Question about LF DoF

[raucousimages]

I was using an on line dof calculator and it shows a difference in dof with different formats (4X5 and 8X10) with the same lens and focal distance. Why?
I don't see why the DoF would change if the only thing I do is change my 8X10 back for a 4X5 back without changing lens, F-stop or focus unless the exrta depth is in the outer edge of the image circle. I am shooting portraits at about 8 to 16 Ft with lenses between 150mm and 480mm so I am working with very shallow DoF and I thought a DoF calculator would help but this info is confusing.

Thanks John

 

[David A. Goldfarb]

DOF is a function of focal length, subject distance and the acceptable circle of confusion for the format. If you move to a larger format, you calculate the DOF with a larger value for acceptable circle of confusion.

In other words, DOF is about print sharpness, not neg sharpness, and presumes certain things about how much you are going to enlarge and normal viewing distance.

For particular purposes (like big enlargements), you might have to deviate from standard values for acceptable circle of confusion (or you might do so indirectly, say by stopping down one or two stops from the recommended aperture on the DOF table or scale).

 

[bob01721]

"... DOF is about print sharpness, not neg sharpness, and presumes certain things about how much you are going to enlarge and normal viewing distance..."

Over the years, I've heard many explanations about DoF, but none as simple and insightful as that statement. Thank you, David.

 

[DrPablo]

The articles in View Camera from Sept/Oct and Nov/Dec on print sharpness are extremely thorough explanations of this. The circle of confusion, of course, is the minimum size at which an in-focus detail can be projected onto the film. So the lenses with the highest resolution will produce the smallest CoC (and therefore finest detail). This matters at your enlargement size, because as you enlarge an image your CoC becomes correspondingly enlarged. To take a fairly reasonable example, a 4x5 image needs to be enlarged only 2x to produce an 8x10, but a 35mm image needs to be enlarged 8.6x to produce an 8x10. So the finest details resolved on 35mm are enlarged more than 4-fold more than on a 4x5 (and obviously a contact-printed 8x10 doesn't need to be enlarged at all). So the CoC that produces an acceptably sharp enlargement (from a reasonable viewing distance) will vary from format to format. Correspondingly, how you define the near and far borders of acceptable sharpness are derived from the CoC, and this is what constitutes the depth of field.

 

[Rolleiflexible]

What you really need to know is this: It's real shallow in large format. And it gets a lot shallower when you move from 4x5 to 8x10. Figure on needing a couple more stops of light to keep the same DOF in 8x10, that you need in 4x5.

A few years ago I toyed with the idea of moving from 5x7 to 8x10. But then I realized I didn't have enough light for it, and sold the bigger camera.

Sanders

 

[David A. Goldfarb]

Hey, Sanders, why don't you sell one or two of your Rolleiflexes and buy some strobes?

 

[DrPablo]

And it gets a lot shallower when you move from 4x5 to 8x10.

I've only shot 4x5 myself, but I've always had the feeling that's true. When I first got my view camera I stopped down a lot more than I needed to. But now even for macros and close portraits I seldom need to stop down narrower than f/11 to get my subject in focus with an OOF background.

 

[eddym]

In case you'd like to do the math, here are the formulas. You have to start with the hyperfocal distance:
Hfd=(FL)squared/f times d
where Hfd=hyperfocal distance; FL=focal length; f= f stop; and d=diameter of circle of confusion.
Once you know hyperfocal distance you can calculate depth of field for any lens:

dn=(Hfd)times(D)/(Hfd+D), where dn=near depth (the distance in front of the plane of focus that appears sharp); and D=distance. All measurements must of course be in the same units; milliemeters are easiest to work with.

df=(Hfd)times(D)/(Hfd-D), where df=far depth (the distance behind the plane of focus, as above)

dn+df= total depth of field.

