Re: F-Stop Number relative to the sensor coverage of a lens.

[abreu-canedo]

Interesting addition, harddrive.

Will this difference in photons captured per-pixel given differently sized pixels actually produce a brighter image? Could I see this in a waveform monitor? I don't currently own cameras in all of these varieties of sensors to be able to run the tests.

I have no no reason or data to argue agains your point. The only sticking point for me is, with the varieties of sensor sizes and resolutions (affecting pixel size), how is it that we are able to maintain industry standard readings a cross T-markings in different lenses. After all, a s35 sensor can have pixels of varying sizes. If what you say is true, should I expose an fS700 1.5 stop or so higher than an FS100? To get an equivalently bright image? (Both s35 sensors, but different pixel counts). Am I understanding you correctly?

I can't do the tests for that, but I'm intrigued

 

[M. Gilden]

So much compounding confusion!

I really wish we'd stop talking about lens equivalency and things like "FF sensor sees more light". While on a technical level the latter statment is true, it doesn't mean what people think it does. Those concepts confuse the heck out of people who don't grasp the inner workings here. The pixels in the middle don't see more light, lets all agree on that shall we? And the mm and f-stop are calculated on their own merits, regardless of sensor size, so let's please throw the whole "equivalence" thing out the window.

A speedbooster DOES do something rather different by focusing all the photons into a smaller area (less scatter). Unlike all this lens equivalency garbage, the booster actually changes the FOV, DOF, and intensity of light, as if the lens were actually an F-stop faster. The explanation above with the projector on the wall is a good one. I usually bring a simpler one- take out your phone and activate the flashlight app. Make a circle on the wall of light, then bring the phone closer to the wall. The circle of light gets smaller AND BRIGHTER as a result of less scatter. Therefore, the inner APS-C sized area sees MORE intense light, and the exposure will need to be adjusted according.

If you have an A7s, this all very easy to see with your own eyes.
1)Pair the A7S with a 50mm Canon lens at 2.8 on a standard adapter and expose properly.
2)Swap to APS-C mode (which just uses the inner APS-C area of pixels), and you'll see the exposure is EXACTLY THE SAME. It didn't get darker.
3)Now put a speedbooster in place of the standard adapter. On full frame mode, you'll get a horrible vignette, but switch to APS-C mode, and you'll get roughly the same image as step 1, but with one major exception: The image is now a stop overexposed. You now need to adjust accordingly to bring it back down.

I think that covers everything. Any questions?

 

[Samuel H]

No, the intensity doesn't change, but double the area of the photosite and twice as much light (twice as many photons) will impact.

Think of amount of light as "total number of photons", and intensity as "no of photons per unit area". Hence you can increase the amount of light (per photosite) by either increasing the intensity or increasing the area.

This difference between intensity and amount is crucial to understanding - they are different things. If you want the scientific version, try http://en.wikipedia.org/wiki/Luminous_flux#Relationship_to_luminous_intensity . "Intensity" corresponds to "luminous intensity", "amount of light" corresponds to "luminous flux".

I like that explanation.

In terms of modern digital cameras. intensity would define how bright an image is (so, your ISO), and amount of light would affect how much noise you get at that ISO.

 

[M. Gilden]

Interesting addition, harddrive.

Will this difference in photons captured per-pixel given differently sized pixels actually produce a brighter image? Could I see this in a waveform monitor? I don't currently own cameras in all of these varieties of sensors to be able to run the tests.

I have no no reason or data to argue agains your point. The only sticking point for me is, with the varieties of sensor sizes and resolutions (affecting pixel size), how is it that we are able to maintain industry standard readings a cross T-markings in different lenses. After all, a s35 sensor can have pixels of varying sizes. If what you say is true, should I expose an fS700 1.5 stop or so higher than an FS100? To get an equivalently bright image? (Both s35 sensors, but different pixel counts). Am I understanding you correctly?

