A couple of questions for Great Bustard…

Joe Pineapples II said:

olliess said:

Joe Pineapples II said:

Maybe I should have said "Hey! What about Green's functions?"

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Well, they don't amount to much in isolation.

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In the sense that they allow the solution to be expressed in terms of a superposition of elementary sources; which seems to me to contradict the original statement that you can't consider the light as coming from a set of point sources. Of course, the resulting field will have a more complex dependence than a simple ISL.

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If you look at the so called Kirchhoff's diffraction formula, you will see U (the field) multiplied by the normal derivative of the Green's function G, not by G itself; and the normal derivative of U multiplied by G. Both G and its normal derivatives can ne considered as point sources but that sum may provide cancelations in some directions. It is not the same as the naïve formulation of the Huygen's principle, which implies integral of UG (no derivatives, no sum).

Note that the wave is implicitly assumed to be outgoing (satisfying the Sommerfeld radiation conditions). Incoming would mean i=sqrt(-1) there replaced by -i. Then the normal derivative of U, which is a priori unknown can be expressed on the surface through U itself. The relation is given by a non-local operator, meaning that to know dU/dn at some point, you need to know U everywhere on that surface. At high frequencies however, that operator becomes almost local (we call it pseudo-local). It can be well approximated by the so-called Kirchoff's approximation. You can see some of it here:

https://en.wikipedia.org/wiki/Kirchhoff's_diffraction_formula,

look for the cosine terms. So in the end, you do get a superposition of point sources (approximately) but the amplitude carries information about the angle of incidence of the incoming rays, and it is dependent of the direction of the propagation in after it passes the surface, as well. This is what is missed in the naïve formulation of the Huygens' principle which creates the impression that the wave stats to propagate from each point in all directions. The directional dependence is totally ignored, which is the whole point of a lens.

Now, all that is about the so-called complex field U (think of it as scalar components of either the electric or the magnetic filed in Maxwell). The light intensity is actually the modulus of it squared. That is phaseless quantity and interpreting the Huygens' principle through it is even more questionable. For acoustic waves however, what we measure is U itself (pressure).

EDIT: Bottom line: stick with the model of photons moving along lines for the purpose of this discussion.

Karl Persson said:

Andre Affleck said:

Ziggie said:

Great Bustard said:

Karl Persson said:

I felt this overlooked the effect of the Inverse Square Law on the cone of light the lens projects onto the camara’s image surface. In the case of the 50mm lens the image is projected twice the distance of the 25mm lens. The projection is subject to the Inverse Square Law that says that at double the distance the light intensity has dissipated to only ¼ of the original intensity, effectively nullifying any perceived advantage of the larger image area.

This is similar to the behavior of your old slide projector. The image got brighter and smaller as you moved the projector closer to the wall it was projected onto and dimmer and larger as the projector was moved further away from the wall. Are you not overlooking this?

Click to expand...

The ISL (Inverse Square Law) does not apply to focused light. Your projector analogy is a good one. It doesn't matter how far away the screen is so long as the screen captures all of the light coming from the projector.

Click to expand...

The Inverse Square Law does in fact apply to focused light as my projector example illustrates. If I move the projector further away from the wall the projected image becomes dimmer specifically due to the effects of Inverse Square Law.

Karl Persson said:

Karl Persson said:

2.

You responded by supplying a link to a section on your website as a means of explanation. That section, however, did not address this specific issue at all. The only reference to the Inverse Square Law was in relation to scene to camera distance, i.e. in front of the lens as opposed to behind the lens. You state:

“The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval.”

I find this statement a bit confusing as it seems to suggest that a person standing 10ft away from me should appear much brighter to my eyes than a person standing 20ft away which in reality is of course not the case. Can you clarify what you consider the role of the Inverse Square Law to be with regard to scene to camera distance?

Click to expand...

If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

Click to expand...

Are you saying that the scene is emitting light?

Karl Persson said:

However, if the light is focused, such as the light in the lens being projected on the sensor, then, as your projector analogy wonderfully demonstrates, the distance is irrelevant (so long as the sensor is large enough to capture all the light being projected on it).

Click to expand...

The distance is not irrelevant. It will affect the light intensity. With greater distance the image becomes dimmer unless compensated for by using a larger lens aperture.

Click to expand...

If what you say is true, then if I take a picture of a featureless, homogeneous white wall from 6 feet away and again from 12 feet away, using the same FL, SS, ISO, and aperture, the RGB values should be vastly different, but in fact they will be exactly the same. Give it a try.

