No, it's based on actual photography... and doing it for many years. If you use a 28mm f/2 and a 50mm f/2 at the same aperture (obviously towards the wide open end of the spectrum), you'd have to be blind to not notice the difference. The Q has a 28mm lens and cropping it does not change the fact that it is a 28mm lens with 28mm lens characteristics.
ramosa
B&W
I don't think this question would require empirical testing. jsrockit is spot on in that, a 50-ish crop of a 28mm lens would retain its innate 28mm DoF properties.
Daneinbalto
Established
Empirical data
Empirical data
This was a quick and dirty experiment with my digital camera zoom lens. One shot at focal length 35 mm (56 mm equivalent). The other at 17 mm (28 mm equivalent), which was then cropped and enlarged. Focus is on object on the grass. Due to the slowness of the lens at 35 mm focal length the best I could do was f/5. Which is which? I realize the limitation of doing this experiment at f/5. Would anybody care to try at f/2 or the like? My interchangeable lenses are for film.
Empirical data
This was a quick and dirty experiment with my digital camera zoom lens. One shot at focal length 35 mm (56 mm equivalent). The other at 17 mm (28 mm equivalent), which was then cropped and enlarged. Focus is on object on the grass. Due to the slowness of the lens at 35 mm focal length the best I could do was f/5. Which is which? I realize the limitation of doing this experiment at f/5. Would anybody care to try at f/2 or the like? My interchangeable lenses are for film.
Attachments
mrmeadows
Established
Here we go yet again ... another confused thread regarding depth of field (DOF). What's a person to think about these apparently contradictory comments?
The OP has asked: "...what happens to DOF if you apply a 50 mm crop to a shot captured with a 28 mm lens? If you compare two shots taken of the same subject, at the same distance, with the same f/stop, one with a native 50 mm lens and the other with a 28 mm lens but cropped to correspond to the 50 mm shot, will the DOF be different?"
As with most DOF questions, the OP's question is not well-defined, not complete. Thus the question can have different correct answers whenever different responders complete the question with different unstated assumptions in getting their answer. But not all proposed answers are correct, however, and figuring out which is correct is another problem.
DOF is simple and well-defined and always yields a unique answer to any well-defined DOF question. The following is my attempt to explain it with very transparent concepts and very minimal mathematics. I've been working on this explanation for some time and now seems like a good time to try it out on you guys. Please let me know what you think of it.
Anyone who understands my explanation will be able to determine when DOF questions are well-defined and will be able to determine/calculate the unique answer to the well-defined ones. After presenting my explanation, I will use it to treat the OP's question and will present four examples that show how the answer can change dramatically as unstated assumptions change.
--- Mike
What IS depth of field? DOF exists only as a perception by a human viewer of a photographic image (negative or slide or print or electronic display). Consider a film negative, for simplicity. The objects recorded in the negative will be perfectly in-focus for only the single plane at the focus distance from the camera. Objects at all other distances from the camera will be out of focus, more so as their distance differs from the plane of focus. The image on the negative of every out-of-focus point will be blurred, becoming a blur circle instead of a point image, and will have a diameter that grows as the point is further from the plane of focus. Large out-of-focus objects will consist of a myriad of out-of-focus points which overlap and interfere with one another to make such large objects blurry. This much is a physical consequence of the way that photographic lenses work and films record lens images.
Human vision is limited in resolution; there's a limit to the size of things that one can see, blur circles included. So if a negative is viewed by a human from the closest distance at which he can focus his eyes, then the human will be able to resolve the large blur circles of some out-of-focus objects, and these objects will be perceived as out-of-focus. All the objects whose blur circles the human cannot resolve will be perceived as in-focus and the distances from the camera of these in-focus objects are within the DOF. The DOF thus is defined as the difference in distance from the camera between the nearest and the farthest objects which are perceived to be in-focus. If the human moves back from the negative, he will not be able to resolve some of the smaller blur circles that he could previously see, and the DOF will increase.
Now enlarge the negative to make a print. The blur circles will also be enlarged. This makes them easier for a human to resolve from the same close viewing distance, so the DOF will decrease. If the viewer moves back from the print, he will again not resolve some of the smaller blur circles and the DOF will increase.
