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#1
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Image circle versus stopping down?
The way I understood it stopping down increased the image circle because it improved the quality of the image. But then I found this quote "4. Why does the size of the image circle in a Large Format lens change as the lens is stopped down? Well . . . it doesn't actually. What actually happens is that when the lens is wide open, the center of the image is very bright and the corners of the image are VERY dark. So dark in fact that they do not expose the film (When the film is properly exposed for the center of the image). As the lens is stopped down, uniformity of illumination improves and therefore the circle of usable illumination increases." That's from here http://www.schneideroptics.com/info/...mat_lenses/#q3 So if I understand this. You could use a lens wide open and get the same image circle if you could use something like a centre filter? In other words the edges shouldn't be soft. Does this make sense to anybody? Nick |
#2
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Image circle versus stopping down?
Nick Zentena wrote in message ...
The way I understood it stopping down increased the image circle because it improved the quality of the image. But then I found this quote "4. Why does the size of the image circle in a Large Format lens change as the lens is stopped down? Well . . . it doesn't actually. What actually happens is that when the lens is wide open, the center of the image is very bright and the corners of the image are VERY dark. So dark in fact that they do not expose the film (When the film is properly exposed for the center of the image). As the lens is stopped down, uniformity of illumination improves and therefore the circle of usable illumination increases." That's from here http://www.schneideroptics.com/info/...mat_lenses/#q3 So if I understand this. You could use a lens wide open and get the same image circle if you could use something like a centre filter? In other words the edges shouldn't be soft. Does this make sense to anybody? Nick A center filter does not perform the same function as the stop. The explanation on the Schndeider site is incomplete and oversimplified. The main aberrations which vary with the stop are spherical aberration and coma. Spherical exists all over the image, coma is not present at the center but increases with the image angle. Spherical aberration causes a hazy surround to highlights. Enough spherical causes an overall haze and a blury image, it is the cause of the softness of soft focus lenses. Coma is related to spherical but is asymmetrical. It causes smeared, somewhat tear-drop-shaped blur spots along radiuses toward the center. They can point toward or away from the center of the image depending on where they mainly occur in the lens. Coma varies rapidly with the stop. Both of these aberrations vary with the angle the light rays take through the lens. Other factors being equal they increase with the speed of the lens, and coma especially increases with the angle of coverage, i.e., its harder to correct for wide angle lenses. These aberrations become smaller as the light going through the lens is concentrated toward the center of the lens. This is what stopping down accomplishes. Uniformity of illumination is not affected by the stop except where there is some mechanical vignetting from the lens mount. For most lenses this is gone when stopping down about two stops. Some wide angle lenses are designed to reduce this mechanical vignetting (the Angulon is an example). One of the inherent properties of lenses which produce orthographic images is that the illumination falls off with image angle. For a "normal" design lens this fall off is approximately proportional to cos^4 theta where theta is the "half angle" of the image point. There are designs of lenses which have improved illumination but these are not better than about cos^3 theta, better, but there is still fall off. A non-orthograpic lens, like a fish-eye lens, can have better illumination because the light is more concentrated near the margins. A center filter acts to compensate, at least in part, for the fall off by letting more light in at the margins of the image. The filter is not exactly a stop but rather a tapered obstruction. The lens projects a very blury, out of focus, image of the center filter onto the film. One reason a center filter must be used with the lens stopped down is that the range of angles the light going through the filter must be limited for it work. It is possible that a center filter _may_ act, to some extent, as a obstructive stop. An obstructive stop limits the rays of light to the margins of the lens rather than the center, as does a normal stop. An obstructive filter changes the balance of all the aberrations and has an effect on the MTF curves. In comparison to a standard stop the obstrucive stop can raise the edge contrast at the price of lowering the resolution. The Schneider explanation about the balance of aberrations to diffraction is correct. As the stop size is reduced the aberrations are reduced but the diffraction is increased. At some point the two curves cross. That is the "optimum stop". Actually, there isn't a single optimum stop. Since coma and sphrical abberation play a part in the image quality, and, since both vary with image angle, the optmum stop will vary with the image angle wanted from the lens. For instance, a Dagor, which has relatively low coma (as do all symmetrical lenses) has a circle of illumination of nearly 90 degrees. At about f/11 it has a coverage with good image quality of around 60 degrees. To get the maximum coverage angle it must be stopped down to f/45. The image quality at the margins of the image at 87 degrees will be best at f/45 but not at the center, there it will be best at around f/11 or f/16. Richard Knoppow Los Angeles, CA, USA |
#3
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Image circle versus stopping down?