Click to expand...

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

Click to expand...

Correct. As GB pointed out, I was talking in front if the lens and Ziggie (and you) are talking after the lens, so we are not talking about the same thing. However, in line with Ziggie's original post, your extension tube would be accompanied by a change in sensor size (larger), which would then render the same RGB value. So once again, even though the ISL is taking place, the drop in illuminance is accounted for by the area change.

antoineb said:

What will any answer change in the way you're taking photographs, or the type of system you use?

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Yes.

antoineb said:

In our day and age with sensors that are clean up to easily ISO 1600 even on m43 and much higher on FF, pretty much the only remaining important factor is how much DOF control we get. Medium format gets very very shallow DOF, FF is also often too shallow to handly, APS-C is reasonable, m43 is deeper.

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There is no such thing as too high an ISO.

FF can achieve any DOF that M43rds can--perhaps more.

FF or even MF doesn't have shallow DOF when taking photos of large objects (low magnification.)

Cell phones don't have enough DOF when doing some macros (high magnification.)

To know what to use, you have to understand the issues.

Great Bustard said:

The projected image does become dimmer per square millimeter on the surface it is projected on, but not dimmer per proportion of the photo. For example, one millionth of the projected photo would be made from the same amount of light no matter how far the screen was from the source.

Click to expand...

OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

Great Bustard said:

Great Bustard said:

Great Bustard said:

If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

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Are you saying that the scene is emitting light?

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Emitting and/or reflecting. Usually reflecting. Neither here nor there, really.

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This is a bit vague. I agree with usually reflecting, though. Where does the light usually originate from in those cases?

Andre Affleck said:

Ziggie said:

A 25mm lens will basically project a brighter image than a 50mm lens unless compensated for by using a larger aperture.

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Which it does by design in order to keep the exposure the same at a particular f stop regardless of FL. The physical aperture changes with FL in order to keep relative aperture (f-stop) the same.

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Yes of course. I agree fully.

Andre Affleck said:

Andre Affleck said:

This is why large format view cameras shooters have always had to figure in the bellows extension factor when shooting close-ups and using a hand-held light meter.

As the lens gets racked forward for close focusing it moves further away from the image plane and the image gets progressively darker requiring the photographer to adjust the exposure accordingly. While this also holds true for the smaller format SLR or mirrorless cameras most people don’t notice because the built-in light meter automatically compensates for this while the view cameras of course don’t have built-in meters.

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Not quite. As mentioned, the physical aperture increases with FL to maintain the same F-stop. So considering f2 on a 25mm lens, the physical aperture is smaller than f2 on a 50mm lens.

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I’m not referring to lenses of different focal length here. With any one lens the projected image on the camera's ground glass becomes progressively dimmer as you move the lens further away from the image plane for close focusing.

Andre Affleck said:

Andre Affleck said:

I have encountered this numerous times in real life scenarios when photographing with a view camera.

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Sure, but It's a different scenario for view cameras. They were more primative, and physical aperture was not linked to bellows travel.

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Not at all. All cameras function according to the same basic principles. And of course the the physical aperture is not linked to bellows travel. Not any more than the physical aperture is linked to the focusing mechanism of any other camera. Why would it?

Ziggie said:

Great Bustard said:

The projected image does become dimmer per square millimeter on the surface it is projected on, but not dimmer per proportion of the photo. For example, one millionth of the projected photo would be made from the same amount of light no matter how far the screen was from the source.

Click to expand...

OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

Click to expand...

For the sake of clarity can you explain precisely how you are applying the inverse square law? What is the origin of the distance measurement?

Is this for a point source imaged by the lens? Here the intensity varies approximately inversely as the square of the distance from the image plane, sufficiently far from the image. The approximation fails for distances less than the diameter of the Airy disk divided by the lens working numerical aperture - or the aberration-limited image diameter divided by NA if this is larger.

For an extended scene imaged by a small aperture lens, with aperture diameter much less than both image width and focal length, the intensity will decrease with distance from the lens exit pupil. For large distances, the relationship is approximately inverse square - as in your example of a projector illuminating a wall. Again, there is a near field regime where the inverse square relationship clearly fails. For a 5 mm diameter aperture, the intensity does not drop to 1/4 if the distance from the aperture is doubled from 1 mm to 2 mm.

