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10-31-2011, 03:17 AM   #1
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Question about DOF

Hi!

I am a beginner in DSLR field. So may be my question is silly a little bit. I have read some texts (Wikipedia, etc.) about DOF and even carried out some DOF calculations in Excel on my own. However, I do not feel very strong and knowing in the concept of DOF.
As I understand, in theory, the length of DOF is determined only by four factors – focal length of lens, distance to object, aperture and circle of confusion. In turn, the circle of confusion is determined by the size of the camera’s sensor. For example, it is 0.02 mm for Pentax K-x (my camera) and for cameras with sensor that have the same crop factor. This theory is very nice, straightforward and even quite simple. Just input four figures in Excel and you get the result .
So theoretically the length of DOF for different lenses with the same focal length and the aperture (if the same camera is used) will be equal because all four parameters will be the same. For example, I have a standard kit lens DA L 18-55mm F3.5-5.6 AL. If I open aperture to F3.5 (maximum for my lens) and set focal length to 18 mm, the hyperfocal distance will be 4.56 m, according to theoretic formulae. If I focus camera to the object 5 m away, I should get sharp image from 2.39 m away from camera to infinity. Theoretically I can achieve the same DOF by more expensive and high quality lens which has maximal aperture smaller that F3.5 (e.g. F2.8 or even F1.4) if I set its aperture to F3.5. However, I am not sure that these both lenses will perform equally in practice .
The second point is the real sharpness of image. Theoretically, I can shoot landscape pictures at F3.5 with my standard kit lens and I should get sharp image due to large DOF. However, I am not sure that I can really take sharp landscape images at aperture F3.5 with my lens.
I would like to know from more experienced members how real DOF differs from theoretical. I presume that higher quality lenses perform better and that the real DOF might not be equal for different lenses although the aperture settings, focal length, etc. is equal. I also presume that real sharpness at relatively wide apertures (e.g., F3.5) might not be as good as at smaller aperture (e.g., F11) although the object is located beyond the hyperfocal distance.

I will highly appreciate opinions.

10-31-2011, 04:14 AM   #2
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QuoteOriginally posted by Alberts Quote
Hi!

I am a beginner in DSLR field. So may be my question is silly a little bit. I have read some texts (Wikipedia, etc.) about DOF and even carried out some DOF calculations in Excel on my own. However, I do not feel very strong and knowing in the concept of DOF.
As I understand, in theory, the length of DOF is determined only by four factors – focal length of lens, distance to object, aperture and circle of confusion. In turn, the circle of confusion is determined by the size of the camera’s sensor. For example, it is 0.02 mm for Pentax K-x (my camera) and for cameras with sensor that have the same crop factor. This theory is very nice, straightforward and even quite simple. Just input four figures in Excel and you get the result .................I presume that higher quality lenses perform better and that the real DOF might not be equal for different lenses although the aperture settings, focal length, etc. is equal. I also presume that real sharpness at relatively wide apertures (e.g., F3.5) might not be as good as at smaller aperture (e.g., F11) although the object is located beyond the hyperfocal distance.

I will highly appreciate opinions.
Your understanding is good.

I think some of the uncertainty regarding Circle of Confusion can be reduced by requiring that an infinitesimal point on the subject be rendered no larger than a pixel on the display. Mapping a typical computer display pixel to a APS-C camera's sensor results in a Circle of Confusion of about 0.02mm.

You say..." I also presume that real sharpness at relatively wide apertures (e.g., F3.5) might not be as good as at smaller aperture (e.g., F11) although the object is located beyond the hyperfocal distance."

That's correct in practice - further, there's an optimum f-stop range because at large f-stops the diffraction spot for even a perfect lens becomes larger than the circle of confusion and lenses are imperfect at small f-stops.

For an important simplification note that for small depths of field (away from hyperfocal conditions), focal length doesn't matter much - f-stop dominates.

Last edited by newarts; 10-31-2011 at 04:32 AM.
10-31-2011, 04:41 AM   #3
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Hi Alberts. Other people are better at camera maths, so I'll just add some general thoughts. Depth of Field is subjective. Just because a formula suggests that you should get a sharp image from 2.39m to infinity, it doesn't automatically follow that the level of sharpness will meet your expectations. I also believe that such formulas assume a perfect lens, whereas you are shooting at the limits of your lens (F3.5 / 18mm). Shooting at the limits is bound to show the optical compromises that the lens designers built into your lens.

From a personal perspective, if I was focusing at an object 5m away and I wanted to maintain acceptable sharpness from 2.39m to infinity, I wouldn't be shooting at f3.5 (Despite what a formula suggests). I'd be using at least F8. I'd even try F11 and F16.