I can't do the tests for that, but I'm intrigued

This is easily explained. Understand that ISO values are artificial on digital sensors. A sensor with larger photosites will natively see more intense light per-pixel than one with smaller ones. But for the sake of standards, they increase or decrease gain in order to hit standardized ISO numbers. They are not necessarily starting at the same point- Let's say for argument's sake that a Rebel sees a native ISO of 320, and a 5D sees a native ISO of 800. That means that at ISO 400, the 5D is turning down the gain, whereas the Rebel is increasing it, from their respective native values. That means the Rebel sensor is going to have more noise at the same ISO. But the exposure values are standardized in order to be useful.

So, in a nutshell, an f/2 at a shutter speed of 50 and ISO of 400 should have the same exposure across any camera, regardless of sensor size and regardless of what size frame the lens was made for. The only thing that throws a wrench into this calculation is a speedbooster, but we aren't discussing that.

 

[Joshua_G]

No, that formula never changes directly - not if you simply work with ISO numbers - but the media size does come into play because **WITH ALL ELSE EQUAL** a bigger sensor will have a higher base ISO number.

Wrong!
The sensitivity (ISO number) in digital cameras is dialed in by the user, within certain limits that each camera has.

There are full frame cameras that has lower maximum enabled ISO than some other APS-C and pocket (point and shoot) cameras. For instance, Canon 5D MKI (FF) has max available ISO of 3,200, while Canon 7D MKII (APS-C) has max available ISO of 16,000 and Canon PowerShot SX60 HS (pocket) has max available ISO of 3,200.

With digital cameras that have the same sensor technology and the same processor, the larger the sensor, the higher ISO is enabled. However this is because of noise - it has nothing to do with exposure, or with the amount of light hitting the entire sensor.

Yes, you can set most digital cameras to the ISO setting of choosing, but I referred to base ISO.

There is no consistent relation between sensor size and base setting. All 3 cameras in the above example, Canon 5D MKI (FF), Canon 7D MKII (APS-C) and Canon PowerShot SX60 HS have base ISO of 100.

Increasing any cameras ISO setting above the base will inevitably adversely affect noise and/or DR. That's why your example of choosing the same film speed in your examples isn't helpful. The digital analogy to your examples would be for the 35mm sensor to be (say) 12Mp, the 5x4 camera to be something like 200Mp to make the photosites the same size.

You'd then expect a similar base ISO for each, and with an f2 lens for each to get the same exposure. I'm sure you agree with that? But you then end up with a 12Mp image in one case, a 200Mp image in the other. And (assuming the same angle of view) the 5x4 camera lens must be a longer focal length, so if both are to be f2, simple optics says that it must be a much bigger diameter.

But the two facts are inextricably linked. In a digital camera it's pretty obvious by what happens when you raise the ISO setting. For a given shutter speed, you'll have to close the iris to maintain exposure, or reduce the incident light level. But the ISO increase will inevitably mean a drop in quality - more noise etc. You are trading exposure versus quality.

All the above relate to image quality and noise, not to exposure and amount of light hitting the media.

And if you start to bring sensor size into the equation it becomes a three way tradeoff …

Sensor size isn't existing in the exposure formula.
One cannot change the laws of physics at will.

… (**WITH ALL ELSE EQUAL** - "yawn", I'm getting tired of saying that.....

) as a bigger sensor will have a higher ISO for the same quality.

1
You are wrong that bigger sensor will have higher available ISO as a rule. See my above examples.
2
You are bringing in image quality and mix it with amount of light hitting the entire sensor.
Those are two separate issues.

And let's remind ourselves what the original question was:

No, that formula never changes directly - not if you simply work with ISO numbers - but the media size does come into play because **WITH ALL ELSE EQUAL** a bigger sensor will have a higher base ISO number. Yes, you can set most digital cameras to the ISO setting of choosing, but I referred to base ISO. Increasing any cameras ISO setting above the base will inevitably adversely affect noise and/or DR. That's why your example of choosing the same film speed in your examples isn't helpful. The digital analogy to your examples would be for the 35mm sensor to be (say) 12Mp, the 5x4 camera to be something like 200Mp to make the photosites the same size.