Are you considering the inverse square law as it applies to a fixed optical configuration, or are you adjusting focus or even focal length between comparisons? Are you only interested in the image plane intensity for different focal lengths, but the same physical aperture diameter (exit pupil)? Alternatively for different focal length but the same relative aperture or F-number?

It is straightforward to calculate the result in each case, but you need to specify the question sufficiently precisely. As others have said, it may be easier to work with conservation of energy or conservation of étendue and luminance, rather than the inverse square law .

Ziggie said:

This is a bit vague. I agree with usually reflecting, though. Where does the light usually originate from in those cases?

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Andre Affleck said:

Ziggie said:

The distance is not irrelevant. It will affect the light intensity. With greater distance the image becomes dimmer unless compensated for by using a larger lens aperture.

Click to expand...

If what you say is true, then if I take a picture of a featureless, homogeneous white wall from 6 feet away and again from 12 feet away, using the same FL, SS, ISO, and aperture, the RGB values should be vastly different, but in fact they will be exactly the same. Give it a try.

Click to expand...

You seem to misunderstand what I am saying here. I am referring to the distance behind the lens not in front of it. The distance in front of the lens is irrelevant as you correctly state. The distance behind the lens varies with the focal length of the lens. A 100mm lens will project a sharp image at twice the distance of a 50mm lens.

Ziggie said:

OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

Click to expand...

It's a simple geometric optics ray tube - there's no mystery involved; it is all perfectly clearly defined. What's the point of repeating the same question over?

Joe

Karl Persson said:

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

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Extension tubes move the lens further away from the camera’s image plane making the projected image dimmer.

Anyone can easily verify that with a simple experiment:

Stick your camera on a tripod and point it at a uniform and evenly lit surface. A featureless, homogeneous wall will do fine. In manual exposure mode adjust the exposure so that the light meter indicates correct exposure. Make sure your settings are fixed (no auto ISO or auto anything). Now stick some extension tubes between the lens and camera body. The meter will indicate a drop in light intensity corresponding to the increased distance between lens and image plane.

Ziggie said:

Karl Persson said:

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

Click to expand...

Extension tubes move the lens further away from the camera’s image plane making the projected image dimmer.

Anyone can easily verify that with a simple experiment:

Click to expand...

Exactly, that was my point.

And this leads to the need for some clarification regarding "equivalence":

one has to account for the difference between, in theory, a FF DSLR and a FF mirrorless, when used with lenses optimised for respective back focus. And whether the eq. theory is correct when you use, say, a FF zoom lens on both FF and mFT (with adapter) bodies or only when you use a mFT lens on the latter; it can´t be both.

Andre Affleck said:

Karl Persson said:

Andre Affleck said:

Ziggie said:

Great Bustard said:

Andre Affleck said:

I felt this overlooked the effect of the Inverse Square Law on the cone of light the lens projects onto the camara’s image surface. In the case of the 50mm lens the image is projected twice the distance of the 25mm lens. The projection is subject to the Inverse Square Law that says that at double the distance the light intensity has dissipated to only ¼ of the original intensity, effectively nullifying any perceived advantage of the larger image area.

This is similar to the behavior of your old slide projector. The image got brighter and smaller as you moved the projector closer to the wall it was projected onto and dimmer and larger as the projector was moved further away from the wall. Are you not overlooking this?

Click to expand...

The ISL (Inverse Square Law) does not apply to focused light. Your projector analogy is a good one. It doesn't matter how far away the screen is so long as the screen captures all of the light coming from the projector.

Click to expand...

The Inverse Square Law does in fact apply to focused light as my projector example illustrates. If I move the projector further away from the wall the projected image becomes dimmer specifically due to the effects of Inverse Square Law.

Andre Affleck said:

Andre Affleck said:

2.

You responded by supplying a link to a section on your website as a means of explanation. That section, however, did not address this specific issue at all. The only reference to the Inverse Square Law was in relation to scene to camera distance, i.e. in front of the lens as opposed to behind the lens. You state:

“The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval.”

I find this statement a bit confusing as it seems to suggest that a person standing 10ft away from me should appear much brighter to my eyes than a person standing 20ft away which in reality is of course not the case. Can you clarify what you consider the role of the Inverse Square Law to be with regard to scene to camera distance?

Click to expand...

If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

Click to expand...

Are you saying that the scene is emitting light?

Andre Affleck said:

However, if the light is focused, such as the light in the lens being projected on the sensor, then, as your projector analogy wonderfully demonstrates, the distance is irrelevant (so long as the sensor is large enough to capture all the light being projected on it).