By using a smaller aperture, the overall sharpness will improve (especially in the edges). Shooting at F3.5 at 18mm is shooting at your lens' maximum aperture. I don't have any experience with your lens, but most lenses aren't optimised for best results at their widest aperture, especially in the corners. This is easy to test yourself - try different apertures and see if there is a difference.

You're right - you should be able to achieve the same DOF as more expensive lenses at the same parameters. But, this goes back to the point above. An F1.4 lens is already stopped down at F3.5 and is likely to be in its optimal shooting zone. Compared with your lens, the point of focus is likely to sharper. The corners will also be sharper and possibly show less vignetting. But the DOF should be similar.

If you were shooting at F5.6, there would only be a negligible difference between your lens and higher quality one. At F8, most people wouldn't be able to positively identify a photo from your lens compared with a more expensive lens.
10-31-2011, 05:58 AM   #4
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there is something missig in the OP's initial summary, which is, aside from this omission rather good.

you need to consider not just the size of the sensor but the size of the print ior viewed image, and the viewing distance.

the reason for this is that the circle of confusion is based upon perceiving a point as a point, and not as a blurry out of focus circle.

therefore the larger you print (and also the closer in you crop), or the closer you view from does impact the perceived sharpness.

In fact, the circle of confusion is based upon an 8 x 10 inch print. There are very few of us who actually view images today on the equivelent 13 inch monitor, and don't zoom in!

As I am writing this in front of my 22 inch monitor (13 x 19 inch print) for example, clearly I am not one of them either.

The unfortunate reality is DOF calculations are totally useless unless you know before hand what your final magnification ratio will be.

So, the question back to the OP is what do you really want to achieve?

10-31-2011, 07:26 AM   #5
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Going to this page, changing the camera to your camera and lens and reviewing the data may help with the mystery.

For me the only thing I can recall every time is the general rule mentioned by newarts above, emphasis mine:

QuoteQuote:
For an important simplification note that for small depths of field (away from hyperfocal conditions), focal length doesn't matter much - f-stop dominates.
10-31-2011, 08:42 AM   #6
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My take: DOF, the range of distance that appears to be acceptably sharp, is a complex product of photography, presentation, and perception.
  • Perception: different people 'see' images in different ways depending on vision, attention, sanity, etc. Photographers can't control this very well.
  • Presentation: how an image is displayed greatly affects what seems to be acceptably sharp. Shown small enough, almost anything looks good.
  • Photography: You can control frame size, distance (both subject-to-lens and subject-to-background), aperture, and focal length. You can also affect DOF with lighting. A brightly-lit subject within a darker context will just seem sharper.
Any image is only really sharp at the focus distance. Other parts of the image field may seem acceptably sharp -- to somebody. Or maybe not. But everywhere but that focus distance is just approximately sharp. You get to figure out what's close enough for your purposes.

The evil twin of DOF is diffraction. You can stop-down the aperture to get thick DOF, but somewhere beyond the diffraction limit, the image gets fuzzy. Diffraction limit is a function of frame size. f/64 is great with an 8x10in frame but pretty bad with APS-C, where the limit is about f/9. Yes, you can stop-down to f/16 or beyond without seeing diffraction. The diffraction limit applies to perfectly-focused tripodded shots; handholding shake (even with SR) swamps-out any diffraction fuzziness. For absolute best pixel-peeping sharpness, use f/11 and a tripod.

Does lens quality (and cost) affect DOF? Low edge resolution and aberrations *could* make image margins look OOF (out-of-focus), sure, especially when the aperture is wide-open. We can exploit that if we keep the subject centered and sharp. But again, no matter what you think the DOF should be, it's only sharpest at the focus distance; objects much nearer or further will appear a bit OOF, and that can't be helped. So, be paranoid: If your calculator says that f/8 will give acceptable DOF, use f/11 instead, or f/16.

Last edited by RioRico; 11-01-2011 at 02:48 AM.
11-01-2011, 01:37 AM   #7
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Hi!

Thanks to everyone for comprehensive information. I tried to summarize my conclusions below.
First, the only real sharp image is at focus distance. In other distances there will be only acceptably sharp image not real sharp one.
Second, as I understand now, the circle of confusion that is used in standard formulae refers only to 8 x 10 inch print or 13 inch monitor. If, for example, I watch photos on my notebook’s 15.4” monitor or even on larger my workstation’s monitor, I should use much smaller circle of confusion to get more accurate DOF calculations. And even the result will be only theoretical. May be there are some formulae how to recalculate the circle of confusion if larger print is used instead of 8 x 10 inch print? It would be interesting to calculate DOF with decreased circle of confusion.
Third, if there are two lenses with different widest aperture, for example, lens Nr1 one with F3.5 (a standard kit lens) and lens Nr.2 with F2.8, then at aperture F3.5 lens Nr.2 will perform better and give sharper real DOF that lens Nr1. However, at apertures that smaller the difference will not be so visible. So at apertures F8, F11 and so both lenses will perform quite similar.
Last, at very small apertures the IQ worsens due to diffraction, especially if tripod is used. So very small apertures like F32 and so is not the best solution to improve the real DOF.