So the DOF value depends on which blur circles in a photographic image a human viewer can perceive. Thus DOF is not a fixed property of a negative (or digital image). DOF is somehow related to the camera, a lens and an image (film or digital) they have produced. But DOF is also related to the human viewer and the viewing conditions. How to sort all this out?
We can identify exactly six independent factors which TOGETHER determine the DOF in a given situation. The six independent factors are the following, and we also list the INDIVIDUAL effect of changes to each factor on DOF when all the others are kept fixed.
1. focal length of the lens; longer focal length yields smaller DOF
2. f-stop of the lens; larger f-stop (smaller f-number) yields smaller DOF
3. focus distance setting of the lens; nearer focus distance yields smaller DOF
4. magnification of the viewed image; higher magnification yields smaller DOF
5. viewing distance to the viewed image; closer viewing distance yields smaller DOF
6. visual acuity of the viewer's eyes; people with 20/20 vision will experience a smaller DOF than all people with poorer vision and greater DOF than all people with better vision
The DOF involves all six of these factors. Change any one of them alone and the DOF changes. When you change more than one factor, the DOF can either increase or decrease depending on the degree to which the individual changes complement or oppose one another. Multiple changes can be complicated to treat. In many instances one cannot simply reason with just English (or any other language) to get a verifiably correct answer. Instead, one MUST use a mathematical treatment, a key result of which is given by this equation.
DOF = (2 * F^2 * f * D^2 * M * d *a) / (F^4 * M^2 - (f * D * d * a)^2)
where the value of each of the six factors is represented by a letter
F is the focal length of the lens
f is the f-number (1.4 or 8, for example)
D is the focus distance to the plane in focus (in the same units as F)
M is the magnification of the viewed image (M=1 for just a negative)
d is the viewing distance
a is human visual acuity (commonly about 0.00058, corresponding to 1/30°)
and * means multiply while x^2 = x * x, and x^4 = x * x * x * x
and (f * D * d * a)^2 means to multiply the four together and then square that
This equation is derived using simple geometrical optics. You can solve most common DOF questions by putting the appropriate values into this equation and calculating the result. But you must know the numerical values of all six factors to get any result at all; the question at issue must be well-defined.
Comments about the equation:
1. The equation applies to general photography for which the focus distance is much greater than the lens focal length (D>>F); the correct equation for macro work is slightly but importantly different and a bit more complicated.
2. The equation doesn't handle D = infinity very transparently; besides, if D=infinity then DOF=infinity, too, and no equation is needed.
3. If the calculated result of the equation is negative, that means that the DOF stretches to infinity.
4. Other DOF factors have been defined and used in other DOF equations, but the equation shown can accomplish everything that those others can and with fewer misunderstandings. For example DOF equations are often written in terms of something called the hyperfocal distance H, but the hyperfocal distance is just a special combination of our factors: H = F^2*M/(f*d*a). Similarly, the oft-defined circle of confusion c is just c = d*a/M.
Now apply this equation to the OP's question. We're going to calculate two values of DOF, one for the 50mm lens and one for the 28mm lens and then compare them. The OP told us that the f-number and focus distance are to be the same for both lenses but we weren't given values. Let's take f=4 and D = 3meter.
We already know that we need more information, and that the results of the comparison will depend on it. Let's assume a full frame film/sensor and magnify both of the negatives eight times (M=8) to make two prints that are 0.192m x 0.288m and we agree to view both from a distance equal to their common diagonal (d=0.346m). Further, we'll use the "standard" visual acuity (a=0.00058). Then our equations become
DOF(50) = (2 * 0.050^2 * 4 * 3^2 * 8 * 0.346 * 0.00058) / (0.050^4 * 8^2 - (4 * 3 * 0.00058 * 0.346)^2)
DOF(28) = (2 * 0.028^2 * 4 * 3^2 * 8 * 0.346 * 0.00058) / 0.028^4 * 8^2 - (4 * 3 * 0.346 * 0.00058)^2
and we can calculate the results which are
DOF(50) = 0.733 meter
DOF(28) = 2.08 meter
Comparing the two results, we find that the DOF of the 28mm lens is much greater than that of the 50mm lens. That is expected, because the only difference is focal length, and we already know generally how that affects DOF. Cropping in this example is just a matter of either masking the projected 28mm negative in the easel or trimming the print to show just the part of the 28mm negative that is included as the entire 50mm negative. After cropping, the 28mm print will be about 146mm x 219mm, which is smaller than the 50mm print. It's smaller, because the portion of a 28mm negative that matches an entire 50mm negative is only about 76% of each side of the 28mm negative (and 0.76^2 = 58% of the 28mm negative's area). But we still view it from 0.346m.
