Great Bustard
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With all due respect, yes it is. If you were a man of your word, you wouldn't try to backpeddle out of it, either.No it isn't!That clip is BS and you know it.
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Exposure.
- Thread starter Great Bustard
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You're right, we should just quit this nonsense.With all due respect, yes it is. If you were a man of your word, you wouldn't try to backpeddle out of it, either.No it isn't!That clip is BS and you know it.
Oh, and by the way, YOUR WRONG!
Andre Affleck
Senior Member
It was about corner/edge falloff due the break down of the inverse square law that started this part of the debate. On one of your posts you basically paraphrased a section from your essay here:
First of all, why must you assume a uniformly lit scene , and is that what you really meant to say? A uniformly lit scene is one where the light source itself is emitting light uniformly. It says nothing about the characteristic of the light entering the lens because you have not characterized how the light interacts with the scene (object color, surface reflectance), hence no need for uniformity in the first place. Or did you mean to say the the light entering the lens was uniform? They are completely different situations. In either case you are only talking about ratios anyway, what does it matter?
Then you allude to a fall off at wide angles when the distance from the corners is not similar to the distant at the center. What evidence do you have that it has to do with the ISL? How do you know it isn't just the angle if incidence as it strikes the sensor at those wide angles? Or does it have anything to do with rectilinearity? You also say the distance to the corners are close to the distance to the center, but the corners are actually 35% further, which is significant yet the fall off is not that severe. Or are you not talking about a planer FOV like the surface of a wall? There are too many unknowns in your premise.
Great Bustard
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Exposure is the density of light on the sensor. If the scene is not uniformly lit, the density of the light will vary. However, if we are only considering the portion of the scene that the photos have in common, then this condition is not necessary.
Consider a wall made of nothing but LEDs. We can consider each LED a point source of light. We take a pic of this wall 5 ft away from the center of the wall at 25mm with the line of sight perpendicular to the wall.
The LEDs in the center are 5 ft away, the LEDs in the corners are 6.6 ft away. So, less light from the corner LEDs will reach the aperture than from the center LEDs, per the ISL (I'm assuming that light from LEDs is not collimated).
I'm certainly not saying that there are not other variables. I'm just saying that the ISL is one of them.
Andre Affleck
Senior Member
But the density will vary anyway due to differences in reflectance.
However, if we are only considering the portion of the scene that the photos have in common, then this condition is not necessary.
It's not necessary in either case.
Correct at the aperture. Incorrect at the sensor, because the lens represents that reduced light the same as the same the center one (assume non-rectilinear lenses for now). The density is maintained. For rectilinear lenses one could argue that the representation due to distortion caused it, not the break down of the ISL. In, fact, the ISL was maintained in your premise anyway (further away, less light).
By the way, would you consider the distance of those corner LEDs to be close to the distance of the center ones?
Here, I am just saying that the way you described your premise, it's a little difficult to nail down exactly what you are trying to lay down because there are still some aspects floating around. Just an opinion.
Great Bustard
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I don't know what you mean by that.
Why not?
Please explain in more detail. Specifically, why the same light per area falls on the sensor from an LED in the corner as from an LED in the center, given that the corner LED is 6.6 ft away and the center LED is 5 ft away.
Also, please limit the discussion to rectilinear lenses at this point, since rectilinear lenses, by far, are the most common.
Please explain in more detail.
I don't know what you mean. The corner LEDs are 6.6 ft away in the example, the center LED is 5 ft away in the example. Are you asking me if 6.6 ft is "close to" 5 ft? If so, then, yes, as I said in the OP, the differential represents only 1/4 of a stop.
If you could say what you think my premise is, that would help me a lot to be more specific.
Andre Affleck
Senior Member
If you have a uniformly lit scene, but are illuminating black objects and white object, the light reaching the sensor is no longer uniform (black objects reflect less light than white).
It's no different than having a non-uniform light source illuminating a white wall of homogeneous reflectance. Furthermore, it doesn't matter because you are talking about ratios in your premises, not absolute measurements.
Because it is represented that way (and I'm not trying to be smart a$$, it really is, right?). The lens changes the representation by making the light converge to a smaller area when it reaches the sensor. Therefore, at the focal point, the density is the same even though the total light is reduced for the corner LED. Now the LEDs are small but not really point sources. If you compared the areas of each LED on the sensor you will find the corner one to have less area than the center, which again exactly offsets the loss in light.