Click to expand...

The distance is not irrelevant. It will affect the light intensity. With greater distance the image becomes dimmer unless compensated for by using a larger lens aperture.

Click to expand...

If what you say is true, then if I take a picture of a featureless, homogeneous white wall from 6 feet away and again from 12 feet away, using the same FL, SS, ISO, and aperture, the RGB values should be vastly different, but in fact they will be exactly the same. Give it a try.

Click to expand...

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

Click to expand...

Correct. As GB pointed out, I was talking in front if the lens and Ziggie (and you) are talking after the lens, so we are not talking about the same thing. However, in line with Ziggie's original post, your extension tube would be accompanied by a change in sensor size (larger), which would then render the same RGB value.

Click to expand...

?

The RGB values (per pixel) won´t change because you use a larger sensor.
Sensors sample the intensity at each receptor, not the total light.

Let´s make it simple: with a quite common 105/2,8 FF macro lens you will loose 2 stops going from infinity to MFD, despite the fact that such modern lenses will severely affect the focal length (to shorter) as you close the distance.

It still wouldn´t fit a MF camera

alanr0 said:

Ziggie said:

Great Bustard said:

The projected image does become dimmer per square millimeter on the surface it is projected on, but not dimmer per proportion of the photo. For example, one millionth of the projected photo would be made from the same amount of light no matter how far the screen was from the source.

Click to expand...

OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

Click to expand...

For the sake of clarity can you explain precisely how you are applying the inverse square law? What is the origin of the distance measurement?

Is this for a point source imaged by the lens? Here the intensity varies approximately inversely as the square of the distance from the image plane, sufficiently far from the image. The approximation fails for distances less than the diameter of the Airy disk divided by the lens working numerical aperture - or the aberration-limited image diameter divided by NA if this is larger.

For an extended scene imaged by a small aperture lens, with aperture diameter much less than both image width and focal length, the intensity will decrease with distance from the lens exit pupil. For large distances, the relationship is approximately inverse square - as in your example of a projector illuminating a wall. Again, there is a near field regime where the inverse square relationship clearly fails. For a 5 mm diameter aperture, the intensity does not drop to 1/4 if the distance from the aperture is doubled from 1 mm to 2 mm.

Click to expand...

If he means as the inverse square of the distance to the exit pupil (as a plane), when you approach it, the intensity would be infinite there along the whole pupil, and that is a lot of light. Our noise problems would be solved then.

Ziggie said:

Karl Persson said:

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

Click to expand...

Extension tubes move the lens further away from the camera’s image plane making the projected image dimmer.

Click to expand...

Actually, they make it blurrier (front focused).

Karl Persson said:

Andre Affleck said:

Karl Persson said:

Andre Affleck said:

Ziggie said:

Great Bustard said:

Karl Persson said:

I felt this overlooked the effect of the Inverse Square Law on the cone of light the lens projects onto the camara’s image surface. In the case of the 50mm lens the image is projected twice the distance of the 25mm lens. The projection is subject to the Inverse Square Law that says that at double the distance the light intensity has dissipated to only ¼ of the original intensity, effectively nullifying any perceived advantage of the larger image area.

This is similar to the behavior of your old slide projector. The image got brighter and smaller as you moved the projector closer to the wall it was projected onto and dimmer and larger as the projector was moved further away from the wall. Are you not overlooking this?

Click to expand...

The ISL (Inverse Square Law) does not apply to focused light. Your projector analogy is a good one. It doesn't matter how far away the screen is so long as the screen captures all of the light coming from the projector.

Click to expand...

The Inverse Square Law does in fact apply to focused light as my projector example illustrates. If I move the projector further away from the wall the projected image becomes dimmer specifically due to the effects of Inverse Square Law.

Karl Persson said:

Karl Persson said:

2.

You responded by supplying a link to a section on your website as a means of explanation. That section, however, did not address this specific issue at all. The only reference to the Inverse Square Law was in relation to scene to camera distance, i.e. in front of the lens as opposed to behind the lens. You state:

“The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval.”

I find this statement a bit confusing as it seems to suggest that a person standing 10ft away from me should appear much brighter to my eyes than a person standing 20ft away which in reality is of course not the case. Can you clarify what you consider the role of the Inverse Square Law to be with regard to scene to camera distance?

Click to expand...

If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

Click to expand...

Are you saying that the scene is emitting light?

