Doug Kerr
Well-known member
There has recently been considerable discussion here of the implication of what is sometimes called the "chimney effect" in digital cameras. This refers to the fact that, from a performance standpoint, we can think of the detectors of a digital sensor as lying at the bottom of tiny "chimneys". Thus, if we consider a tiny region of the object lying on the axis, and consider the "cone" of light emerging from the lens that will be converged at points across the corresponding region of the sensor, the light in the outermost portion of the cone will not be as effectual upon the photodetectors as the light in the central part of the cone. That's because the "more oblique" light is less able to "fully drop down the chimneys" and fall on the photodetectors.
One consequence of this is that as we increase the aperture of the lens (decrease the f-number), and consider a fixed luminance of a scene patch, the "impact" on the sensor increases more slowly than the area of the lens' entrance pupil - that is more slowly than the decreasing f-number would suggest. A recent report in Luminous Landscape presents some data developed by DxO Labs that shows the degree of this affect for various actual camera sensors. With an f/1.2 lens, the increase in exposure effect on the photodetectors over that from, say, an f/8 lens is about 1/2 stop less than the f-number would lead us to expect.
Now, those familiar with basic photometric theory may say:
In fact, that sounds credible. But it doesn't really work quite that way. Nevertheless, there is a real effect that story would seem to suggest.
Let's first look at the meaning of the thing about the cosine of the angle of incidence. Suppose we have a small "pencil" of light (a bundle of parallel rays within a tubular region of a certain diameter) and we consider it first striking the film "head on" and then we consider it striking the film at an angle of 30° from head on.
It will indeed cause an illuminance on the film less in the second case than the first, in the amount of the cosine of 30°.
And it also illuminates, with those two values of illuminance, a larger area in the second case than the first. (Imagine a circular wood dowel, first cut square across and then cut at an angle of 30° to straight across. The cut surface in the second case will be larger.)
In fact this is really why the illuminance is lower in the second case. There, the same amount of light (that contained in our hypothetical "pencil") lands over a larger area of the film. Since illuminance is the ratio of the amount of light to the area over which it lands, the illuminance is lower in the second (oblique arrival) case. (Note that this fundamental definition of illuminance contains no reference to a cosine; that has already done its work in our example by affecting what the area is.)
All this would fits well with the little story stated above.
But we might then ask, "what affect does this have on the behavior of a lens". More specifically:
In fact, answering that question directly is for more complicated than we might at first think. To do it, we must have in hand a model of what happens between the exit pupil and the film. This is very complicated. In fact, even respected optical textbooks often bungle analysis of this.
But we can get at the answer another way, not requiring us to have a model for "image space" photometrics.
Her is the basic plan of attack:
• We consider some tiny patch on the subject, having a certain luminance. (We cannot consider a point, since no light emerges from a point - it has zero area for any light to come from. We will use a point on the axis, since the off-axis situation introduces other complications, which will obscure the basic story here.)
• We consider how much light from that tiny patch enters the lens through the lens' entrance pupil. (We note that the f-number of the lens tells us, if we know the focal length) the diameter of the entrance pupil.)
• Assuming that the lens has 100% transmission (and if it doesn't, the final result will still come out the same way), then all the light from the subject patch entering ("captured by") the lens will be deposited on the film over a small region that is the image of the object region.
• We know the area of the "image patch"; its dimensions are just the dimensions of the assumed object patch times the magnification of the cameras in this situation.
• We know the amount of light that is deposited on this area, and the area of the region.
• Thus we can determine the illuminate on the film in that region.
Not a cosine of the angle of incidence in sight!
So, with this outlook, we might expect that, if we double the diameter of the entrance pupil, thus quadrupling its area (as would happen when we cut the f-number in half), then for an object region of a certain luminance we would get, on the image on the film, exactly twice the illuminance.
And that is very close to being true.
However, if we look at the matter in very fine detail, we find that this is not quite so.
Let's consider two small parts of the area of the entrance pupil, with the same area (perhaps 1 mm square), one at the center of the entrance pupil and one at the outer edge. We'll call these "windows".
Will those each collect the same amount of light from our hypothetical small region on the object. Not quite. Why?
• The distance from the subject patch to these two windows is slightly different (the "path" to the outer window is at a slight angle). Thus the luminous flux density from the object patch will be less at the outer window than at the central one (yes, its that old inverse square law). My God, is it very much smaller? No. Just smaller.
• The amount of light that will pass through a window bathed in light of a certain luminous flux density depends on the area of the window as observed from the direction of arrival of the light of interest (the "projected area" of the window in that direction). In the case of our outer window, at a (slight) angle to the path of the arriving light, its projected area is less than that of the central window (which the light strikes head-on). My God, is it very much smaller? No. Just smaller.
Thus the light collected from a patch on the object by a windows of a certain area in the entrance pupil is less as that area is farther from the center of the pupil.
So, as we add area to the pupil by decreasing the f-number and thus increasing the diameter of the pupil, the amount of light collected from some patch on the object does not increase quite as fast as the area increases. If we cut the f-number in half, the amount of light collected from the patch does not (quite) quadruple.
Thus, as we cut the f-number in half, the illuminant on the film for that object patch does not (quite) quadruple. By very much? Certainly not, except perhaps for high-magnification macro work, where the rim of the entrance pupil falls a substantial distance off the axis as seen from the object. In most cases, the effect is so small that it is masked by other complications in "real" lens behavior.
But note that in explaining this, I made no mention of the angle at which rays from the outer regions of the exit pupil strike the lens, or the cosines of those angles. I didn't have to. Nor do I know, precisely, how to do that. Nor, apparently, really do the authors of respected optical text books.