Next consider making the just cropped portion of the 28mm print the same size as the 50mm print when the 50mm is magnified 8 times. This will require a magnification for the 28mm of 8/0.76 = 10.5; all other factors retain the same values. In this case, the equation give
DOF(50) = 0.733 meter
DOF(28) = 1.919 meter
The 28mm print still has much greater DOF than the 50mm print.
Next suppose we magnify the 28mm negative by 20.8 while keeping all the other factors the same as before. Then the results are
DOF(50) = 0.733 meter
DOF(28) = 0.733 meter
Now the 50mm DOF is the same as the 28mm DOF. Again, cropping amounts to masking or trimming the 28mm image to match the extent of the 50mm image, and the print size will be about 0.380m x 0.570m, almost twice as large as the 50mm print but showing the same scene. Again we view both from 0.346m.
If we magnify the 28mm negative by 25 while keeping all else constant, the results will be
DOF(50) = 0.733 meter
DOF(28) = 0.608 meter
and the 50mm DOF is greater than the DOF of the 28mm.
To make the 28mm print have the same DOF as the 50mm, one must magnify the 28mm negative 2.6 times more than the 50mm negative. The mathematical reason why one must magnify the 28mm negative so much more is that the DOF depends on F^2 * M and (F^2 * M)^2. Thus if the focal length changes by a factor of 50/28 = 1.79 as in the OP's question, then the magnification must change by a factor of 1.79^2=3.19 to keep the product of these two factors constant. But since the denominator of the equation includes a second, negative term not involving F and M, this calculated 3.19 factor actually only needs to be 2.6 to accomplish equal DOFs.
So, in summary, the OP's question was ambiguous and the DOF of the 28mm can be greater than, the same as, or smaller than the DOF of the 50mm, depending on the values of the factors which the OP didn't specify but which are necessary for a well-posed question that this or any other DOF equation can answer. One can calculate comparisons for many other interesting situations involving these two lenses or for any other well-defined situation using the equation given.
The OP has asked: "...what happens to DOF if you apply a 50 mm crop to a shot captured with a 28 mm lens? If you compare two shots taken of the same subject, at the same distance, with the same f/stop, one with a native 50 mm lens and the other with a 28 mm lens but cropped to correspond to the 50 mm shot, will the DOF be different?"
As with most DOF questions, the OP's question is not well-defined, not complete. Thus the question can have different correct answers whenever different responders complete the question with different unstated assumptions in getting their answer. But not all proposed answers are correct, however, and figuring out which is correct is another problem.
DOF is simple and well-defined and always yields a unique answer to any well-defined DOF question. The following is my attempt to explain it with very transparent concepts and very minimal mathematics. I've been working on this explanation for some time and now seems like a good time to try it out on you guys. Please let me know what you think of it.
Anyone who understands my explanation will be able to determine when DOF questions are well-defined and will be able to determine/calculate the unique answer to the well-defined ones. After presenting my explanation, I will use it to treat the OP's question and will present four examples that show how the answer can change dramatically as unstated assumptions change.
--- Mike
What IS depth of field? DOF exists only as a perception by a human viewer of a photographic image (negative or slide or print or electronic display). Consider a film negative, for simplicity. The objects recorded in the negative will be perfectly in-focus for only the single plane at the focus distance from the camera. Objects at all other distances from the camera will be out of focus, more so as their distance differs from the plane of focus. The image on the negative of every out-of-focus point will be blurred, becoming a blur circle instead of a point image, and will have a diameter that grows as the point is further from the plane of focus. Large out-of-focus objects will consist of a myriad of out-of-focus points which overlap and interfere with one another to make such large objects blurry. This much is a physical consequence of the way that photographic lenses work and films record lens images.
Human vision is limited in resolution; there's a limit to the size of things that one can see, blur circles included. So if a negative is viewed by a human from the closest distance at which he can focus his eyes, then the human will be able to resolve the large blur circles of some out-of-focus objects, and these objects will be perceived as out-of-focus. All the objects whose blur circles the human cannot resolve will be perceived as in-focus and the distances from the camera of these in-focus objects are within the DOF. The DOF thus is defined as the difference in distance from the camera between the nearest and the farthest objects which are perceived to be in-focus. If the human moves back from the negative, he will not be able to resolve some of the smaller blur circles that he could previously see, and the DOF will increase.
