 - sqrt(d * (d/4 - f)) for magnification > 1.

It computes the subject-to-lens distance g from the subject-to-image distance d and the focal length f. Please note that the distance d cannot be smaller than 4 * f. Lenses cannot cast real images at distances shorter than 4 * f; that's a physical law.

In case you're wondering why the formula takes f as one of the input parameters even though we stated that, at large magnifications, depth-of-field does not depend on focal length then please note that f gets cancelled out of the equation at high magnification but not at low magnification.

-- Olaf

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