luigi bertolotti Posted May 5, 2008 Share #1 Posted May 5, 2008 Advertisement (gone after registration) Yesterday night I wanted to take & post a pic of an old lens of mine (it's in the historical section... Summaron 35...) ; I was uncertain between using my favorite macro set (Tele Elmar 135 head on Visoflex) , or the quickier Summicron Dual Range: I decide for this because the subject was inclined, 3-4 cm of effective "visible depth" and I wanted a decent depth of field... but, afterthat, I was caught by a tech doubt... is it really so ? I never went in depth with the math of DOF... question is: if one achieves the SAME field of view (say, a 36 x 54 rectangle, ratio 1:2 on M8) with a 50 and a 135 lens, of course with diaphragm closed the same, is the effective depth of focus different ? I'd say yes,but, not knowing the math at the base, am not at all sure... Link to post Share on other sites More sharing options...
Advertisement Posted May 5, 2008 Posted May 5, 2008 Hi luigi bertolotti, Take a look here Tech question about DOF - macro environment. I'm sure you'll find what you were looking for!
giordano Posted May 5, 2008 Share #2 Posted May 5, 2008 The (Ilford) Manual of Photography gives this formula for depth of field at near distances: Total Depth = 2*c*N*(m+1)/m^2 where c is circle of confusion, N is the numerical f/stop and m is the magnification ratio. Focal length isn't a factor. Link to post Share on other sites More sharing options...
ghammer Posted May 5, 2008 Share #3 Posted May 5, 2008 The (Ilford) Manual of Photography gives this formula for depth of field at near distances: Total Depth = 2*c*N*(m+1)/m^2 where c is circle of confusion, N is the numerical f/stop and m is the magnification ratio. Focal length isn't a factor. This is an approximation. The exact expression is: 2*c*N*(m+1)/(m^2-(c*N/f)^2) with f the focal length. So there is really a difference for all ratios m (Luigi is right), but the difference is very small for c approaching 1. For Luigis case: c=0.33, m=0.5, and N=16 it is 6.403mm versus 6.400mm. Link to post Share on other sites More sharing options...
jaapv Posted May 5, 2008 Share #4 Posted May 5, 2008 As DOF is a function of magnification there is a difference in DOF between an image of the same size taken by different lenses. The magnification of fore- and background is different, even if the subject is the same size on the sensor. However, in macro work, DOF is so narrow that it does not really make a lot of difference. Link to post Share on other sites More sharing options...
luigi bertolotti Posted May 5, 2008 Author Share #5 Posted May 5, 2008 Thanks for the quick answers, straight & clear... but... you say c is the circle of confusion (I remember to have often seen the term "CoC" for it)... is it correct an estimate value of 0,33 mm for M8 ? I know it's a questionable issue, but seem to remember to have seen, somewhere, a magnitude-different value, like 0,023 or 0,027... maybe I'm missing something... I checked some Net resource... there are lot of... but the values are typically in the range 0,02x mm Link to post Share on other sites More sharing options...
luigi bertolotti Posted May 5, 2008 Author Share #6 Posted May 5, 2008 As DOF is a function of magnification there is a difference in DOF between an image of the same size taken by different lenses. The magnification of fore- and background is different, even if the subject is the same size on the sensor. However, in macro work, DOF is so narrow that it does not really make a lot of difference. That is the problem... the above posted formulas include "magnification"... has it to be considered as the "magnification ratio" I achieve (the 1:2 of the example) or the magnification of the lens in itself, clearly related to focal length ? Link to post Share on other sites More sharing options...
giordano Posted May 5, 2008 Share #7 Posted May 5, 2008 Advertisement (gone after registration) That is the problem... the above posted formulas include "magnification"... has it to be considered as the "magnification ratio" I achieve (the 1:2 of the example) or the magnification of the lens in itself, clearly related to focal length ? m is the ratio between object size and image size: in your example, 0.5. As ghammer says, the formula I posted is an approximation. It's good enough for everyday "macro" work. Link to post Share on other sites More sharing options...