Now enlarge the negative to make a print. The blur circles will also be enlarged. This makes them easier for a human to resolve from the same close viewing distance, so the DOF will decrease. If the viewer moves back from the print, he will again not resolve some of the smaller blur circles and the DOF will increase.
So the DOF value depends on which blur circles in a photographic image a human viewer can perceive. Thus DOF is not a fixed property of a negative (or digital image). DOF is somehow related to the camera, a lens and an image (film or digital) they have produced. But DOF is also related to the human viewer and the viewing conditions. How to sort all this out?
We can identify exactly six independent factors which TOGETHER determine the DOF in a given situation. The six independent factors are the following, and we also list the INDIVIDUAL effect of changes to each factor on DOF when all the others are kept fixed.
1. focal length of the lens; longer focal length yields smaller DOF
2. f-stop of the lens; larger f-stop (smaller f-number) yields smaller DOF
3. focus distance setting of the lens; nearer focus distance yields smaller DOF
4. magnification of the viewed image; higher magnification yields smaller DOF
5. viewing distance to the viewed image; closer viewing distance yields smaller DOF
6. visual acuity of the viewer's eyes; people with 20/20 vision will experience a smaller DOF than all people with poorer vision and greater DOF than all people with better vision
The DOF involves all six of these factors. Change any one of them alone and the DOF changes. When you change more than one factor, the DOF can either increase or decrease depending on the degree to which the individual changes complement or oppose one another. Multiple changes can be complicated to treat. In many instances one cannot simply reason with just English (or any other language) to get a verifiably correct answer. Instead, one MUST use a mathematical treatment, a key result of which is given by this equation.
DOF = (2 * F^2 * f * D^2 * M * d *a) / (F^4 * M^2 - (f * D * d * a)^2)
where the value of each of the six factors is represented by a letter
F is the focal length of the lens
f is the f-number (1.4 or 8, for example)
D is the focus distance to the plane in focus (in the same units as F)
M is the magnification of the viewed image (M=1 for just a negative)
d is the viewing distance
a is human visual acuity (commonly about 0.00058, corresponding to 1/30°)
and * means multiply while x^2 = x * x, and x^4 = x * x * x * x
and (f * D * d * a)^2 means to multiply the four together and then square that
This equation is derived using simple geometrical optics. You can solve most common DOF questions by putting the appropriate values into this equation and calculating the result. But you must know the numerical values of all six factors to get any result at all; the question at issue must be well-defined.
Comments about the equation:
1. The equation applies to general photography for which the focus distance is much greater than the lens focal length (D>>F); the correct equation for macro work is slightly but importantly different and a bit more complicated.
2. The equation doesn't handle D = infinity very transparently; besides, if D=infinity then DOF=infinity, too, and no equation is needed.
3. If the calculated result of the equation is negative, that means that the DOF stretches to infinity.
4. Other DOF factors have been defined and used in other DOF equations, but the equation shown can accomplish everything that those others can and with fewer misunderstandings. For example DOF equations are often written in terms of something called the hyperfocal distance H, but the hyperfocal distance is just a special combination of our factors: H = F^2*M/(f*d*a). Similarly, the oft-defined circle of confusion c is just c = d*a/M.
Now apply this equation to the OP's question. We're going to calculate two values of DOF, one for the 50mm lens and one for the 28mm lens and then compare them. The OP told us that the f-number and focus distance are to be the same for both lenses but we weren't given values. Let's take f=4 and D = 3meter.