The major variable is of course d, the diameter of the circle of confusion, and it is defined differently by different sources, such as lens manufacturers. Rollei, for example, gives a "general use" d for 6x6 as .084 mm; for "critical use" it is .056 mm. In other words, critical use implies greater enlargement of the negative.

 

[Sirius Glass]

The circle of confusion is part of the story. If you take a camera, any format camera, and have an image size for that format film, then regardless of focal length, the depth of field will be the same! I found this out with a 35mm when I wanted to photograph an object filling the frame => I would change to a wide angle lens and have the same DOF ... I tired all my lens and had the same DOF [Then 20mm, 24mm, 28mm, 35mm, 50mm, 58mm, 85-205mm zoom]. I tried this many times and always had the same result and could not understand why. :confused:

Many years later, I worked at Kodak and I was taking optics classes. I asked about this conundrum. The instructor started with the focal length equations and with some algebra proved that with a constant image size, there is an ideal DOF for an image size and one can't do better than that.

How does this apply? There is a myth that the larger the format the less DOF there will be. This is not true. If you took your LF [or even MF] lenses and projected an image of say someones face onto a 24mm X 36 mm frame you would get a great depth of field. Now repeat that with a 6 cm X 6 cm frame first with the same image size as before => same DOF, the fill the 6 X 6 frame => less DOF for the same image size. This can be repeated for 4" X 5" and 8" X 10". The DOF is the same for the same image size, but increase the image size and the DOF decreases!

Your problem is relative, if you want a greater DOF, backup and accept a smaller image size. The result will be that you start with a smaller image on the film but you will have a greater DOF. The enlarger will produce the product you want with the DOF that you want!

Or you could use an emulsion that was faster ... but you went to LF [MF] to get ride of grain and improve the resolution so that is not a satisfactory solution!

References with equations:
http://www.luminous-landscape.com/tutorials/understanding-series/dof.shtml
http://www.luminous-landscape.com/tutorials/dof2.shtml
http://www.photozone.de/3Technology/demos/DOFbutton.htm
http://www.vanwalree.com/optics/dof.html
http://www.normankoren.com/Tutorials/MTF6.html#DOFsharpness

Maybe this should be cross posted on the 35mm and MF forums too.

Steve

 

[David A. Goldfarb]

Your problem is relative, if you want a greater DOF, backup and accept a smaller image size. The result will be that you start with a smaller image on the film but you will have a greater DOF. The enlarger will produce the product you want with the DOF that you want!

This is true, but stepping back so that the image is smaller on film and cropping at the enlarging stage is essentially the same as just shooting a smaller format. Indeed, if you want more DOF with the same framing as you are using on a larger format, and stopping down is not possible or desirable, then you should shoot a smaller format.

 

[DrPablo]

I could probably figure this out were I energetic enough to look at the equations, but maybe one of you knows this.

F/stop, of course, is the quotient of focal length divided by pupil diameter.

Insofar as aperture is a major determinant of DOF, is it the focal length or the absolute diameter that influences DOF more?

Obviously if you held all variables equal except focal length, a plot of DOF versus focal length would show an inverse relationship.

But if you plotted DOF versus pupil diameter would you find a different curve?

In other words does aperture affect DOF in a way that is quantitatively independent of focal length?

 

[John Koehrer]

I have seen in a couple of sources that say "regardless of focal length, at the same PHYSICAL aperture the dof is the same". 10MM=10MM.

 

[DrPablo]

Just to demonstrate what a flipping geek I am, I had to try it for myself.

The DOF was calculated using dofmaster.com, and the parameters were an APS camera 2 feet from the subject. I needed to use this in order to avoid dealing with infinite depths of field.

It's easy to envision a fixed pupil diameter. Just select a constant ratio of focal length to f/stop and you'll get it. So I chose a diameter of 2mm, which occurs when the focal length is 2x the f/stop.

As we're all comfortable with, a diminishing f/stop (all else equal) should increase DOF, and an increasing focal length (all else equal) should decrease DOF.