You'd then expect a similar base ISO for each, and with an f2 lens for each to get the same exposure. I'm sure you agree with that? But you then end up with a 12Mp image in one case, a 200Mp image in the other. And (assuming the same angle of view) the 5x4 camera lens must be a longer focal length, so if both are to be f2, simple optics says that it must be a much bigger diameter.

But the two facts are inextricably linked. In a digital camera it's pretty obvious by what happens when you raise the ISO setting. For a given shutter speed, you'll have to close the iris to maintain exposure, or reduce the incident light level. But the ISO increase will inevitably mean a drop in quality - more noise etc. You are trading exposure versus quality.

And if you start to bring sensor size into the equation it becomes a three way tradeoff (**WITH ALL ELSE EQUAL** - "yawn", I'm getting tired of saying that.....

) as a bigger sensor will have a higher ISO for the same quality. And let's remind ourselves what the original question was:

To which the answer is an unequivocal yes. And not just theoretically. If you put a SpeedBooster type adaptor (but APS-C *TO* FF) the effective aperture for exposure purposes would indeed be very close to f3. The adaptor would have the effect of making the image dimmer, but covering the bigger area.



This is where you are not recognising the significance of *EVERYTHING ELSE BEING EQUAL*, because one of the key factors in "everything" is resolution. So increase sensor size and the assumption is you don't get more photosites - you get bigger ones. So in fact the image doesn't necessarily get better (same resolution), but bigger photosites = higher base ISO. So the exposure is exactly what will change.


It will give the same BRIGHTNESS of image - which will indeed be the relevant factor for exposure - but that is simply NOT the same as AMOUNT of light reaching the media. The amount is brightness x area and I am really running out of ways to say that. Put a reflector behind a bulb and you won't change the total AMOUNT of light it emits, but you will make it's effect BRIGHTER in front of the beam.


Well, I also have a degree in electronics originally, and I think I'm pretty clear that electrical power is given by I*V - current times voltage. In this case, the total amount of incident light is proportional to brightness * area. The f stop will directly determine the brightness, so in each case you're looking for something to multiply by to get the answer. What's "nonsense" about that?



But they are all tied in together. If ALL that happened was a decrease in effective focal length and increase in aperture, the image would still cover the original area, be a wider angle of view and it would mean light had somehow appeared out of nowhere, breaking the Laws of Physics!

It's the change in coverage area that is key to understanding why this is not the case. The principle is exactly the same as a projector with a zoom lens. Turn the ring and the image shrinks and gets brighter, but you still see the same image - just smaller. The projector still emits the same *AMOUNT* of light, but the smaller image means it is therefore brighter.

…
Assuming "f2" is describing the APS-C lens itself. And assuming the crop factor is 1.5x. Then is it true that the AMOUNT of light being let in at f2 would be equivalent to 1.5 x f2 = f3 in Full Frame currency?
…

To which the answer is an unequivocal yes.

To which the answer is definitely no.
The amount of light hitting the entire sensor is determined by 2 things and only 2 things:
1
The amount of light being reflected by the subject.
2
The f stop being set on the lens.

The sensor size has no effect here.

And not just theoretically. If you put a SpeedBooster type adaptor (but APS-C *TO* FF) …

This is purely theoretical, since there is no such SpeedBooster.

… the effective aperture for exposure purposes would indeed be very close to f3. The adaptor would have the effect of making the image dimmer, but covering the bigger area.

Again, SpeedBooster increases the effective f stop because of it decreases the effective focal length. It has nothing to do with sensor size.

This is where you are not recognising the significance of *EVERYTHING ELSE BEING EQUAL*, because one of the key factors in "everything" is resolution. …

Resolution isn't a part of the exposure formula and it has nothing to do with the amount of light hitting the entire sensor.
One cannot change the laws of physics at will.