We already know that we need more information, and that the results of the comparison will depend on it. Let's assume a full frame film/sensor and magnify both of the negatives eight times (M=8) to make two prints that are 0.192m x 0.288m and we agree to view both from a distance equal to their common diagonal (d=0.346m). Further, we'll use the "standard" visual acuity (a=0.00058). Then our equations become
DOF(50) = (2 * 0.050^2 * 4 * 3^2 * 8 * 0.346 * 0.00058) / (0.050^4 * 8^2 - (4 * 3 * 0.00058 * 0.346)^2)
DOF(28) = (2 * 0.028^2 * 4 * 3^2 * 8 * 0.346 * 0.00058) / 0.028^4 * 8^2 - (4 * 3 * 0.346 * 0.00058)^2
and we can calculate the results which are
DOF(50) = 0.733 meter
DOF(28) = 2.08 meter
Comparing the two results, we find that the DOF of the 28mm lens is much greater than that of the 50mm lens. That is expected, because the only difference is focal length, and we already know generally how that affects DOF. Cropping in this example is just a matter of either masking the projected 28mm negative in the easel or trimming the print to show just the part of the 28mm negative that is included as the entire 50mm negative. After cropping, the 28mm print will be about 146mm x 219mm, which is smaller than the 50mm print. It's smaller, because the portion of a 28mm negative that matches an entire 50mm negative is only about 76% of each side of the 28mm negative (and 0.76^2 = 58% of the 28mm negative's area). But we still view it from 0.346m.
Next consider making the just cropped portion of the 28mm print the same size as the 50mm print when the 50mm is magnified 8 times. This will require a magnification for the 28mm of 8/0.76 = 10.5; all other factors retain the same values. In this case, the equation give
DOF(50) = 0.733 meter
DOF(28) = 1.919 meter
The 28mm print still has much greater DOF than the 50mm print.
Next suppose we magnify the 28mm negative by 20.8 while keeping all the other factors the same as before. Then the results are
DOF(50) = 0.733 meter
DOF(28) = 0.733 meter
Now the 50mm DOF is the same as the 28mm DOF. Again, cropping amounts to masking or trimming the 28mm image to match the extent of the 50mm image, and the print size will be about 0.380m x 0.570m, almost twice as large as the 50mm print but showing the same scene. Again we view both from 0.346m.
If we magnify the 28mm negative by 25 while keeping all else constant, the results will be
DOF(50) = 0.733 meter
DOF(28) = 0.608 meter
and the 50mm DOF is greater than the DOF of the 28mm.
To make the 28mm print have the same DOF as the 50mm, one must magnify the 28mm negative 2.6 times more than the 50mm negative. The mathematical reason why one must magnify the 28mm negative so much more is that the DOF depends on F^2 * M and (F^2 * M)^2. Thus if the focal length changes by a factor of 50/28 = 1.79 as in the OP's question, then the magnification must change by a factor of 1.79^2=3.19 to keep the product of these two factors constant. But since the denominator of the equation includes a second, negative term not involving F and M, this calculated 3.19 factor actually only needs to be 2.6 to accomplish equal DOFs.
So, in summary, the OP's question was ambiguous and the DOF of the 28mm can be greater than, the same as, or smaller than the DOF of the 50mm, depending on the values of the factors which the OP didn't specify but which are necessary for a well-posed question that this or any other DOF equation can answer. One can calculate comparisons for many other interesting situations involving these two lenses or for any other well-defined situation using the equation given.
tunalegs
Pretended Artist
Even though I know it's pointless to experiment with this idea, let me fetch my tripod. It may be amusing.
tunalegs
Pretended Artist
Daneinbalto
Established
The difference is obvious.Well would you look at that!
See if you can tell which one was taken with a 50mm lens, and which one was taken with a 28mm lens.
Sorry for persevering about empirical evidence, but before Galileo a lot of smart people would tell you that a heavy ball drops faster than a light ball.
I realize that mrmeadows' post is based on theories that are firmly rooted in empirical evidence but sometimes "a picture says more than a thousand words."
Especially if you want to know what you can get from the 50 mm crop in the Q compared to another camera with a native 50 mm lens, which was the motivation behind my question. Thank you to all who helped clear this up.
Daneinbalto
Established
Is it correct to assume that the DOF in 50 mm crop mode on the Q corresponds to that of a camera with a smaller sensor (crop factor 50/28 = 1.8), equipped with a 28 mm lens. Assuming same distance to subject, same f/stop, same print size?