As you can see here, despite the fact that the physical aperture is identical for all combinations, DOF decreases dramatically as you increase focal length, and this is not mitigated by higher f/stops.

As you can see when you plot DOF against focal length (with physical aperture held constant) DOF asymptotically decreases towards 0 with increasing focal length, and it asymptotically increases towards infinity with decreasing focal length.

So I can conclude that physical aperture in and of itself is not an important determinant of DOF except in relation to a certain focal length. In other words 10mm is not 10mm. Its effect changes depending on the focal length.

 

[Sirius Glass]

So I can conclude that physical aperture in and of itself is not an important determinant of DOF except in relation to a certain focal length. In other words 10mm is not 10mm. Its effect changes depending on the focal length.

You missed my point. Reread my post. If you work the equations, you will see that:
1) depth of DOF increases as the f/stop decreases,
2) for the same image size on the focal plane focal length drops out of the equation and is not a factor for DOF.

Therefore, increase the film speed, increase the f/stop, reduce the image size on the focal plane, or reduce the film size. For this problem only the last two are effective. [You can argue with me, but you will never win an argument with physics. Have you ever noticed that we have thick books on how photons behave and yet those photons know exactly what to do without ever spending any time studing those books. ]

You really need to do the algebra on this one. See the referenced web sites for the equations to start from. Typing equations does not work well in these types of forum so this is left as an exercise to the reader.

Unfortunately, diffaction starts to dominate and therefore the theorectical best DOF turns out to be:

~f/8 for 35mm
~f11 for MF
~f/16 for LF
Your mileage may vary.​

Steve

 

[DrPablo]

I didn't misread your post, I was just responding to John's and wanted to work it out for myself. And I'm better at Excel than I am at algebra, so I chose to do it that way instead

 

[Justin Cormack]

The amount of diffraction does depend on the pupil diameter, but then the further the pupil is from the film the more this diffraction is enlarged, so effectively aperture is all you need to know, not pupil size.

 

[argus]

To take a fairly reasonable example, a 4x5 image needs to be enlarged only 2x to produce an 8x10

I'm sorry to correct you but that should read 4 times if you take the square size of the format into account.

G

 

[Sirius Glass]

Ok campers, here is a website that clearly demonstrates that for a given image size the DOF is independent of focal length:

http://www.luminous-landscape.com/tutorials/dof2.shtml

and

from
http://www.luminous-landscape.com/tutorials/understanding-series/dof.shtml

DOF / Focal Length & Image Size
At the beginning of this tutorial I wrote, "Most people also believe that wide angle lenses have more depth of field than telephoto lenses (false)." Why is this such a common misconception?

It seems logical. Wide angle lenses appear to have more depth of field than long lenses, don't they? Yes, they appear to but only when you don't take subject size into account.

Imagine a person in a photograph holding a flower in front of them, and with a mountain range in the background. Now set up the shot so that the person's head fills about half the frame. Take a series of shots with every lens in your arsenal, from a 14mm ultra wide to a 600mm ultra telephoto. As long as the head is the same size in each frame the Depth of Field will be the same. The flower and the mountains will be equally in focus or out of focus as you change lenses.

Though I don't use a model with a flower and a mountain, here is a link to a page that demonstrates a small test that anyone can do themselves so that they can see the validity of my assertion.

Steve

 

[Helen B]

The Luminous Landscape article seems a little restricted in its treatment of the subject - it doesn't consider enough conditions. The link Steve gave earlier to http://www.vanwalree.com/optics/dof.html goes into the subject in greater detail, and considers a variety of cases.

From that site:
"Generally, when two lenses are compared at the same image magnification and F-number one must discern between two cases. When for both lenses the object distance is much smaller than the hyperfocal distance, the depths of field are essentially the same. On the other hand, when one or both object distances is not small with respect to the hyperfocal distance, the lens with the shorter focal length brings more depth of field. "

There are a few such discussions here on APUG.

