A couple of questions for Great Bustard…

In a couple of threads in the Micro 4/3 forum last week I asked for an explanation of some issues being discussed. GB did not have an opportunity to fully answer before the threads filled but suggested I ask him here:

1.

In the threads you stated that the FF image area receives four times the light of an mFT image area due to the fact that it has four times the surface area. This is assuming lenses of the same FOV (50mm for FF and 25mm for mFT) and identical physical aperture diameter and shutter speed in both cases.

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?

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?

Ziggie said:

In a couple of threads in the Micro 4/3 forum last week I asked for an explanation of some issues being discussed. GB did not have an opportunity to fully answer before the threads filled but suggested I ask him here:

1.

In the threads you stated that the FF image area receives four times the light of an mFT image area due to the fact that it has four times the surface area. This is assuming lenses of the same FOV (50mm for FF and 25mm for mFT) and identical physical aperture diameter and shutter speed in both cases.

Click to expand...

Identical f-number, not identical physical aperture diameter - read the thread more carefully.

Ziggie 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?

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.”

Click to expand...

Of course. Someone shines a flashlight at you from 10 metres, producing a beam of a certain diameter at your location. Now they move back to 20 metres. The beam diameter is doubled, so the optical power is spread out over four times the area, hence the intensity in Watts per square metre is reduced by a factor of four. It would be a very odd world if that wasn't true.

Ziggie said:

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...

Think about two people holding 1 m^2 reflectors on a sunny day standing at different distances from you...

Joe

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

In the case of the distance from the front of the lens to the sensor, even if the light is from a point source, it is not that small distance but rather the distance to the source that obeys the inverse square law.

A flash is approximately a point source, so it obeys the inverse square law.

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

Click to expand...

In this example, your spotmeter is accepting light from different areas of the object when you take readings from close and far. Put another way, you are collecting from more points when you measure from further away.

olliess said:

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

Click to expand...

In this example, your spotmeter is accepting light from different areas of the object when you take readings from close and far. Put another way, you are collecting from more points when you measure from further away.

Click to expand...

If what you say is true, why is it that when I take a reading close of a middle tone (Zone V) and then use that to set the exposure on a camera well away from the object, I get the "correct" exposure? This is a technique used since the early days of film.

The point source for an object reflecting light in sunlight is the sun. The delta in distance from the point source when I move from the object to the camera is so small that it would make so slight a difference in light level that you could not measure it.

edhannon said:

olliess said:

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

Click to expand...

In this example, your spotmeter is accepting light from different areas of the object when you take readings from close and far. Put another way, you are collecting from more points when you measure from further away.

Click to expand...

If what you say is true, why is it that when I take a reading close of a middle tone (Zone V) and then use that to set the exposure on a camera well away from the object, I get the "correct" exposure? This is a technique used since the early days of film.

Click to expand...

I am not sure I'm understanding your question. Are you asking why your spotmeter gives the correct reading from afar?

edhannon said:

The point source for an object reflecting light in sunlight is the sun. The delta in distance from the point source when I move from the object to the camera is so small that it would make so slight a difference in light level that you could not measure it.

Click to expand...

For general objects, the distance from the sun is already accounted for in the intensity of light illuminating the object.

edhannon said:

--
Ed Cannon
http://www.pbase.com/edhannon

Click to expand...

olliess said:

edhannon said:

olliess said:

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

Click to expand...

In this example, your spotmeter is accepting light from different areas of the object when you take readings from close and far. Put another way, you are collecting from more points when you measure from further away.

Click to expand...

If what you say is true, why is it that when I take a reading close of a middle tone (Zone V) and then use that to set the exposure on a camera well away from the object, I get the "correct" exposure? This is a technique used since the early days of film.

Click to expand...

I am not sure I'm understanding your question. Are you asking why your spotmeter gives the correct reading from afar?

Click to expand...