"Richard Knoppow" wrote in message om... [...] One of the inherent properties of lenses which produce orthographic images is that the illumination falls off with image angle. For a "normal" design lens this fall off is approximately proportional to cos^4 theta where theta is the "half angle" of the image point. There are designs of lenses which have improved illumination but these are not better than about cos^3 theta, better, but there is still fall off. [...] Care to help an innumerate? Taking the 38mm Biogon as an example, what's the light fall-off in terms of F-stops? |
#4
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Image circle versus stopping down?
"jjs" wrote in message ... "Richard Knoppow" wrote in message om... [...] One of the inherent properties of lenses which produce orthographic images is that the illumination falls off with image angle. For a "normal" design lens this fall off is approximately proportional to cos^4 theta where theta is the "half angle" of the image point. There are designs of lenses which have improved illumination but these are not better than about cos^3 theta, better, but there is still fall off. [...] Care to help an innumerate? Taking the 38mm Biogon as an example, what's the light fall-off in terms of F-stops? If you consider the light at the center of the image circle to have a magnitude of 1 (units do not matter because we will be dealing with ratios), the number of stops of fall off at an angle Theta degrees off of the center of the center of the image would be: Fall-off in Stops = 3.322 x Log(1 / Cos^4 of the angle Theta) If you assume the fall off of the lens design is Cos^3 instead, the formula would obviously be: Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta) For instance, the fall off at a point 30 degrees off of the center, considering a lens with Cos^4 fall off, would be: Fall-off in Stops = 3.322 x Log(1 / Cos^4 ( 30 )) Fall-off in Stops = 3.322 x Log(1 / (0.866^4)) Fall-off in Stops = 3.322 x Log(1 / 0.5625 ) Fall-off in Stops = 3.322 x Log(1.77777) Fall-off in Stops = 3.322 x 0.2498 Fall-off in Stops = 0.83 stops Guillermo |
#5
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Image circle versus stopping down?
"f/256" wrote in message ogers.com... "jjs" wrote: Care to help an innumerate? Taking the 38mm Biogon as an example, what's the light fall-off in terms of F-stops? If you consider the light at the center of the image circle to have a magnitude of 1 (units do not matter because we will be dealing with ratios), the number of stops of fall off at an angle Theta degrees off of the center of the center of the image would be: Fall-off in Stops = 3.322 x Log(1 / Cos^4 of the angle Theta) If you assume the fall off of the lens design is Cos^3 instead, the formula would obviously be: Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta) For instance, the fall off at a point 30 degrees off of the center, considering a lens with Cos^4 fall off, would be: Fall-off in Stops = 3.322 x Log(1 / Cos^4 ( 30 )) Fall-off in Stops = 3.322 x Log(1 / (0.866^4)) Fall-off in Stops = 3.322 x Log(1 / 0.5625 ) Fall-off in Stops = 3.322 x Log(1.77777) Fall-off in Stops = 3.322 x 0.2498 Fall-off in Stops = 0.83 stops Very helpful, Guillermo. I'm a bit closer to understanding. So, finding the Cos power of the lens-design remains problematic. Here's the lens I'm working with: http://course1.winona.edu/jstafford/...s1/index2.html There is very little light fall of even in the corners. In fact there is so little I can hardly find any. Note the construction of the lens: it covers 4x5" (actually more than 5x5") and the rear lens is 4.5" in diameter. |
#6
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Image circle versus stopping down?