Best,
Helen

 

[Sirius Glass]

There are a few such discussions here on APUG.

Best,
Helen

I am new to this forum so please be patient with me so that I get this right ... you are saying that we should have more discussions like this or that we should not have more discussions like this?

Steve

 

[Helen B]

I am new to this forum so please be patient with me so that I get this right ... you are saying that we should have more discussions like this or that we should not have more discussions like this?

Steve

Hi Steve,

Welcome to APUG. I was trying not to express an opinion, only to offer a piece of information. If you want my opinion, it is that discussions such as this can't do any harm. They may cover a lot of old ground, but there's generally some different slant, or some different way of looking at things. Always something to learn, and no harm learning it. There's value in looking through the archives, and there's value in raising the issue afresh. That's my opinion. As to what other people 'should' do, that's up to them, not me.

Best,
Helen

 

[DrPablo]

I'm sorry to correct you but that should read 4 times if you take the square size of the format into account.

G

While I appreciate your input, and have no argument with the math, I disagree with the rationale behind your correction. Enlargement factors are nearly always discussed in terms of linear dimension, not area. Area is simply not a very useful way to discuss the subject, and this is for several reasons.

1) Because we discuss resolved detail in terms of linear parameters. LPM is a linear parameter, i.e. line pairs per millimeter -- NOT square millimeters. Or circle of confusion, which is the diameter of a circle, not the area of the circle. If you want to determine how resolution changes with successive enlargements, you are using linear measurements. For instance, 50 lpm goes to 25 lpm when you enlarge from 4x5 to 8x10. That's a 2x enlargement, not a 4x enlargement. The articles about print sharpness in the Jul/Aug and Sept/Oct issues of View Camera are pretty clear about this.

2) Because our convention for describing sizes uses a linear nomenclature. We talk about 4x5 and 8x10 inch prints. We don't talk about 20 square inch or 80 square inch prints. And using a linear nomenclature allows us to avoid the synonymity of different aspect ratios that have the same area. So a 2x10 image could have the same area as a 4x5, but they're vastly different things. If you start from 4x5, then you contact print a 4x5 with no enlargement. But the only way to go from 4x5 to 2x10 would be to make an enlargement to 8x10 and crop 75% of it away. The area stays identical, but you have been forced to make an enlargement to achieve the new image.

3) When discussing crop factors between different film formats, you ALSO discuss this in terms of linear dimensions. Going from 24x36 to 15x24 film necessitates a 1.6 correction factor. This is because the linear dimensions of a 35mm film are 1.6x longer than APS film. You don't apply an area correction, even though indeed the larger frame has 2.56x the area.

So while it's true that an 8x10 has 4 times the area of 4x5, it is not very helpful to discuss enlargements as multiplications of surface area -- and that's why you don't see enlargement factor discussed that way very often.

 

[kirkfry]

Come on, don't make this hard. My 300 mm lens produces exactly the same image on 4X5 film as it does on 8X10 at the same f stop with exactly the same DOF. There is just less picture on the 4X5. In contact prints they will have exactly the same DOF. It is only after you start mucking about with the enlargement that it is different. So if you enlarge fuzzy it looks fuzzier. It is not rocket science.

 

[Rolleiflexible]

Kirk, I agree with you. It's not a hard subject. And of course you are right, that the DOF of any lens does not change based on the size of the film behind it. The last chapter of the story is that, to make the same image in 8x10 that you made with the 300 in 4x5, you have to move up to a 600 mm lens. And everybody knows that the DOF in a longer lens is shallower than in a shorter one.

Simple stuff. No need for quadratic equations, just go shoot with longer and shorter lenses and see what happens. The bigger the format, the longer the lens, the more light needed to maintain any given DOF. That's all a photographer needs to know. Leave the numbers to the engineers.

Sanders

 

[argus]

Thanks Paul.
I appreciate your explanation. I think it's clear and settled for me.

G