Yes. To give a specific example (one used by portrait photographers from the advent of reflecting meters), if I stand next to a model and measure her the skin on her face and then place that reading on Zone VI ( if Caucasian) I will get the same brightness level of her face no mater how far back the camera is and what focal length lens I use. If the inverse square law applied to reflected/scattered light that technique would not work. It would underexpose.

olliess said:

olliess said:

The point source for an object reflecting light in sunlight is the sun. The delta in distance from the point source when I move from the object to the camera is so small that it would make so slight a difference in light level that you could not measure it.

Click to expand...

For general objects, the distance from the sun is already accounted for in the intensity of light illuminating the object.

Click to expand...

(edit typo)

If the light illuminating the object is reflected then the point source is not the reflection but the original source.

--
Ed Hannon
http://www.pbase.com/edhannon

edhannon said:

olliess said:

edhannon said:

olliess said:

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

Click to expand...

In this example, your spotmeter is accepting light from different areas of the object when you take readings from close and far. Put another way, you are collecting from more points when you measure from further away.

Click to expand...

If what you say is true, why is it that when I take a reading close of a middle tone (Zone V) and then use that to set the exposure on a camera well away from the object, I get the "correct" exposure? This is a technique used since the early days of film.

Click to expand...

I am not sure I'm understanding your question. Are you asking why your spotmeter gives the correct reading from afar?

Click to expand...

Yes. To give a specific example (one used by portrait photographers from the advent of reflecting meters), if I stand next to a model and measure her the skin on her face and then place that reading on Zone VI ( if Caucasian) I will get the same brightness level of her face no mater how far back the camera is and what focal length lens I use. If the inverse square law applied to reflected/scattered light that technique would not work. It would underexpose.

Click to expand...

So then it's what I said in my first post. If you spotmeter from a greater distance, then your 1deg spot measures more points or, in this case, more area of the model's face.

In other words , the inverse square law always "applies," but you don't apply it in your calculation because the metered area conveniently increases by a square law.

edhannon said:

edhannon said:

edhannon said:

The point source for an object reflecting light in sunlight is the sun. The delta in distance from the point source when I move from the object to the camera is so small that it would make so slight a difference in light level that you could not measure it.

Click to expand...

For general objects, the distance from the sun is already accounted for in the intensity of light illuminating the object.

Click to expand...

The light illuminating the object is reflected then the point source is not the reflection but the original source.

Click to expand...

As I said, the falloff from the original point source is incorporated into the illumination on the object.

Here's an interesting thought: take the example of a tiny matte white bead or spot, illuminated by the sun, that fits entirely inside your spotmeter circle when you meter it. For simplicity also assume the background is black. Now back up to 2x the distance. Would an inverse square law apply then?

edhannon said:

--
Ed Hannon
http://www.pbase.com/edhannon

Click to expand...

olliess said:

So then it's what I said in my first post. If you spotmeter from a greater distance, then your 1deg spot measures more points or, in this case, more area of the model's face.

In other words , the inverse square law always "applies," but you don't apply it in your calculation because the metered area conveniently increases by a square law.

Click to expand...

However, the exposure I set from the up close meter reading gives the same brightness of the face no matter how far back I move the camera. If the light from the model obeyed the inverse square law of light, then as I moved back, the light reflected from her face would decrease with the inverse square law and her face would get darker as I moved the camera back.

olliess said:

As I said, the falloff from the original point source is incorporated into the illumination on the object.

Click to expand...

If I understand what you are saying correctly, reflected light behaves as though it were being emitted from a point source. But it does not. The inverse square law of light works because, as the wave front from the point source expands outward from the source, the number of photons per unit area decreases proportional to the increase in area - or proportional to the square of the distance.

olliess said:

Here's an interesting thought: take the example of a tiny matte white bead or spot, illuminated by the sun, that fits entirely inside your spotmeter circle when you meter it. For simplicity also assume the background is black. Now back up to 2x the distance. Would an inverse square law apply then?

Click to expand...