Karl Persson said:

However, if the light is focused, such as the light in the lens being projected on the sensor, then, as your projector analogy wonderfully demonstrates, the distance is irrelevant (so long as the sensor is large enough to capture all the light being projected on it).

Click to expand...

The distance is not irrelevant. It will affect the light intensity. With greater distance the image becomes dimmer unless compensated for by using a larger lens aperture.

Click to expand...

If what you say is true, then if I take a picture of a featureless, homogeneous white wall from 6 feet away and again from 12 feet away, using the same FL, SS, ISO, and aperture, the RGB values should be vastly different, but in fact they will be exactly the same. Give it a try.

Click to expand...

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

Click to expand...

Correct. As GB pointed out, I was talking in front if the lens and Ziggie (and you) are talking after the lens, so we are not talking about the same thing. However, in line with Ziggie's original post, your extension tube would be accompanied by a change in sensor size (larger), which would then render the same RGB value.

Click to expand...

?

The RGB values (per pixel) won´t change because you use a larger sensor.
Sensors sample the intensity at each receptor, not the total light.

Let´s make it simple: with a quite common 105/2,8 FF macro lens you will loose 2 stops going from infinity to MFD, despite the fact that such modern lenses will severely affect the focal length (to shorter) as you close the distance.

It still wouldn´t fit a MF camera

Click to expand...

RGB values are derived from normalized light intensity. Let's take a simple example to illustrate what is happening.

Let's say 1000 photons exit the back of a lens towards sensor. As the light moves towards the sensor it expands, and therefore light intensity drops (as Ziggie correctly stipulates). Let's assume we have a 10 x 10 pixel sensor that captures all of the 1000 photons, so 10 photons are captured per pixel, and 10 photons represents an RGB value of 255 (white). Now move that same sensor back so that only half of the light is captured by that sensor. Each pixel is now receiving only five photons so the RGB value drops by 1/2. Let's instead replace that sensor with one that is twice as large, but with the same 10 x 10 pixel resolution. This time, all of the light is captured at that further distance, and each pixel is again receiving 10 photons representing the same RGB value of 255. Notice however that the light intensity at the sensor has dropped by one half, but normalization has maintain the same RGB value.

Andre Affleck said:

Karl Persson said:

Andre Affleck said:

Karl Persson said:

Andre Affleck said:

Ziggie said:

Great Bustard said:

Andre Affleck said:

I felt this overlooked the effect of the Inverse Square Law on the cone of light the lens projects onto the camara’s image surface. In the case of the 50mm lens the image is projected twice the distance of the 25mm lens. The projection is subject to the Inverse Square Law that says that at double the distance the light intensity has dissipated to only ¼ of the original intensity, effectively nullifying any perceived advantage of the larger image area.

This is similar to the behavior of your old slide projector. The image got brighter and smaller as you moved the projector closer to the wall it was projected onto and dimmer and larger as the projector was moved further away from the wall. Are you not overlooking this?

Click to expand...

The ISL (Inverse Square Law) does not apply to focused light. Your projector analogy is a good one. It doesn't matter how far away the screen is so long as the screen captures all of the light coming from the projector.

Click to expand...

The Inverse Square Law does in fact apply to focused light as my projector example illustrates. If I move the projector further away from the wall the projected image becomes dimmer specifically due to the effects of Inverse Square Law.

Andre Affleck said:

Andre Affleck said:

2.

You responded by supplying a link to a section on your website as a means of explanation. That section, however, did not address this specific issue at all. The only reference to the Inverse Square Law was in relation to scene to camera distance, i.e. in front of the lens as opposed to behind the lens. You state:

“The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval.”

I find this statement a bit confusing as it seems to suggest that a person standing 10ft away from me should appear much brighter to my eyes than a person standing 20ft away which in reality is of course not the case. Can you clarify what you consider the role of the Inverse Square Law to be with regard to scene to camera distance?

Click to expand...

If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

Click to expand...

Are you saying that the scene is emitting light?

Andre Affleck said:

However, if the light is focused, such as the light in the lens being projected on the sensor, then, as your projector analogy wonderfully demonstrates, the distance is irrelevant (so long as the sensor is large enough to capture all the light being projected on it).

Click to expand...

The distance is not irrelevant. It will affect the light intensity. With greater distance the image becomes dimmer unless compensated for by using a larger lens aperture.

Click to expand...