So, finding the Cos power of the lens-design remains problematic. Here's the lens I'm working with: http://course1.winona.edu/jstafford/...s1/index2.html There is very little light fall of even in the corners. In fact there is so little I can hardly find any. Note the construction of the lens: it covers 4x5" (actually more than 5x5") and the rear lens is 4.5" in diameter. That's a big piece of glass! As Richard said, and who am I to argue with him!, designs give you not better than Cos^3, that lens 3" FL give you and angle of 46 degrees at the corner of a 4x5 negative, still 1.5 stops according to Cos^3 fall off, so it may not look like there is any fall off but likely there is. Here is an explanation to the Cosine^4 and how to find the fstops of fall off, in case it is helpful for you to understand what's going on: The way Cosine law works is like this: because a point off the center is farther away by 1/Cosine of the angle, light at that point is dimmer than at the center, because the InverseSquareLaw light is dimmer by 1/Cos^2 at that point (that account for 2 times Cos in the cos^4 law), because light falls on an angle on the film plane, it covers an area 1/cos bigger that at the center of the film, more coverage with the same light means dimmer light per unit of area (another Cos in the Cos^4 law) and finally because the aperture is no seen as a circle but as an oval that has Cosine of the angle less area, it allows less light thru by a factor of -you guessed it!- cosine of the angle (yet another Cosine in the cos^4 law), the whole thing combines to Cosine^4. If we assume a light intensity of 1 at the center, the fall off 60 degrees off of the center would be Cos^4 (60 degrees) = 0.0625, in other words if you have 100 units of light at the center, at the corners you would have only 6.25 units of light !! How do you calculate the numbers of stops between the 6.25 and 100 ? You double 6.25 until you get 100, like this: 6.25 x 2 = 12.5 one stop 12.5 x 2 = 25 two stops 25 x 2 = 50 three stops 50 x 4 = 100 four stops There are 4 stops fall off with respect to the center at a point 60 degrees from the center. Guillermo |
#7
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Image circle versus stopping down?
Here is an example based on the Biogon 38mm:
Assuming it has a fall off as good as Cos^3 The size of a 6x6 negative is actually 55mmx55mm, its diagonal is 77.77mm and there is arcTan(77.77/(2x38)) degrees from the center to the corners of the film, that is 45.66 degrees or a total angle of view of 91.32 degrees from corner to corner. The fall off at the corners would be: Theta = 45.66 Cos(45.66) = 0.6989 Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta) Fall-off in Stops = 3.322 x Log(1 / (0.6989)^3) Fall-off in Stops = 3.322 x Log(1 / 0.3414) Fall-off in Stops = 3.322 x Log(2.929) Fall-off in Stops = 3.322 x 0.46672 Fall-off in Stops = 1.55 There are ONE and a HALF stops of fall off at the corners of the film with respect to the center of the film!! Guillermo |
#8
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Image circle versus stopping down?
"f/256" wrote in message . rogers.com... Here is an example based on the Biogon 38mm: Assuming it has a fall off as good as Cos^3 [...] There are ONE and a HALF stops of fall off at the corners of the film with respect to the center of the film!! Funny, but I don't find the 38mm Biogon that bad in real life. |
#9
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Image circle versus stopping down?
Richard Knoppow wrote:
One of the inherent properties of lenses which produce orthographic images is that the illumination falls off with image angle. For a "normal" design lens this fall off is approximately proportional to cos^4 theta where theta is the "half angle" of the image point. There are designs of lenses which have improved illumination but these are not better than about cos^3 theta, better, but there is still fall off. A non-orthograpic lens, like a fish-eye lens, can have better illumination because the light is more concentrated near the margins. What about retro focus wide angle lenses? While not common on LF cameras, how does that design compare to "normal" wide angle lenses as far as light fall off? I have noticed almost no light fall off on the 30mm medformat fisheye and even the regular wide angle SLR lenses don't seem to be as affected by fall off as some of the wide LF lenses I've used. -- Stacey |
#10
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Image circle versus stopping down?
Stacey wrote in message ...
Richard Knoppow wrote: One of the inherent properties of lenses which produce orthographic images is that the illumination falls off with image angle. For a "normal" design lens this fall off is approximately proportional to cos^4 theta where theta is the "half angle" of the image point. There are designs of lenses which have improved illumination but these are not better than about cos^3 theta, better, but there is still fall off. A non-orthograpic lens, like a fish-eye lens, can have better illumination because the light is more concentrated near the margins. What about retro focus wide angle lenses? While not common on LF cameras, how does that design compare to "normal" wide angle lenses as far as light fall off? I have noticed almost no light fall off on the 30mm medformat fisheye and even the regular wide angle SLR lenses don't seem to be as affected by fall off as some of the wide LF lenses I've used. Extreme examples of reversed telephotos can have zero falloff *and* zero distortion. The problem is that they are really really big. Way too big for large format. See the simple two-element example I posted above. Systems like this are actually used for high performance DMD and LCD projections systems which require perfect telecentricity, extremely large working distance, and zero distortion. Zero falloff means cos^0. Actually, its possible for the corner illumination to be slightly *greater* than the center illumination. Brian www.caldwellphotographic.com |
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