ghammer Posted May 5, 2008 Share #8 Posted May 5, 2008 Thanks for the quick answers, straight & clear... but... you say c is the circle of confusion (I remember to have often seen the term "CoC" for it)... is it correct an estimate value of 0,33 mm for M8 ? I know it's a questionable issue, but seem to remember to have seen, somewhere, a magnitude-different value, like 0,023 or 0,027... maybe I'm missing something... I checked some Net resource... there are lot of... but the values are typically in the range 0,02x mm My typing error: The »classical« circle of diffusion is 1/30 mm = 0.033. This is the value the markings on the lenses are based on. Normally it is better to use a smaller value, especially for digital. Link to post Share on other sites More sharing options...
jaapv Posted May 5, 2008 Share #9 Posted May 5, 2008 Formally it should be 0.023. However, given the more precise nature of a sensor, resulting in a more pronounced DOF gradient, a circle of 0.02 would be more appropriate. Link to post Share on other sites More sharing options...
giordano Posted May 5, 2008 Share #10 Posted May 5, 2008 This is an approximation. The exact expression is: 2*c*N*(m+1)/(m^2-(c*N/f)^2) with f the focal length. So there is really a difference for all ratios m (Luigi is right), but the difference is very small for c approaching 1. For Luigis case: c=0.33, m=0.5, and N=16 it is 6.403mm versus 6.400mm. I hope you meant c=1/30mm (approx 0.03333). The simpler formula I posted gives 6.4mm DoF for either focal length, so the difference between the two formulas is negligible here. But a 1/30 mm CoC for 35mm equipment dates from the 1920s and 30s . Modern lenses, films and sensors are capable of much better performance and it's often appropriate to work from a much smaller CoC - especially with a "cropped" sensor like the M8's. At 1/50mm CoC and 1:2 magnification and f/16, total DoF is less than 4mm. I'll leave it to someone else to calculate the effect of the lens extension on f/stop on DoF! Link to post Share on other sites More sharing options...
luigi bertolotti Posted May 5, 2008 Author Share #11 Posted May 5, 2008 ..... I'll leave it to someone else to calculate the effect of the lens extension on f/stop on DoF! Eh eh, I'm very happy to have caused a deep technical discussion from a innocent question... And this last statement is intriguing... I know that with significan "bellows extension" (or Viso + rings and so...) one has to take into account an "EV factor" (figures exist on bellows specs...) : this means that the "real f/stop" is closer than the nominal... but DOES this affect DOF ? My guess is "no"... but, as you say, stand to be corrected by someone else... Reasoning "crudely" on the lens wide open : the "EV factor" becomes accountable at significant extensions for not all the light that passes through the lens systems reaches the film/sensor... the effective angle decreases... a certain amount of lightrays are "lost"... so, one can guess that for some "small" f stops closing the real exposure doesn't change... imagine a f 2,5 lens, very extended by bellows... if i close to 2,8 only, maybe the lightrays blocked by the diaphragm are anyway "unuseful" due to the extension... oh well, let's wait for some real math from someone... Link to post Share on other sites More sharing options...
luigi bertolotti Posted May 5, 2008 Author Share #12 Posted May 5, 2008 Sorry,,, I'm thinking my "crude" thinkings at the end of my previous post can be totally nonsense... Link to post Share on other sites More sharing options...
jaapv Posted May 5, 2008 Share #13 Posted May 5, 2008 Yes, lens extension will influence DOF. The light loss you mention is caused by the lengthening of the focal length of the lens. The aperture is a mathematical function of the effective diameter of the lens and the focal length, so you are reducing the "real" aperture, effectively influencing DOF. The relation between the influence of the focal length/magnification shift by the extension and the shrinking of the aperture I leave to somebody else... Link to post Share on other sites More sharing options...
jank Posted May 5, 2008 Share #14 Posted May 5, 2008 This si all true when one considers an "ideal lens". The coc may also get defined -enlarged by the resolution of the lens , not byt the resolution of the sensor/ film only. That would be also true for the size of the apreture, the lens when stopped down also looses its performance, therefore apparent depth of field increases. The older , poorer lens might exhibit, or rather make impression having larger depth of field. Jan Link to post Share on other sites More sharing options...