If what you say is true, then if I take a picture of a featureless, homogeneous white wall from 6 feet away and again from 12 feet away, using the same FL, SS, ISO, and aperture, the RGB values should be vastly different, but in fact they will be exactly the same. Give it a try.

Click to expand...

And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

Click to expand...

Correct. As GB pointed out, I was talking in front if the lens and Ziggie (and you) are talking after the lens, so we are not talking about the same thing. However, in line with Ziggie's original post, your extension tube would be accompanied by a change in sensor size (larger), which would then render the same RGB value.

Click to expand...

?

The RGB values (per pixel) won´t change because you use a larger sensor.
Sensors sample the intensity at each receptor, not the total light.

Let´s make it simple: with a quite common 105/2,8 FF macro lens you will loose 2 stops going from infinity to MFD, despite the fact that such modern lenses will severely affect the focal length (to shorter) as you close the distance.

It still wouldn´t fit a MF camera

Click to expand...

RGB values are derived from normalized light intensity. Let's take a simple example to illustrate what is happening.

Let's say 1000 photons exit the back of a lens towards sensor. As the light moves towards the sensor it expands, and therefore light intensity drops (as Ziggie correctly stipulates). Let's assume we have a 10 x 10 pixel sensor that captures all of the 1000 photons, so 10 photons are captured per pixel, and 10 photons represents an RGB value of 255 (white). Now move that same sensor back so that only half of the light is captured by that sensor. Each pixel is now receiving only five photons so the RGB value drops by 1/2. Let's instead replace that sensor with one that is twice as large, but with the same 10 x 10 pixel resolution. This time, all of the light is captured at that further distance, and each pixel is again receiving 10 photons representing the same RGB value of 255. Notice however that the light intensity at the sensor has dropped by one half, but normalization has maintain the same RGB value.

Click to expand...

A 4 liter canister holds more water than a 1 liter canister.
Very enlightening.....

:~|

Karl Persson said:

Andre Affleck said:

Karl Persson said:

Andre Affleck said:

Karl Persson said:

Andre Affleck said:

Ziggie said:

Great Bustard said:

Karl Persson said:

I felt this overlooked the effect of the Inverse Square Law on the cone of light the lens projects onto the camara’s image surface. In the case of the 50mm lens the image is projected twice the distance of the 25mm lens. The projection is subject to the Inverse Square Law that says that at double the distance the light intensity has dissipated to only ¼ of the original intensity, effectively nullifying any perceived advantage of the larger image area.

This is similar to the behavior of your old slide projector. The image got brighter and smaller as you moved the projector closer to the wall it was projected onto and dimmer and larger as the projector was moved further away from the wall. Are you not overlooking this?

Click to expand...

The ISL (Inverse Square Law) does not apply to focused light. Your projector analogy is a good one. It doesn't matter how far away the screen is so long as the screen captures all of the light coming from the projector.

Click to expand...

The Inverse Square Law does in fact apply to focused light as my projector example illustrates. If I move the projector further away from the wall the projected image becomes dimmer specifically due to the effects of Inverse Square Law.

Karl Persson said:

Karl Persson said:

2.

You responded by supplying a link to a section on your website as a means of explanation. That section, however, did not address this specific issue at all. The only reference to the Inverse Square Law was in relation to scene to camera distance, i.e. in front of the lens as opposed to behind the lens. You state:

“The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval.”

I find this statement a bit confusing as it seems to suggest that a person standing 10ft away from me should appear much brighter to my eyes than a person standing 20ft away which in reality is of course not the case. Can you clarify what you consider the role of the Inverse Square Law to be with regard to scene to camera distance?

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If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

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Are you saying that the scene is emitting light?

Karl Persson said:

However, if the light is focused, such as the light in the lens being projected on the sensor, then, as your projector analogy wonderfully demonstrates, the distance is irrelevant (so long as the sensor is large enough to capture all the light being projected on it).

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The distance is not irrelevant. It will affect the light intensity. With greater distance the image becomes dimmer unless compensated for by using a larger lens aperture.

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If what you say is true, then if I take a picture of a featureless, homogeneous white wall from 6 feet away and again from 12 feet away, using the same FL, SS, ISO, and aperture, the RGB values should be vastly different, but in fact they will be exactly the same. Give it a try.

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And if you put an extension tube between lens and body, i.e. putting the lens further from the sensor (using same SS and aperture).
What happens to your RGB values, still the same?No.