… So increase sensor size and the assumption is you don't get more photosites - you get bigger ones. …

This is correct only for sensors of the same technology. Sensors of different technologies have different amount of photsites per sensor area.

So in fact the image doesn't necessarily get better (same resolution), but bigger photosites = higher base ISO. So the exposure is exactly what will change.

Wrong. See my above examples.

It will give the same BRIGHTNESS of image - which will indeed be the relevant factor for exposure - but that is simply NOT the same as AMOUNT of light reaching the media. The amount is brightness x area and I am really running out of ways to say that. Put a reflector behind a bulb and you won't change the total AMOUNT of light it emits, but you will make it's effect BRIGHTER in front of the beam.

Wrong. I related to it few times in this reply and in previous replies.

Well, I also have a degree in electronics originally, and I think I'm pretty clear that electrical power is given by I*V - current times voltage. …

This is correct.

… In this case, the total amount of incident light is proportional to brightness * area.

Wrong.
The amount of light hitting the entire sensor is determined by 2 things and only 2 things:
1
The amount of light being reflected by the subject.
2
The f stop being set on the lens.

One cannot change the laws of physics at will.

The f stop will directly determine the brightness, so in each case you're looking for something to multiply by to get the answer. What's "nonsense" about that?

The nonsense is having analogy of exposure to electric power.

But they are all tied in together. If ALL that happened was a decrease in effective focal length and increase in aperture, the image would still cover the original area, be a wider angle of view and it would mean light had somehow appeared out of nowhere, breaking the Laws of Physics!

Review your physical knowledge.
The focal length is part of the amount of light formula. For a given entrance pupil, the smaller the focal length the more open the f number is.

It's the change in coverage area that is key to understanding why this is not the case.

You are confusing coverage area with exposure and the amount of light hitting the entire sensor. Coverage area with exposure and the amount of light hitting the entire sensor are two entirely different matters.

The principle is exactly the same as a projector with a zoom lens. Turn the ring and the image shrinks and gets brighter, but you still see the same image - just smaller. The projector still emits the same *AMOUNT* of light, but the smaller image means it is therefore brighter.

This is due to two reasons:
1
When you zoom out, the focal length decreases, hence the effective f number increase (when f/2 is higher than f/4).

2
It is the amount of light emitted by the projector's lamp spread over larger or smaller area.


Since we are going round in circles over and over, this will be my last reply to you in this thread.

You are welcomed to make any statements you may wish to.

 

[Joshua_G]

I like the distinction between quantity and intensity. I also find it helpful to think in a per-pixel basis.
…

The number of photons hitting each pixel, or photo-site, affects the amount of noise, it has nothing to do with exposure.

…

While it's true that larger pixels are more sensitive and thus produce less noise at higher ISO's, this does not affect the intensity of light being let in by a lens at a given aperture. It simply affects our own decisions of how to expose it.

This is correct.

Only when we're talking about optical conversions with the speed booster are we then increasing the intensity of light per-pixel, and thus, affecting exposure and f-stop.

…

The increase in f stop enabled by using speed booster is due to the decrease in the effective focal length while maintaining the same entrance pupil, it has nothing to do with the number of photons per pixel.

 

[Joshua_G]

…
So, in a nutshell, an f/2 at a shutter speed of 50 and ISO of 400 should have the same exposure across any camera, regardless of sensor size and regardless of what size frame the lens was made for.

This is correct.

The only thing that throws a wrench into this calculation is a speedbooster, but we aren't discussing that.

SpeedBooster increases the effective f number (when f/2 is higher than f/4) because of it decreases the effective focal length.