Best regards,
Doug
One consequence of this is that as we increase the aperture of the lens (decrease the f-number), and consider a fixed luminance of a scene patch, the "impact" on the sensor increases more slowly than the area of the lens' entrance pupil - that is more slowly than the decreasing f-number would suggest. A recent report in Luminous Landscape presents some data developed by DxO Labs that shows the degree of this affect for various actual camera sensors. With an f/1.2 lens, the increase in exposure effect on the photodetectors over that from, say, an f/8 lens is about 1/2 stop less than the f-number would lead us to expect.
Now, those familiar with basic photometric theory may say:
That's very interesting, but don't forget we have a similar phenomenon already at work even with film [where there is nothing equivalent to the "chimney effect"]. That is, the light in the outer portions of the cone [that is, coming from the outer portion of the exit pupil], by virtue of its arriving obliquely on the film, contributes less to the illuminance on the film than the same amount of light in the center of the cone, which arrives "straight on". This is recognized by the fact that the illuminance caused by arriving light varies as the cosine of the angle of incidence.
In fact, that sounds credible. But it doesn't really work quite that way. Nevertheless, there is a real effect that story would seem to suggest.
Let's first look at the meaning of the thing about the cosine of the angle of incidence. Suppose we have a small "pencil" of light (a bundle of parallel rays within a tubular region of a certain diameter) and we consider it first striking the film "head on" and then we consider it striking the film at an angle of 30° from head on.
It will indeed cause an illuminance on the film less in the second case than the first, in the amount of the cosine of 30°.
And it also illuminates, with those two values of illuminance, a larger area in the second case than the first. (Imagine a circular wood dowel, first cut square across and then cut at an angle of 30° to straight across. The cut surface in the second case will be larger.)
In fact this is really why the illuminance is lower in the second case. There, the same amount of light (that contained in our hypothetical "pencil") lands over a larger area of the film. Since illuminance is the ratio of the amount of light to the area over which it lands, the illuminance is lower in the second (oblique arrival) case. (Note that this fundamental definition of illuminance contains no reference to a cosine; that has already done its work in our example by affecting what the area is.)
All this would fits well with the little story stated above.
But we might then ask, "what affect does this have on the behavior of a lens". More specifically:
We normally expect that is we cut the f-number of a lens in half, the illuminance on the film (for a given scene luminance) will quadruple. Won't the "cosine of the angle of incidence" matter diminish this increase.
In fact, answering that question directly is for more complicated than we might at first think. To do it, we must have in hand a model of what happens between the exit pupil and the film. This is very complicated. In fact, even respected optical textbooks often bungle analysis of this.
But we can get at the answer another way, not requiring us to have a model for "image space" photometrics.
Her is the basic plan of attack:
• We consider some tiny patch on the subject, having a certain luminance. (We cannot consider a point, since no light emerges from a point - it has zero area for any light to come from. We will use a point on the axis, since the off-axis situation introduces other complications, which will obscure the basic story here.)
• We consider how much light from that tiny patch enters the lens through the lens' entrance pupil. (We note that the f-number of the lens tells us, if we know the focal length) the diameter of the entrance pupil.)
• Assuming that the lens has 100% transmission (and if it doesn't, the final result will still come out the same way), then all the light from the subject patch entering ("captured by") the lens will be deposited on the film over a small region that is the image of the object region.
• We know the area of the "image patch"; its dimensions are just the dimensions of the assumed object patch times the magnification of the cameras in this situation.
• We know the amount of light that is deposited on this area, and the area of the region.
• Thus we can determine the illuminate on the film in that region.
Not a cosine of the angle of incidence in sight!
So, with this outlook, we might expect that, if we double the diameter of the entrance pupil, thus quadrupling its area (as would happen when we cut the f-number in half), then for an object region of a certain luminance we would get, on the image on the film, exactly twice the illuminance.
And that is very close to being true.
However, if we look at the matter in very fine detail, we find that this is not quite so.
Let's consider two small parts of the area of the entrance pupil, with the same area (perhaps 1 mm square), one at the center of the entrance pupil and one at the outer edge. We'll call these "windows".
Will those each collect the same amount of light from our hypothetical small region on the object. Not quite. Why?
• The distance from the subject patch to these two windows is slightly different (the "path" to the outer window is at a slight angle). Thus the luminous flux density from the object patch will be less at the outer window than at the central one (yes, its that old inverse square law). My God, is it very much smaller? No. Just smaller.
• The amount of light that will pass through a window bathed in light of a certain luminous flux density depends on the area of the window as observed from the direction of arrival of the light of interest (the "projected area" of the window in that direction). In the case of our outer window, at a (slight) angle to the path of the arriving light, its projected area is less than that of the central window (which the light strikes head-on). My God, is it very much smaller? No. Just smaller.
Thus the light collected from a patch on the object by a windows of a certain area in the entrance pupil is less as that area is farther from the center of the pupil.
So, as we add area to the pupil by decreasing the f-number and thus increasing the diameter of the pupil, the amount of light collected from some patch on the object does not increase quite as fast as the area increases. If we cut the f-number in half, the amount of light collected from the patch does not (quite) quadruple.
Thus, as we cut the f-number in half, the illuminant on the film for that object patch does not (quite) quadruple. By very much? Certainly not, except perhaps for high-magnification macro work, where the rim of the entrance pupil falls a substantial distance off the axis as seen from the object. In most cases, the effect is so small that it is masked by other complications in "real" lens behavior.
But note that in explaining this, I made no mention of the angle at which rays from the outer regions of the exit pupil strike the lens, or the cosines of those angles. I didn't have to. Nor do I know, precisely, how to do that. Nor, apparently, really do the authors of respected optical text books.
Best regards,
Doug