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Correct. As GB pointed out, I was talking in front if the lens and Ziggie (and you) are talking after the lens, so we are not talking about the same thing. However, in line with Ziggie's original post, your extension tube would be accompanied by a change in sensor size (larger), which would then render the same RGB value.

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?

The RGB values (per pixel) won´t change because you use a larger sensor.
Sensors sample the intensity at each receptor, not the total light.

Let´s make it simple: with a quite common 105/2,8 FF macro lens you will loose 2 stops going from infinity to MFD, despite the fact that such modern lenses will severely affect the focal length (to shorter) as you close the distance.

It still wouldn´t fit a MF camera

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RGB values are derived from normalized light intensity. Let's take a simple example to illustrate what is happening.

Let's say 1000 photons exit the back of a lens towards sensor. As the light moves towards the sensor it expands, and therefore light intensity drops (as Ziggie correctly stipulates). Let's assume we have a 10 x 10 pixel sensor that captures all of the 1000 photons, so 10 photons are captured per pixel, and 10 photons represents an RGB value of 255 (white). Now move that same sensor back so that only half of the light is captured by that sensor. Each pixel is now receiving only five photons so the RGB value drops by 1/2. Let's instead replace that sensor with one that is twice as large, but with the same 10 x 10 pixel resolution. This time, all of the light is captured at that further distance, and each pixel is again receiving 10 photons representing the same RGB value of 255. Notice however that the light intensity at the sensor has dropped by one half, but normalization has maintain the same RGB value.

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A 4 liter canister holds more water than a 1 liter canister.
Very enlightening.....

:~|

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And yet after normalization, both are equally as full....

alanr0 said:

Ziggie said:

Great Bustard said:

The projected image does become dimmer per square millimeter on the surface it is projected on, but not dimmer per proportion of the photo. For example, one millionth of the projected photo would be made from the same amount of light no matter how far the screen was from the source.

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OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

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For the sake of clarity can you explain precisely how you are applying the inverse square law? What is the origin of the distance measurement?

Is this for a point source imaged by the lens? Here the intensity varies approximately inversely as the square of the distance from the image plane, sufficiently far from the image. The approximation fails for distances less than the diameter of the Airy disk divided by the lens working numerical aperture - or the aberration-limited image diameter divided by NA if this is larger.

For an extended scene imaged by a small aperture lens, with aperture diameter much less than both image width and focal length, the intensity will decrease with distance from the lens exit pupil. For large distances, the relationship is approximately inverse square - as in your example of a projector illuminating a wall. Again, there is a near field regime where the inverse square relationship clearly fails. For a 5 mm diameter aperture, the intensity does not drop to 1/4 if the distance from the aperture is doubled from 1 mm to 2 mm.

Are you considering the inverse square law as it applies to a fixed optical configuration, or are you adjusting focus or even focal length between comparisons? Are you only interested in the image plane intensity for different focal lengths, but the same physical aperture diameter (exit pupil)? Alternatively for different focal length but the same relative aperture or F-number?

It is straightforward to calculate the result in each case, but you need to specify the question sufficiently precisely. As others have said, it may be easier to work with conservation of energy or conservation of étendue and luminance, rather than the inverse square law .

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Thank you for saying that. It strikes me that a lot of these discussions would be greatly simplified by simply talking about the etendue or luminance. I'm not enough of an expert to know if that deals with all of the issues raised by JACS, especially in the diffraction limit.

But at the limit of my understanding, in for the case of lens with perfect refractions, then these conservation principles tell us really only need to think about what happen on the subject side of the lens. Isn't that correct?

alanr0 said:

alanr0 said:

This is a bit vague. I agree with usually reflecting, though. Where does the light usually originate from in those cases?

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Jeff said:

alanr0 said:

Ziggie said:

Great Bustard said:

The projected image does become dimmer per square millimeter on the surface it is projected on, but not dimmer per proportion of the photo. For example, one millionth of the projected photo would be made from the same amount of light no matter how far the screen was from the source.

Click to expand...

OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

Click to expand...

For the sake of clarity can you explain precisely how you are applying the inverse square law? What is the origin of the distance measurement?

Is this for a point source imaged by the lens? Here the intensity varies approximately inversely as the square of the distance from the image plane, sufficiently far from the image. The approximation fails for distances less than the diameter of the Airy disk divided by the lens working numerical aperture - or the aberration-limited image diameter divided by NA if this is larger.