 

[abreu-canedo]

Joshua_G,

The speedbooster takes the amount of light that a lens would allocate to a 35mm sensor, and focuses it into the smaller image circle appropriate for, let's say, a s35mm sensor. In affect, it's focusing light in a similar way some people used to do as children, by focusing sunlight with their glasses into a small point, to et things on fire. Sorry for the poor example, but in essence, this should increase the intensity of light when focused into a smaller area...does it not?

Ah, and thanks fr the clarification on the gain adjustment each camera makes for hitting ISO standards. I suspected this was the case, but thought I'd ask since my technical knowledge is limited.

 

[M. Gilden]

SpeedBooster increases the effective f number (when f/2 is higher than f/4) because of it decreases the effective focal length.

Which has as a side effect, the focusing of photons to a smaller area, thus filling the photosites with more intense light. You're agreeing with us and yet arguing at the same time, which is madness.

 

[M. Gilden]

Joshua_G,

The speedbooster takes the amount of light that a lens would allocate to a 35mm sensor, and focuses it into the smaller image circle appropriate for, let's say, a s35mm sensor. In affect, it's focusing light in a similar way some people used to do as children, by focusing sunlight with their glasses into a small point, to et things on fire. Sorry for the poor example, but in essence, this should increase the intensity of light when focused into a smaller area...does it not?

^ this guy gets it.

 

[Joshua_G]

Joshua_G,

The speedbooster takes the amount of light that a lens would allocate to a 35mm sensor, and focuses it into the smaller image circle appropriate for, let's say, a s35mm sensor. In affect, it's focusing light in a similar way some people used to do as children, by focusing sunlight with their glasses into a small point, to et things on fire. Sorry for the poor example, but in essence, this should increase the intensity of light when focused into a smaller area...does it not?

…

Please look at what I wrote above few times, in few replies:

The amount of light hitting the entire sensor is determined by 2 things and only 2 things:
1
The amount of light being reflected by the subject.
2
The f stop being set on the lens.

…
The focal length is part of the amount of light formula. For a given entrance pupil, the smaller the focal length the more open the f number is.

…

 

[M. Gilden]

Please look at what I wrote above few times, in few replies:

You can say it as many times as you want. You can stomp your feet and hold your breath until you turn blue. You can't change physics, which you are arguing with.

Refocusing light changes the light formula. We can say it over and over again too.

 

[Joshua_G]

Which has as a side effect, the focusing of photons to a smaller area, thus filling the photosites with more intense light. You're agreeing with us and yet arguing at the same time, which is madness.

You see agreement on my part where there isn't an agreement.
This is what I wrote few times in this thread:

The exposure is being determined by 3 factors:
1
The amount of light falling on the subject.
2
The sensitivity of the media (film or sensor).
3
The f number (or T number) set on the lens.

…

The case of Speed Booster is different, since the optics in it changes the actual f number by the fact that it decreases the effective focal length.
The maximum f number of a given lens is its' focal length in millimeters divided by the diameter of the front element in millimeters.
For f numbers other than the maximum one, the f number is the focal length in millimeters divided by the diameter of the entrance pupil in millimeters. See: http://en.wikipedia.org/wiki/F-number .
The Speed Booster optics changes the effective focal length, thus increases the f number. See: http://www.metabones.com/products/?c=speed-booster .
…

Neither the f number formula (focal length divided by entrance pupil) nor the exposure formula relate to the media size, or the number of photons hitting each photosite.

The number of photons hitting each photosite affect the amount of noise, not the exposure, which is the brightness of the subject as 'seen' by the sensor.

You can say it as many times as you want. You can stomp your feet and hold your breath until you turn blue. You can't change physics, which you are arguing with.

Refocusing light changes the light formula. We can say it over and over again too.

I stick to physics constantly, some others here confuse different phenomena and mix different phenomena which are not directly related to one another.

What is the formula that determines the exposure? Where is focus in the exposure formula?

 

[harddrive]

Interesting addition, harddrive.

Will this difference in photons captured per-pixel given differently sized pixels actually produce a brighter image?