Last edited:
Harry Caul
Well-known
I just pulled the trigger on a Ricoh GR II as I wanted a big-sensor, jeans-pocketable camera. My ideal FOV would be 40mm -- why no company has filled this niche yet is beyond me. Anyway, me and 28mm aren't always the best match, so I too was trying to figure out what happens as you use the 35/50mm crop modes. The GR has a 18mm lens and APS-C size sensor -- which gives you ~28mm equiv. FOV. As you crop, an easy way to visualize it would be like having a 35mm equiv. FOV using a m43 size sensor, or a 50mm equiv. FOV using a 1" size sensor. Granted those would be using an aperture of F2.8 so the roughly equivalent in FF in total would be (from this thread):
NO CROP -- 28mm FOV, F4 DOF, 16mp, APS-C sensor area
35mm CROP MODE -- 35mm FOV, F5.6 DOF, 10.3mp, m43 sensor area
50mm CROP MODE -- 47mm FOV, F7-ish DOF, 5.6mp, 1" sensor area
That didn't sound terribly impressive compared to FF. However, when compared to the other high end pocketable P&S cameras like the RX100, it was enough to get me to pull the trigger on the GR II yesterday!
Equivalent apertures and FOV of 1" sensor cameras throughout the zoom range (from this DPReview article):
Back of the envelope, yor your FF Leica Q it would look something like:
NO CROP -- 28mm FOV, F1.7 DOF, 24mp (FF sensor area)
35mm CROP MODE -- 35mm FOV, F2.4-ish DOF, 15.3mp (50%>APSC sensor area)
50mm CROP MODE -- 50mm FOV, F4-ish DOF, 8.4mp (between m43/APSC sensor area)
Hopefully that was close and hope that helps
NO CROP -- 28mm FOV, F4 DOF, 16mp, APS-C sensor area
35mm CROP MODE -- 35mm FOV, F5.6 DOF, 10.3mp, m43 sensor area
50mm CROP MODE -- 47mm FOV, F7-ish DOF, 5.6mp, 1" sensor area
That didn't sound terribly impressive compared to FF. However, when compared to the other high end pocketable P&S cameras like the RX100, it was enough to get me to pull the trigger on the GR II yesterday!
Equivalent apertures and FOV of 1" sensor cameras throughout the zoom range (from this DPReview article):
Back of the envelope, yor your FF Leica Q it would look something like:
NO CROP -- 28mm FOV, F1.7 DOF, 24mp (FF sensor area)
35mm CROP MODE -- 35mm FOV, F2.4-ish DOF, 15.3mp (50%>APSC sensor area)
50mm CROP MODE -- 50mm FOV, F4-ish DOF, 8.4mp (between m43/APSC sensor area)
Hopefully that was close and hope that helps
Daneinbalto
Established
So for a given crop, you can calculate an associated increase in apparent f/stop? And presumably the number is constant across the range of f/stops?
Harry Caul
Well-known
Yes, most people estimate the conversions for DOF and FOV as:
m43->FF roughly 2 f-stops DOF, 2x listed focal length -- ex. Olympus 12mm/F2 acts like a 24mm/F4 FF lens.
APS-C->FF roughly 1 f-stop DOF, 1.5x listed focal length -- ex. Fuji 35mm/F1.4 acts like a 50mm/2 FF lens.
I'm not sure about DOF impact of 1 inch sensors, but judging from DPReviews chart the F2.8 RX10 gives a FF-equivalent DOF of roughly F7.7-7.8, or about 2.8 stops above the listed f-stop. The crop factor for 1 inch sensors is 2.7x listed focal length. So a Nikon 10mm/F2.8 would give similar characteristics to a 27mm/F7.7 FF lens.
Roughly
m43->FF roughly 2 f-stops DOF, 2x listed focal length -- ex. Olympus 12mm/F2 acts like a 24mm/F4 FF lens.
APS-C->FF roughly 1 f-stop DOF, 1.5x listed focal length -- ex. Fuji 35mm/F1.4 acts like a 50mm/2 FF lens.
I'm not sure about DOF impact of 1 inch sensors, but judging from DPReviews chart the F2.8 RX10 gives a FF-equivalent DOF of roughly F7.7-7.8, or about 2.8 stops above the listed f-stop. The crop factor for 1 inch sensors is 2.7x listed focal length. So a Nikon 10mm/F2.8 would give similar characteristics to a 27mm/F7.7 FF lens.
Roughly
Rob-F
Likes Leicas
This thread should probably be a sticky. I think Mr. Mike M. has nailed it.
icebear
Veteran
Obviously the general facts are clear but I don't follow the logic why the print size/magnification has any influence on the DOF of the original shot ...???