No. If I measure the spot and set the exposure based on that and move the camera back as above, the brightness of the spot will not change in the image. It will just get smaller. If it were obeying the inverse square law it would get darker as I moved the camera back.

Ziggie said:

[...]

Click to expand...

Ziggie said:

In the threads you stated that the FF image area receives four times the light of an mFT image area due to the fact that it has four times the surface area. This is assuming lenses of the same FOV (50mm for FF and 25mm for mFT) and identical physical aperture diameter and shutter speed in both cases.

Click to expand...

No, when dealing with equivalence we are talking about identical exposure; that means, not "physical aperture" but "the same f/ value". for the same shutter interval you referred to.

Ziggie 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.

[...]

Click to expand...

But, in the case at hand, all the sensor elements receive the same light (otherwise they would not be equally well exposed). But the total light in the case of the 4x bigger sensor will be 4x greater.

The best way to resolve which model matches reality better is to devise an experiment where there is measurable differences in the predictions of the models. Then run the experiment.

What I propose is setting up two objects.

First a small light (without reflector) at night. Then standing 10 feet away from it, measure the light and set it for mid tone. Take an image. Then take another image 20 feet away at the same camera settings. I believe that both models will predict that the brightness will decrease by 2 stops.

Second, place a small gray patch (same size as bulb of light) and take images at same distances. I believe that your model predicts that the brightness of the patch will decrease by the same two stops. My model says that there will be no change in brightness of the patch.

Regarding the inverse square law, it's not applicable inside a lens, where the all photons are focused to fall on the sensor. In this case, it's similar to a laser beam, collimated.

Pedagydusz said:

Ziggie said:

[...]

In the threads you stated that the FF image area receives four times the light of an mFT image area due to the fact that it has four times the surface area. This is assuming lenses of the same FOV (50mm for FF and 25mm for mFT) and identical physical aperture diameter and shutter speed in both cases.

Click to expand...

No, when dealing with equivalence we are talking about identical exposure; that means, not "physical aperture" but "the same f/ value". for the same shutter interval you referred to.

Click to expand...

If Ziggie has been talking to Great Bustard about equivalence, the meaning is not "same light per unit area of sensor".

GB's definition of equivalence between images taken with different sensors requires:

• Same shutter speed
• Same diagonal field of view
• Same perspective (i.e. taken from the same location)
• Same depth of field and diffraction blur
• Same total light on the sensor

This requires same physical aperture, rather than same f-stop.

You will find "1/125 s at f/8" and "1/500 at f/4" described as "equivalent exposures". I don't believe that is the intent here.

edhannon said:

What I propose is setting up two objects.

First a small light (without reflector) at night. Then standing 10 feet away from it, measure the light and set it for mid tone. Take an image. Then take another image 20 feet away...

Second, place a small gray patch (same size as bulb of light) and take images at same distances. I believe that your model predicts that the brightness of the patch will decrease by the same two stops. My model says that there will be no change in brightness of the patch.

Click to expand...

That is not what I said above. The key point is whether the measurement circle covers the entire object, regardless of whether it is a source or a reflector.

If so, and the object shrinks in apparent size but maintains the same intensity, then is the spot meter going to measure a smaller 1 degree averaged intensity for the smaller-appearing (further) object?

Ziggie said:

In a couple of threads in the Micro 4/3 forum last week I asked for an explanation of some issues being discussed. GB did not have an opportunity to fully answer before the threads filled but suggested I ask him here:

1.

In the threads you stated that the FF image area receives four times the light of an mFT image area due to the fact that it has four times the surface area. This is assuming lenses of the same FOV (50mm for FF and 25mm for mFT) and identical physical aperture diameter and shutter speed in both cases.

Click to expand...

Not the "identical physical aperture" but the same relative aperture. Please see this bit on the various uses of the term "aperture".

Ziggie 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.

Ziggie 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. 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).

olliess said:

edhannon said:

What I propose is setting up two objects.