No - and I think the fact you are posing the question shows there is still confusion between BRIGHTNESS and AMOUNT OF LIGHT. As I've been trying to explain, they are related, but different.

Let me try another way. Imagine you are a few miles out to sea and can see a lighthouse beam as it sweeps across. You are holding a sheet of paper in front of you, and every few seconds the lighthouse beam illuminates it. You perceive (and could measure) it at a certain "brightness", yes? Then in the intervening seconds you have a torch which you turn on. You move that backwards and forwards until there is no perceptible change in how bright the paper is lit whether it is lit from the lighthouse or your torch.

In this case, both sources produce the same brightness - or intensity of light - at the paper. So can we conclude the same AMOUNT of light is being emitted by the torch and the lighthouse? The answer must be a very definite "no", and hopefully that's obvious - the lighthouse is emitting a far greater amount overall. If you had such a piece of paper, it could illuminate a very big area to that brightness which your torch couldn't!

Really, the whole subject is a reworking of the old "if it takes two men 15 days to dig a trench, how long....." maths questions, and photographically is very similar to the whole concept of reciprocity. You can either use a bright light for a short time, or dimmer for longer. (So 1/125 at f11 and 1/500 at f5.6 both give the same exposure.) In terms of your question, replace time with photosite area. A bright light in a small area is equivalent to a dimmer light in a bigger area.

The only sticking point for me is, with the varieties of sensor sizes and resolutions (affecting pixel size), how is it that we are able to maintain industry standard readings a cross T-markings in different lenses. After all, a s35 sensor can have pixels of varying sizes. If what you say is true, should I expose an fS700 1.5 stop or so higher than an FS100? To get an equivalently bright image? (Both s35 sensors, but different pixel counts). Am I understanding you correctly?

No, you're not understanding me correctly! (See above comments.)

What we are all agreed upon is that regardless of sensor size etc, if you set ANY lens to a given f stop (or more correctly, T-stop) and point it at a given scene, the resulting image at the sensor will have the same brightness, the same intensity. And *from an operational* point of view, you don't really need to know any more. By giving you an ISO figure for whatever camera or filmstock, they've done a lot of the work for you. For any scene brightness, for any ISO setting, then regardless of camera you set a given shutter speed/f stop combination.

When we get on to real cameras, then the actual results are far more difficult to predict because there are multiple variables - which is why when trying to explain the underlying physics I kept stressing (until it was boring!) **all else equal**. There are different sizes of sensors, but also different resolutions, different technologies (CCD v CMOS? Microlenses? Front or back illuminated?, and differing read systems (eg if line skipping is used). But for straightforward operation, measure light level, set ISO rating according to camera or film stock and expose accordingly.

It's only when you try to work out "what are the implications of.....?" that you need to understand more about the underlying fundamentals.

Maybe most important is to realise that it follows from basic optics that for a given angle of view, an f2 lens for FF will not only have a different focal length to the comparable f2 lens for APS-C, but it must also have a bigger front element and iris diameter. This follows from the defining equation for the f stop - focal length/iris diameter. So the FF lens gathers more light in total - but distributes it over a wider area whilst the brightness of the image is the same in each case.

 

[M. Gilden]

This is what I wrote few times in this thread:

And it doesn't matter how many times you write something if we don't agree with it. We saw it. We think it is wrong. Does repeating something with information we don't agree with help?

Neither the f number formula (focal length divided by entrance pupil) nor the exposure formula relate to the media size, or the number of photons hitting each photosite.

Media size, I agree with. But as I undertand it, the intensity of light exposed to a particular pixel is a determined by the amount of photons hitting that photosite. That means that a wider aperture, or F number, would allow MORE photons to hit each photosite than a stopped-down one.

The number of photons hitting each photosite affect the amount of noise, not the exposure, which is the brightness of the subject as 'seen' by the sensor.

Huh...I don't even... what?

Where are coming up with this stuff?!