If I capture a scene with a 3D object and there is a 2 inch DOF in sharp focus, then this will not change when the shot is magnified to whatever size ... what am I missing here?
If I capture a scene with a 3D object and there is a 2 inch DOF in sharp focus, then this will not change when the shot is magnified to whatever size ... what am I missing here?
Daneinbalto
Established
It is because, as mmeadows said, "DOF exists only as a perception by a human viewer". If you size down a print and look at it at the same distance its DOF will appear wider. Think cell phone compared to a large screen or print.Obviously the general facts are clear but I don't follow the logic why the print size/magnification has any influence on the DOF of the original shot ...???
If I capture a scene with a 3D object and there is a 2 inch DOF in sharp focus, then this will not change when the shot is magnified to whatever size ... what am I missing here?
With regard to the 50 mm software crop mode on the full-frame sensor, 28 mm lens Q, an apt analogy might be to a camera with a smaller sensor (crop factor 1.8, or somewhere between micro 4/3 and APS-C), which would capture a 50 mm-equivalent field of view with a 28 mm lens. As Harry Caul stated, f/1.7 would appear as if taken somewhere around f/4 with a native 50 mm lens.
Rob-F
Likes Leicas
Obviously the general facts are clear but I don't follow the logic why the print size/magnification has any influence on the DOF of the original shot ...???
If I capture a scene with a 3D object and there is a 2 inch DOF in sharp focus, then this will not change when the shot is magnified to whatever size ... what am I missing here?
DOF is about the degree of acceptable unsharpness. If we start with a 5x7 print that has foreground or background details not in the plane of sharp focus, the unsharpness of those details may not be evident in the 5x7. But if we blow it up large enough, there will be a point where the sharpness difference between the in- and out-of-focus areas becomes visible.
If you want to check a 35mm slide or negative for sharpness, you will need a loupe or slide projector because the slide is too small to see without magnification. It's the same idea. Whether you use a loupe or just make a bigger print, you had to blow up the picture to see what parts are sharp and what parts aren't.
But this isn't about changing the depth of field. It is changing the perception of depth of field. The info was always there... it's just that making the image larger starts to show all limitations of an image (be it sharpness, focus, depth of field, resolution, etc).
icebear
Veteran
If you photograph a box of matches that sits diagonal in the perfect DOF so that the front and back corner of the box will be tack sharp and matches spilled in the foreground and a candle in the background will be OOF then you can change the size of the print or the viewing distance any way you want. The corners of the box will remaining sharp at any size or any viewing distance- given there are no technical screw ups in enlarging the print or screen view. DOF is a function of focal length, aperture and subject distance given the plane of film/sensor is fixed in the camera. The viewing distance or printed size of the captured image is not relevant.
mrmeadows
Established
Klaus:
You said: "If you photograph a box of matches that sits diagonal in the perfect DOF so that the front and back corner of the box will be tack sharp and matches spilled in the foreground and a candle in the background will be OOF then you can change the size of the print or the viewing distance any way you want. The corners of the box will remaining sharp at any size or any viewing distance- given there are no technical screw ups in enlarging the print or screen view."
What you said would be true ... if you could do it. It is always true, as you say, that the objects in a negative that are "tack sharp" will stay tack sharp at all magnifications and viewing distances for all viewers. But the facts of physical optics dictate that you cannot make the front and back corners of the box of matches both tack sharp with a typical camera and lens, if the corners are at different distances from the film plane. For such a camera, there will always be exactly one plane at one distance from and parallel to the the film plane that will be tack sharp. We can choose either of the corners at different distances to be tack sharp, if we place it in the single plane of focus. Then the other corner will always be out-of-focus, just as everything else not in the unique plane of focus will be out-of-focus. To do what you have in mind would require tilting the optic axis of the lens from being perpendicular to the film plane. Tilting the optic axis will tilt the plane of focus, so that the corners of the box of matches can be at different distances yet both still be in the unique plane of focus. This principle is named the Scheimpflug Principle (see Wikipedia for detail). The ability to do this is one of the strengths of view cameras which can tilt and shift their lens relative to the film plane. For negatives made with tilt, the DOF whould still be measured perpendicular to the plane of focus, just as it is when that plane is not tilted.
You then said: "DOF is a function of focal length, aperture and subject distance given the plane of film/sensor is fixed in the camera. The viewing distance or printed size of the captured image is not relevant."