luigi bertolotti Posted May 5, 2008 Author Share #15 Posted May 5, 2008 Yes, lens extension will influence DOF. The light loss you mention is caused by the lengthening of the focal length of the lens. The aperture is a mathematical function of the effective diameter of the lens and the focal length, so you are reducing the "real" aperture, effectively influencing DOF. The relation between the influence of the focal length/magnification shift by the extension and the shrinking of the aperture I leave to somebody else... The secret key can be into ghammer's formula : 2*c*N*(m+1)/(m^2-(c*N/f)^2) N/f = 1/D where D is the nominal Diameter of the lens (front lens in std. lens designs) How one has to treat N and f into an "extension environment" ? Does f must be set at the "lenghtened" value due to extension ? Does N has to be still treated as the nominal Fstop set by user ? Or... do the two values result into an INVARIANT ? ("real" f increases with extension, so as "real" N....the ratio...) Link to post Share on other sites More sharing options...
jaapv Posted May 5, 2008 Share #16 Posted May 5, 2008 I will be happy to be corrected if I am wrong,it is becoming a bit abtruse, but numerical f-stop I think means calculated f-stop, as opposed to the one indicated by the f-scale. So it should be actual lens diameter divided by real (=extended) focal length. The f=extended focal length in that formula afaik. Link to post Share on other sites More sharing options...
giordano Posted May 5, 2008 Share #17 Posted May 5, 2008 I will be happy to be corrected if I am wrong,it is becoming a bit abtruse, but numerical f-stop I think means calculated f-stop, as opposed to the one indicated by the f-scale. So it should be actual lens diameter divided by real (=extended) focal length. The f=extended focal length in that formula afaik. FWIW I think so too. At a magnification of 1:1 the "focal length" is doubled and the numerical aperture halved (requiring two stops exposure compensation if you're not using a TTL meter) so at magnification of 1:2 there would be a difference of just over 1 stop: set the aperture ring to f/11 and get an effective aperture of about f/18. Link to post Share on other sites More sharing options...
ghammer Posted May 5, 2008 Share #18 Posted May 5, 2008 FWIW I think so too. At a magnification of 1:1 the "focal length" is doubled and the numerical aperture halved (requiring two stops exposure compensation if you're not using a TTL meter) so at magnification of 1:2 there would be a difference of just over 1 stop: set the aperture ring to f/11 and get an effective aperture of about f/18. In the formula I gave above f is the focal length, not the extension, and N is the numerical f-stop. So the formula takes care of the problems related to extension. It depends only on geometric reasoning. There are in fact minor deviations depending on the type of lens. So two lenses of 50mm focal length (ceteris paribus) can have slightly different DoF, but without interest for practical purposes. Link to post Share on other sites More sharing options...
jaapv Posted May 5, 2008 Share #19 Posted May 5, 2008 The problem is, the focal length of a lens is fixed at the given value only at infinity and lengthens as it gets focussed closer, creating a smaller aperture of course. So in your formula f is the focal length at infinity and N the aperture at infinity? Link to post Share on other sites More sharing options...
luigi bertolotti Posted May 5, 2008 Author Share #20 Posted May 5, 2008 In the formula I gave above f is the focal length, not the extension, and N is the numerical f-stop. So the formula takes care of the problems related to extension. It depends only on geometric reasoning. There are in fact minor deviations depending on the type of lens. So two lenses of 50mm focal length (ceteris paribus) can have slightly different DoF, but without interest for practical purposes. Ok, question is if f and N have to be taken as the nominal values of the lens (say, a 50 mm lens with the diaph closed to say 16) or the "real" values accordingly to the extension ... sorry... Jaap made the same answer while I was typing... <added> looking at the formula ... N/f is indipendent from extension... there is also the factor N alone... but the "real" N of course is a function of nominal N and m=magnification... so, if math is math, the formula can be written into a form in which N and f are anyway expressed as nominal values. Is in this sense that you say "the formula takes care of the problems related to extension" ? Link to post Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.