We will, but not yet.
Now consider if the LEDs are replaced by disks of light (IOW, no longer point sources). In this scenario, the ISL does break down, because it is no longer a point source, right? The total light from the corner is still less than the center but not quite by the same amount as when they were point sources. However, when the light reaches the sensor, the lens again has changed the representation of the corner disk by reducing the area (appearing smaller in the photo), therefore the density is still the same.
In either case, the density is maintained at the sensor, even though in one case the ISL broke down, so you can't really say it's due to the ISL.
What is different is the incident angle of the light as it reaches the sensor, and perhaps this is responsible for the fall off, but I don't think it is due to the ILS.
For a rectilinear lens, the ILS it still maintained (consider the point source LED example) just as it was for a non-rectilinear lens. The only difference is that the representation of the corner LED shifts further due to the distortion (shift in area). I still don't think you can blame the ILS. The change in representation only further shifts the loss due to the incident angle.
It's just my opinion, but a 35% change in anything is not considered negligible, that's all I'm saying. BTW, it through me off because I thought maybe you were trying to say they were at the same distance as in a non-planer scene. Doesn't matter though.
Well there are several that I can think of. That's the problem. Perhaps if you had been specific about the objects in the scene, or if they were planer or not (which affects corner distance), or are framing a large white wall, where if you step back you are still have the wall fill the AOF, etc. Anyway this is more subjective.
Andre Affleck
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One more thing. If you were to measure the total light intensity using a light meter just in front of the lens, that total would be from contributions of both LED point sources. If we stepped back and took another measurement, that amount would be less but not consistent with the ISL (break down).
Notice though, that at the sensor, this light is kept segregated inherently. Light from one LED never reaches the where the other LED's light strikes the sensor. If we step back, the same relationship holds and the ISL is maintained.
This is another area of confusion for some when it come to photometric light measurement. In one scenario you are measuring Illuminance (the light meter), the other you are measuring Luminance (the sensor). The ISL only relates to Illuminance, not to Luminance.
Notice though, that at the sensor, this light is kept segregated inherently. Light from one LED never reaches the where the other LED's light strikes the sensor. If we step back, the same relationship holds and the ISL is maintained.
This is another area of confusion for some when it come to photometric light measurement. In one scenario you are measuring Illuminance (the light meter), the other you are measuring Luminance (the sensor). The ISL only relates to Illuminance, not to Luminance.
Steen Bay
Veteran Member
It's the inverse square law, so isn't it 74% more light/area at 5 feet? (or 43% less af 6.6 feet)
Andre Affleck
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I was talking about distance, but you're right, it equates to 74% more light as well. That difference, however, is not seen at he sensor, hence it has nothing to with ISL.
Great Bustard
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Ah. Poorly worded on my part. I meant for a scene with a uniform luminance.
In a post above, Steen said that the bricks on a wall would all be represented with the same area on the sensor and in the photo. So which is it?
I'm snipping the rest until that point is resolved -- otherwise the posts get too cumbersome to follow. As I said, one point at a time, just to make the discussion tractable.
Andre Affleck
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That's fine, but this means a completely blank, featureless image (like a white or gray wall). Is that what you want? And again, what does it matter anyway. You are talking ratios (between two different distances). Only if you are comparing actual values does it matter. Just remove the phrase altogether IMO.
He's right (different representations though). Only the corner bricks are stretched to match the same area as the center. Therefore the representation of the corner bricks are different, right? The representation of the center bricks are unchanged.
Ok, to sum up the corner issue. For a non-rectilinear lens, I am proposing that the light density for the corner light (LED or disk) is exactly the same for the center. For a rectilinear lens, the corners are stretched so the area increases (to match the area of the center), making the density decrease (dimmer). The point is, in either of these cases, it has nothing to do due with the ISL. You could say instead that "due to rectilinear correction and steep incident angles at the corners, light falloff is present" or something to that affect. Not sure how to word the incident angle part but it's supposed to be associated with digital sensors only. However I heard that the micro lenses correct for some of this, not sure.