Noise levels are determined by ISO, which as we all know is raised or lowered on account of proper exposure!
If the intensity of light (as determined by how many photons hit each photosite) is not high enough, you may need to bump up the gain by dialing in a higher ISO value. THAT, and THAT ALONE will increase noise in the image, because the exposure is lower. If the photosites are seeing enough photons/light intensity without it, you can turn the gain down and avoid the excess noise. Therefore, photosites, photons, intensity, and exposure are all 100% related! (how did noise get into this?)

What is the formula that determines the exposure? Where is focus in the exposure formula?

By focally reducing the image circle, you are condensing the photons into a smaller area which has the side effect of intensifying the light (hence the magnifying glass, flashlight, etc, analogy). That's why makes the F-stop change actually possible, and it ABSOLUTELY effects exposure!

The proof can be seen for yourself if you have a camera capable of FF / APSC and a speedbooster, like an A7. In FF and APS-C modes, the lens will have the same exposure at the same settings on the lens, shutter and ISO. But with a speedbooster in APS-C mode, the image is suddenly a stop overexposed! The FF version will be a stop of light behind the APS-C boosted one, which means you can turn down the ISO and reduce noise if you wished.

 

[harddrive]

Wrong!
The sensitivity (ISO number) in digital cameras is dialed in by the user, within certain limits that each camera has.

There are full frame cameras that has lower maximum enabled ISO than some other APS-C and pocket (point and shoot) cameras. For instance, Canon 5D MKI (FF) has max available ISO of 3,200, while Canon 7D MKII (APS-C) has max available ISO of 16,000 and Canon PowerShot SX60 HS (pocket) has max available ISO of 3,200.

With digital cameras that have the same sensor technology and the same processor, the larger the sensor, the higher ISO is enabled. However this is because of noise - it has nothing to do with exposure, or with the amount of light hitting the entire sensor.

Sorry Joshua, but it's you who are wrong and I can only suggest that before you do anything else you really think about the difference between intensity and amount (or flux to use the technical term) - because you are confusing the two. It's pointless for me or anyone else to try to say too much more until you grasp the fundamental difference between the two.

I've consistently referred to base ISO as what changing sensor and/or photosite size will change. This will not necessarily bear a direct relationship to minimum or "normal" ISO on any camera - but you can get an idea of what it is by setting ISO high and seeing how much degradation you get. That works well for comparisons at least, if not for establishing absolute values. Base ISO is a value fundamental to the sensor, unlike the set ISO which is trading off quality for exposure.

This {And not just theoretically. If you put a SpeedBooster type adaptor (but APS-C *TO* FF) …} is purely theoretical, since there is no such SpeedBooster.

Well, I said "type" adaptor, to mean a small to large format adaptor, instead of the large to small SpeedBooster. And such certainly does exist in the form of the B4-s35 convertors being now quite widely employed.

Again, SpeedBooster increases the effective f stop because of it decreases the effective focal length. It has nothing to do with sensor size.

If it had nothing to do with sensor size - if all it did was alter the focal length - it would be a wide angle convertor for FF 35mm! The whole point of it (or any other optical format convertor) is that it does what the name says - format convert. So a lens designed for one format has it's coverage area made smaller (as with SpeedBooster) or larger (as with the B4-s35 convertors) whilst keeping the angle of view of the lens. (Which wouldn't happen with a simple non-optical adaptor.) Make the image smaller and it must get brighter, make it larger and it must get dimmer because it's the same total AMOUNT of light in each case. But condensing or enlarging the image, making it brighter or dimmer, is why you get the apparent change in f stop.

Wrong.
The amount of light hitting the entire sensor is determined by 2 things and only 2 things:
1
The amount of light being reflected by the subject.
2
The f stop being set on the lens.

One cannot change the laws of physics at will..