So, if I correctly understand, the preferable aperture (if large DOF is necessary) for most lenses is between F9 and F16. At smaller apertures the IQ becomes worse. And at larger apertures the image becomes less sharp in DOF zone. Therefore, for landscapes and similar photo F11...F16 might the best apertures, of course if lightning conditions allow them.

Now I am ready to improve my own DOF calculator (Excel based). I would like to link the circle of confusion to the print size. So the results will not be limited to 8 x 10 inch print. Unfortunately, I do not know the mathematical relationship between circle of confusion and print size. May be someone can advise me?
11-01-2011, 02:14 AM   #8
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Hi Alberts,

If you interested have a look at this site. It should reinforce some of what you know and give you a nice little DOF calculator. Oh, if you have an iPad or iPhone then there is a free DOF app, it is a bit simple, but handy.

11-01-2011, 03:19 AM   #9
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QuoteOriginally posted by Alberts Quote
Third, if there are two lenses with different widest aperture, for example, lens Nr1 one with F3.5 (a standard kit lens) and lens Nr.2 with F2.8, then at aperture F3.5 lens Nr.2 will perform better and give sharper real DOF that lens Nr1. However, at apertures that smaller the difference will not be so visible. So at apertures F8, F11 and so both lenses will perform quite similar.
You have everything about right except this. DOF is independent of a lens' other qualities, and just results from the aperture USED (and focal length, distances, etc). Some f/3.5 lenses wide-open will look sharper than some f/2 lenses stopped-down to f/3.5. That may be due to production quality, or to their designs, especially with field flatness.

A flatfield lens will seem sharper at the edges than a curved-field lens -- but the latter may seem to have more 'pop'. I'll compare my flatfield Novoflex Noflexar 105/3.5 enlarger lens on bellows with my curved-field Meyer Trioplan 100/2.8. If I use them both at f/3.5 and focus them for identical framing, their DOF's are the same. The Noflexar image will be sharper at the edges, the Trioplan image will have better microcontrast and 'pop'. Both have 15 iris blades but the Trioplan bokeh is a bit creamier, so its DOF seems smoother.

Also, although less-than-superb lenses have better IQ when stopped down, not everything equalizes by f/11. I can use my DA18-55 @50mm (6 iris blades), and my K50/1.2 (8 blades), and stop them both to f/11, and it's still pretty obvious which is which. I'll throw in my little CZJ Tessar 50/2.8 (12 blades) at f/11, and it's also obviously different. Identical DOF's; different contrast and bokeh and rendering; different images.
11-01-2011, 04:32 AM   #10
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Alberts

For circle of confusion to print size the conversion is linear. Circle of confusion is at the film or sensor. Forn35 mm film the film is 24 x 36 mm and for pentax digital the sensor is 16 x 24 mm. If you consider how the C of C scales to an 8 x 10 print then consider it scaling up with the linear increase in print size or the crop ratio if you crop in on an image before printing
11-01-2011, 06:01 AM   #11
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QuoteQuote:
May be there are some formulae how to recalculate the circle of confusion if larger print is used instead of 8 x 10 inch print? It would be interesting to calculate DOF with decreased circle of confusion.
It is helpful to think about circle of confusion in the context of a digital display. A meaningful circle of confusion for a digital display cannot be smaller than the display's pixel pitch (it might be larger if the viewer stands back a bit so his/her visual resolution is less than that of the display). Therefore it makes sense to use the display pixel pitch as an unambiguous basis for CoC.

The display pixel pitch is then linearly mapped to the sensor dimension to get the Circle of Confusion;

CoC = display.pixel.pitch*(sensor.width/display.width)

The same CoC can be used to estimate the f-number at the diffraction limit; ie the f-number at which the Airy disk diameter is equal to the display pixel pitch.
11-01-2011, 06:36 AM   #12
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QuoteOriginally posted by newarts Quote
It is helpful to think about circle of confusion in the context of a digital display. A meaningful circle of confusion for a digital display cannot be smaller than the display's pixel pitch (it might be larger if the viewer stands back a bit so his/her visual resolution is less than that of the display). Therefore it makes sense to use the display pixel pitch as an unambiguous basis for CoC.