First a small light (without reflector) at night. Then standing 10 feet away from it, measure the light and set it for mid tone. Take an image. Then take another image 20 feet away...

Second, place a small gray patch (same size as bulb of light) and take images at same distances. I believe that your model predicts that the brightness of the patch will decrease by the same two stops. My model says that there will be no change in brightness of the patch.

Click to expand...

That is not what I said above. The key point is whether the measurement circle covers the entire object, regardless of whether it is a source or a reflector.

If so, and the object shrinks in apparent size but maintains the same intensity, then is the spot meter going to measure a smaller 1 degree averaged intensity for the smaller-appearing (further) object?

Click to expand...

But if the light reflected from the patch obeys the inverse square law like the flashlight, then the light decreases proportional to the square of the distance and it will get darker in an image exposed with the same settings.

If it does not get darker when we double the distance from the object to the camera, then the object does not obey the inverse square law measured from the object to the camera.

edhannon said:

olliess said:

edhannon said:

What I propose is setting up two objects.

First a small light (without reflector) at night. Then standing 10 feet away from it, measure the light and set it for mid tone. Take an image. Then take another image 20 feet away...

Second, place a small gray patch (same size as bulb of light) and take images at same distances. I believe that your model predicts that the brightness of the patch will decrease by the same two stops. My model says that there will be no change in brightness of the patch.

Click to expand...

That is not what I said above. The key point is whether the measurement circle covers the entire object, regardless of whether it is a source or a reflector.

If so, and the object shrinks in apparent size but maintains the same intensity, then is the spot meter going to measure a smaller 1 degree averaged intensity for the smaller-appearing (further) object?

Click to expand...

But if the light reflected from the patch obeys the inverse square law like the flashlight, then the light decreases proportional to the square of the distance and it will get darker in an image exposed with the same settings.

If it does not get darker when we double the distance from the object to the camera, then the object does not obey the inverse square law measured from the object to the camera.

Click to expand...

This video contains an excellent summary of the essential physics.

Joe

Joe Pineapples II said:

edhannon said:

olliess said:

edhannon said:

What I propose is setting up two objects.

First a small light (without reflector) at night. Then standing 10 feet away from it, measure the light and set it for mid tone. Take an image. Then take another image 20 feet away...

Second, place a small gray patch (same size as bulb of light) and take images at same distances. I believe that your model predicts that the brightness of the patch will decrease by the same two stops. My model says that there will be no change in brightness of the patch.

Click to expand...

That is not what I said above. The key point is whether the measurement circle covers the entire object, regardless of whether it is a source or a reflector.

If so, and the object shrinks in apparent size but maintains the same intensity, then is the spot meter going to measure a smaller 1 degree averaged intensity for the smaller-appearing (further) object?

Click to expand...

But if the light reflected from the patch obeys the inverse square law like the flashlight, then the light decreases proportional to the square of the distance and it will get darker in an image exposed with the same settings.

If it does not get darker when we double the distance from the object to the camera, then the object does not obey the inverse square law measured from the object to the camera.

Click to expand...

This video contains an excellent summary of the essential physics.

Joe

Click to expand...

That would seem to cover it, yes.

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

In the case of the distance from the front of the lens to the sensor, even if the light is from a point source, it is not that small distance but rather the distance to the source that obeys the inverse square law.

Click to expand...

Well said, and so often misunderstood.

J A C S said:

edhannon said:

Inverse square law does not apply in either case.

Inverse square law says that light from a point source decreases as the square of the distance from the source. This is due to the expanding front of the light which causes the same number of photons to be spread over a larger distance.

Light reflected from an object is not a point source so it does not obey the inverse square law. If you measure the light reflected with a spot meter close to an object and then again away from the object you will get the same reading.

In the case of the distance from the front of the lens to the sensor, even if the light is from a point source, it is not that small distance but rather the distance to the source that obeys the inverse square law.

Click to expand...

Well said, and so often misunderstood.

Click to expand...

Including in the posts above.