Joshua - it's you who is not sticking to the Laws of Physics. I don't disagree with those two statements, but you have simply got to include sensor size as well. It's back to what I said at the start, you simply are not distinguishing between AMOUNT of light (no of photons in total) and INTENSITY of light (no of photons per unit area). So it follows that AMOUNT=INTENSITYxAREA.

Think about it. Just think about it.

 

[harddrive]

I find it easier to understand and explain this stuff by thinking in per-pixel terms. If you have a FF sensor and APS-C sensor with the same sized pixel pitch (say, 18mp APS-C vs 26mp FF), a full frame sensor is basically an APS-C with more pixels around it. The extra pixels outside have no effect on how much light the ones in the middle see. Saying that a FF sensor receives "more light" overall is true because of the extra surface area of the lens, as was stated earlier. But it does not receive more light PER PIXEL (again, assuming photosites are the same size- otherwise this gets way more complicated to explain, but the concept is still true). That APS-C sized area in the middle is seeing just as much light as it would on an APS-C camera sensor.
Putting an APS-C f/2.8 lens on a FF body would vignette since the lens won't cover the outer edges of your sensor, but the inner 30% will see the same f/2.8 in terms of exposure and DOF as a full frame would.

Basically, "lens equivalency" is bullocks. Just confuses the heck out of everyone, and I wish we'd stop doing it.

Rewinding a little, I wanted to comment on this before we all got distracted.....

Point by point, what you say is right, but .... is it really fair to compare a 18mp APS-C vs 26mp FF? It'll certainly keep the photosite size constant, but I'd argue it's not what I meant with the all else equal caveat - the caveat was to assume that whatever the sensor size, they both had the same resolution. Hence bigger photosites for the bigger sensor, bigger base ISO, and it all follows from there.

If you assume same size photosites, then fair enough if your end product from the 18Mp APS-C is going to be smaller than from 26Mp FF (say 10x8 prints v 12x10), but not if they are to be enlarged to the same size.

 

[M. Gilden]

is it really fair to compare a 18mp APS-C vs 26mp FF? It'll certainly keep the photosite size constant, but I'd argue it's not what I meant with the all else equal caveat - the caveat was to assume that whatever the sensor size, they both had the same resolution. Hence bigger photosites for the bigger sensor, bigger base ISO, and it all follows from there.

Why would you assume that? Manufacturers almost always cram more pixels into FF bodies because they can and they will sell more. Sony is the only one who took a step backwards with the A7s.

 

[ectobuilder]

This is one of the issues that keeps popping up; in what "currency" do we think of when we talk about lenses, bodies and sensor sizes?
I tend to stick with a 35mm equivalent when talking about focal lengths, although as J Michael very succinctly put it; when we talk about f-stop it's a ratio within the lens itself; nothing to do with the sensor size.

As harddrive also pointed out, the intensity versus the amount comes down to the fact that on a smaller sensor size you are exposing less area to the light:

Note: Luminous flux (amount) is different to lumens. (http://en.wikipedia.org/wiki/Lumen_(unit) versus http://en.wikipedia.org/wiki/Luminous_flux).

As I've understood it, the amount of light varies significantly because it IS a function of area * intensity.

I don't get too technical into this side of things (It looks like I may need to do a dedicated article on the difference), but you can check out one of my comments on low-light performance which has a few more diagrams: http://www.ozirig.com.au/sensor-characteristics-iso-noise-snr/#Low-light_performance

But this is not what I was asking. The example above assumes the lens coverage is constant. In my question I am trying to compare different lens coverage sizes. i.e. EF vs EF-S lenses at the same advertised f-stop.

 

[M. Gilden]

But this is not what I was asking. The example above assumes the lens coverage is constant. In my question I am trying to compare different lens coverage sizes. i.e. EF vs EF-S lenses at the same advertised f-stop.

Coverage makes no difference. The extra light on the outside of the sensor's viewable area just gets cut off and wasted because there aren't any photosites there to "see". The middle portion gets exposed the same as the middle of a FF would.

Equivalency is a nasty word that just confuses people.