For an extended scene imaged by a small aperture lens, with aperture diameter much less than both image width and focal length, the intensity will decrease with distance from the lens exit pupil. For large distances, the relationship is approximately inverse square - as in your example of a projector illuminating a wall. Again, there is a near field regime where the inverse square relationship clearly fails. For a 5 mm diameter aperture, the intensity does not drop to 1/4 if the distance from the aperture is doubled from 1 mm to 2 mm.

Are you considering the inverse square law as it applies to a fixed optical configuration, or are you adjusting focus or even focal length between comparisons? Are you only interested in the image plane intensity for different focal lengths, but the same physical aperture diameter (exit pupil)? Alternatively for different focal length but the same relative aperture or F-number?

It is straightforward to calculate the result in each case, but you need to specify the question sufficiently precisely. As others have said, it may be easier to work with conservation of energy or conservation of étendue and luminance, rather than the inverse square law .

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Thank you for saying that. It strikes me that a lot of these discussions would be greatly simplified by simply talking about the etendue or luminance. I'm not enough of an expert to know if that deals with all of the issues raised by JACS, especially in the diffraction limit.

But at the limit of my understanding, in for the case of lens with perfect refractions, then these conservation principles tell us really only need to think about what happen on the subject side of the lens. Isn't that correct?

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Jeff -- that's a perfectly good place to start.

For a lens with a fixed physical aperture (entrance pupil):

Incident intensity (illuminance) at entrance pupil = source luminance x solid angle subtended by source x cosine(angle of incidence at lens) x aperture area

For a uniform subject and a sufficiently narrow field of view, cos(angle of incidence) is approximately unity, and this becomes a simple multiplication. Otherwise we integrate over the field of view and over the lens entrance pupil.

In any case, for a fixed subject, aperture and field of view, the total incident power is fixed. If there are no losses, all the light is projected onto the image plane, and from simple conservation of energy the intensity at the sensor is inversely proportional to the square of the linear magnification, and so inversely proportional to the distance from lens rear principal point to the sensor.

This appears similar to an inverse square dependence, and seems to risk JACS' infinite intensity singularity as the focal length approaches zero. The problem lies in the assumption of fixed aperture and no losses. In practice a properly corrected lens satisfies the Abbe sine condition, and has a spherical principal surface centred on the focus. A consequence is that the radius of the entrance pupil cannot be larger than the focal length of the lens, so we can't maintain the assumption of fixed aperture and no change in total light captured for the shortest focal lengths.

Another consequence (or independent constraint) is that the numerical aperture must be less than unity for a lens in air (corresponding to working f-number > 0.5).

Diffraction limits the spot size (proportional to wavelength/NA), so avoids infinite intensity at the sensor, even for a point source in the field of view.

HTH

Jeff said:

But at the limit of my understanding, in for the case of lens with perfect refractions, then these conservation principles tell us really only need to think about what happen on the subject side of the lens. Isn't that correct?

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This is a very good way to think about it. Assuming that the lens does what it is supposed to do, you can just look at the light falling on it (to be more precise, on a single element lens with the same aperture). No need to know what is behind the lens or even the FL; except, well, the AOV which is determined by those.

Ziggie said:

Great Bustard said:

The projected image does become dimmer per square millimeter on the surface it is projected on, but not dimmer per proportion of the photo. For example, one millionth of the projected photo would be made from the same amount of light no matter how far the screen was from the source.

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OK, let me rephrase my original question just for the sake of clarity:

Do you think the Inverse Square Law applies to the projected cone of light the lens projects onto the camera’s image plane?

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Not with regards to the amount of light falling on the sensor. Specifically, if you moved the sensor closer to the exit pupil, no more light from the scene would fall on it (light from outside the scene would fall on it, but not more light from the scene). Likewise, if you moved the sensor further back, the size of the sensor would be adjusted so that all light from the scene would still fall on it. Either way, the ISL is neither here nor there with regards to how much light from the scene falls on the sensor.

Ziggie said:

Ziggie said:

Ziggie said:

Ziggie said:

If the light from the scene is not focused, thus, the ISL applies. For example, if you are twice as far from the scene, the intensity of the light reaching you is 1/4 as great.

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Are you saying that the scene is emitting light?

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Emitting and/or reflecting. Usually reflecting. Neither here nor there, really.

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This is a bit vague. I agree with usually reflecting, though. Where does the light usually originate from in those cases?

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Not sure what you're asking. If I'm taking a landscape photo, for example, light emitted by the sun is reflected by the scene.