The display pixel pitch is then linearly mapped to the sensor dimension to get the Circle of Confusion;

CoC = display.pixel.pitch*(sensor.width/display.width)

The same CoC can be used to estimate the f-number at the diffraction limit; ie the f-number at which the Airy disk diameter is equal to the display pixel pitch.
Keep in mind that the CoC is more relevant to the sensor when disucssing pixel pitch. For example an *istD with the 3000 x 2000 pixel format. the 0.020mm CofC is roughly 3 pixels in diameter. for newer cameras like my K5 this is something like 5-6 pixels, but in considering your point, a 0.02mm CoC translates roughly to a .2mm CoC on an 8 x 10 print. .2mm is .008 inch and on a 13 x 19 print (the size of my monitor is about .016 inch, this is about 3 pixels on my monitor when viewing an image at full screen magnification.

the result is that in reality (at least for anyone using more than a 13 inch monitor, a full screen view has more than one pixel in the CofC
11-01-2011, 08:17 AM   #13
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Perhaps the easiest way to handle CoC is as a fraction of the display width; the Zeiss formula uses 1/1730 of the display diagonal - other fractions have been used as well, 1/1250, 1/1000, etc. Defining CoC in this manner implicitly assumes the viewer looks at the display from a distance that works with no pixelation for his/her vision.

I like to use 1/1000 because the CoC in DoF equation is then simply:

C = sensor.width/1000 which corresponds to a display pixel pitch of about 0.25mm for aps-c viewed at 8x10".
11-02-2011, 01:47 AM   #14
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Hi!

I did some investigation about CoC in Wikipedia Circle of confusion - Wikipedia, the free encyclopedia It seams that the most correct formula (if neglect diffraction effects) is the following:

CoC (mm) = viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25

This formula is based on assumption that the closest comfortable viewing distance is 25 cm. I tried to carry out some calculations for my Pentax K-x sensor (as I understand the APS-C sensor is used in many other Pentax cameras, too). I used the following assumptions:
1) the viewing distance – 25 cm;
2) the desired final-image resolution – 5 lp/mm (according to Wikipedia).

The most complicated factor was the enlargement. For the final image I chose the 8x10 inch format that was declared as etalon in the previous posts. I was not sure how to calculate the enlargement correctly. So I decided to calculate it in two ways – according to the diagonal and according to the largest dimension. I presumed that the format of the Pentax APS-C sensor was 15.7 mm × 23.6 mm. So the diagonal was 28.3 mm. The diagonal for the 8x10 inch format was 325.3 mm. According to the diagonals, the enlargement was 11.5. According to the largest dimensions (23.6 m and 10 inch or 254 mm), the enlargement was about 10.8. So the CoC was either

CoC = 25 / 5 / 11.5 / 25 = 0.017 mm

or

CoC = 25 / 5 / 10.8 / 25 = 0.019 mm

The results are quite close or even equal to the values recommended by DOF calculators. DOF calculators usually use value 0.02mm or 0.019mm. Therefore I can conclude that standard CoC and DOF calculations are really based on 8x10 inch print. If we want to view picture on larger formats, for example, on 15.4” monitors or even 21” monitors, we should use smaller CoC as the enlargement will be higher. However, as I understand, the resolution of the monitor might be lower than 5 lp/mm. In turn, this means that CoC will be higher. So the CoC 0.02mm might be quite acceptable approximate value for ordinary photo even if it viewed on computer screen.

I am looking forward to waiting comments on my calculations and further comments and views about DOF and CoC.
11-02-2011, 03:12 AM   #15
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QuoteOriginally posted by Alberts Quote
Hi!

I did some investigation about CoC in Wikipedia Circle of confusion - Wikipedia, the free encyclopedia It seams that the most correct formula (if neglect diffraction effects) is the following:

CoC (mm) = viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25......

I am looking forward to waiting comments on my calculations and further comments and views about DOF and CoC.
Your calculations are correct. However they imply there is a precise value for CoC while there is no real precision inherent in a concept based on "acceptable sharpness". This is especially true when specifications include certain viewing distances, etc because such conditions actually vary in practice - eg a viewer's eyes are seldom exactly 25cm from the display.

One way to enhance precision in the CoC concept is to define it digitally, in terms of display pixel size. A display CoC smaller than a display pixel is meaningless because any image feature smaller than a pixel is displayed as a uniform brightness over the pixel. As a viewer moves his/her head closer to a digital display there comes a point when his/her visual resolution coincides with display's pixel dimension; at that point CoC is a minimum equal to the pixel size. Hence the DoF equation for a digital display can be written as an inequality -

DoF > 2(display.pixel.size)(sensor.width/display.width)(f-stop)[other terms]....

ie. CoC > (display.pixel.size)(sensor.width/display.width)
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