This statement is false. The plane of focus and the sizes of all the BLUR CIRCLES of all the out-of-focus objects in a negative are fixed at the time of exposure by the factors that you cite, but the DOF is never fixed at the time of exposure by these factors. The DOF is a perception issue, not just a hardware issue, and it depends on whether a viewer can detect the blur circles when he views the negative or a magnified print or other image made from the negative. If the viewer cannot detect the blur circles of an object, then the viewer cannot detect that the object is actually out-of-focus and it will appear to him to be in focus and to be within in the DOF. Only those blur circles that the viewer can detect will cause him to recognize that the corresponding object is actually out-of-focus. Detecting the the blur circles depends on how the sizes of the blur circles compare to the acuity of the viewer's vision at the viewing distance, so DOF depends on magnification, viewing distance and visual acuity in addition to the three hardware factors. If your statement were true, then the DOF of EVERY photograph would be ZERO, because only the objects exactly in the plane of focus are tack sharp and a plane has no depth; therefore there couldn't be any DOF at all.
Hope this response helps.
--- Mike
You said: "If you photograph a box of matches that sits diagonal in the perfect DOF so that the front and back corner of the box will be tack sharp and matches spilled in the foreground and a candle in the background will be OOF then you can change the size of the print or the viewing distance any way you want. The corners of the box will remaining sharp at any size or any viewing distance- given there are no technical screw ups in enlarging the print or screen view."
What you said would be true ... if you could do it. It is always true, as you say, that the objects in a negative that are "tack sharp" will stay tack sharp at all magnifications and viewing distances for all viewers. But the facts of physical optics dictate that you cannot make the front and back corners of the box of matches both tack sharp with a typical camera and lens, if the corners are at different distances from the film plane. For such a camera, there will always be exactly one plane at one distance from and parallel to the the film plane that will be tack sharp. We can choose either of the corners at different distances to be tack sharp, if we place it in the single plane of focus. Then the other corner will always be out-of-focus, just as everything else not in the unique plane of focus will be out-of-focus. To do what you have in mind would require tilting the optic axis of the lens from being perpendicular to the film plane. Tilting the optic axis will tilt the plane of focus, so that the corners of the box of matches can be at different distances yet both still be in the unique plane of focus. This principle is named the Scheimpflug Principle (see Wikipedia for detail). The ability to do this is one of the strengths of view cameras which can tilt and shift their lens relative to the film plane. For negatives made with tilt, the DOF whould still be measured perpendicular to the plane of focus, just as it is when that plane is not tilted.
You then said: "DOF is a function of focal length, aperture and subject distance given the plane of film/sensor is fixed in the camera. The viewing distance or printed size of the captured image is not relevant."
This statement is false. The plane of focus and the sizes of all the BLUR CIRCLES of all the out-of-focus objects in a negative are fixed at the time of exposure by the factors that you cite, but the DOF is never fixed at the time of exposure by these factors. The DOF is a perception issue, not just a hardware issue, and it depends on whether a viewer can detect the blur circles when he views the negative or a magnified print or other image made from the negative. If the viewer cannot detect the blur circles of an object, then the viewer cannot detect that the object is actually out-of-focus and it will appear to him to be in focus and to be within in the DOF. Only those blur circles that the viewer can detect will cause him to recognize that the corresponding object is actually out-of-focus. Detecting the the blur circles depends on how the sizes of the blur circles compare to the acuity of the viewer's vision at the viewing distance, so DOF depends on magnification, viewing distance and visual acuity in addition to the three hardware factors. If your statement were true, then the DOF of EVERY photograph would be ZERO, because only the objects exactly in the plane of focus are tack sharp and a plane has no depth; therefore there couldn't be any DOF at all.
Hope this response helps.
--- Mike
icebear
Veteran
LOL, yeah, helps a lot .
For all practical purpose of handheld photography I go by what I tried to explain.
My definition of tack sharp is: sharp at 100% view on my calibrated 27" NEC, that's good enough in my book.
For adcademic purpose of this discussion, your are correct
.For all practical purpose of handheld photography I go by what I tried to explain.
My definition of tack sharp is: sharp at 100% view on my calibrated 27" NEC, that's good enough in my book.
For adcademic purpose of this discussion, your are correct
zuiko85
Veteran
All I know is that this discussion makes me want to run in circles of confusion.