Also, it has nothing to do with physical vignetting which is normally associated with some kind of periphery obstruction/limitation by the lens elements, and cured by aperture stop.
Great Bustard
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I was just trying to eliminate variables that complicate the issue. You see, the exposure (density of light) varies all over the sensor. Instead, I guess I should simply say "same average luminance".
OK, so the corner bricks are stretched so they take the same area on the sensor and the photo as the center bricks. Now, what was the point you were making about that?
Sorry to cut you off. I don't mean to be rude -- seriously. I just want to deal with a rectilinear lens for the time being, since, as I said earlier, almost all lenses are rectilinear.
Once that's worked out, then I'd love to discuss FE and TSE lenses (I don't know much about how they work, to tell the truth).
Andre Affleck
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That might work. In fact you definitely need to say something because in addition, you are not indicating what is filling in the area of the image when you say "frame twice as wide". So you could say:
Assuming the same average luminance, framing twice as wide, for example, will result in four times as much area, and thus four times as much light.
So now whatever is coming into frame, the average is still maintained. I think the premise is locked in now.
Ok, but you are missing the point. You are not analyzing the non-rectilinear case so you know what is happening for those lenses. Understanding the non-rectilinear geometry is the first step to the proof of the rectilinear case.
How else do I prove to you there are no other contributions to the fall off? If you agree that the light densities are equal (no fall off) for the non-rectilinear geometry, and the only difference with rectilinear is the change in representation due to distortion correction, then by definition there can be nothing else responsible for the fall off. ISL cannot apply then. Not sure how else I'm going to prove it without it.
I'm not saying the non-linear case needs to be mentioned in your essay. It's just to satisfy your criticism.
Steen Bay
Veteran Member
The (apparent) area of a flat subject like a brick (or a piece of paper) depends on both the distance (ISL) and the angle from which it's seen, so in the case of shooting a brick wall, then the area of the corner brick, seen from the cameras position, is affected (reduced) by both the increased distance and the increased angle (compared to the center brick). If we instead were comparing the size of two (soccer) footballs, one in the center and one in the corner of the flat focus plane (seen with our own eyes, not through the lens), then only the distance (ISL) would affect the apparent size, and if shot with a fisheye lens, then both footballs would be round (like they actually are), but represented with different areas on the sensor, and if shot with a rectilinear lens, then the corner football would actually have a larger area in the image than the center football, and its shape would be stretched/distorted (I think, this perspective/projection thing can really be a bit confusing ;-)).
Ken Phillips
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... big glass gathers more starlight. A 10" (250mm) Ø mirror at f/10 will show dimmer stars than a 2" (100mm) Ø lens at f/1.0.
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Andre Affleck
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Yes except for the ISL part. The area is not dependent on the ISL, only the distance is. ISL determines the density (brightness) so ISL has no applicability in your sentence.
Agreed.
Yes which is where I got confused. The objects are no longer co-planer so they need not be the same size anymore.
Andre Affleck
Senior Member
Sorry, when I wrote this I meant to say in the context of this thread, the when we speak of the ISL we are talking about the change in density (brightness) with distance. That's all I was saying. Obviously the area is changing with the square of the distance so it applies in that respect. When I say the ISL doesn't apply Im saying, even though the area is following the ISL, the brightness of two objects does not follow the ISL.
Great Bustard
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OK, let's say we shoot a brick wall so that the brick in the corner of the frame is twice as far from the aperture as the brick in the center of the aperture and all bricks have the same luminance.
If x represents the number of photons that arrive at the aperture from the center brick in a given time interval, how many photons will arrive at the aperture from the corner brick?
After we have the answer to that, then the question is what percent of those photons find their way to the sensor (I maintain that it's the same percentage, center or corner).
Lastly, the exposure simply depends on how large the two bricks are represented, since we have already computed the total number of photons from each brick.
Andre Affleck
Senior Member
Twice as far so there are x/4 as many photons coming from the corner.
Not counting negligible losses through the optics, it is 100% for each.
If area A represents the area of the center, then the area of the corner will be A/4+d where d is the amount of area growth due to optical distortion correction.
What's the point of this post? I already went over this with you three times now. The squares cancel (here's where the ISL goes out the window) leaving the residual change in area due to the distortion. Therefore, the fall off is due to the distortion correction.
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