Depth of field FF V Crop

[juggler]

The best focal length for (solo) portraits is the lens that gives enough distance for attractive perspective, say a minimum of 5ft, with the framing you want. There's a big difference between a tight headshot and waist up. Personally, I like to be around 5-6ft - close enough for good communication, without being intimidating. On full-frame, lens could be anywhere between 75-135mm-ish.

My bold.. it very often depends on how much of the background I want in the frame. The communication thing is important but I'm not averse to shooting full length at 150mm and shouting if that's what the shot calls for.

 

[HoppyUK]

My bold.. it very often depends on how much of the background I want in the frame. The communication thing is important but I'm not averse to shooting full length at 150mm and shouting if that's what the shot calls for.

Haha

The point is, people say this or that lens is 'best for portraits' whereas what's important is the shooting distance. And so long as that's say 5ft-ish or more, perspective problems won't be an issue.

 

[wyx087]

Fujifilm, being a APS-C only brand, solved the DoF disparity easy enough. They offer the 56mm f1.2 lens, equivalent to 85mm f1.8 on full frame in both DoF and FoV.

The f1.2 lens allows to shoot with lower ISO than the f1.8 lens while having all of the subject in focus. Win-win IMHO.

 

[HoppyUK]

Perspective really the point here, not focal length. Perspective is a function of distance, not the lens.

People talk about 'perspective distortion' with wide-angles but that's not the right term - the lens doesn't affect perspective, and neither does it distort perspective.

I prefer the term 'exaggerated perspective' with wide-angles (because they allow you to shoot close, while still getting everything in) and 'flattened perspective' with longer lenses (because of the compression effect of greater distance).

 

[G.K.Jnr.]

Fujifilm, being a APS-C only brand, solved the DoF disparity easy enough. They offer the 52mm f1.2 lens,

52mm ? I think you mean the 56mm F1.2.

 

[sk66]

Steven, no again, and you cannot have read the link to DPReview's article. See page 2, Equivalence of Total Light, and page 4, Real World Total Light Demonstration. It's all there

If I might hazard a guess, you seem to be stuck on the fact that the same subject 'radiates the same amount of light' if I can put it like that, and it doesn't change. What we're discussing is the total amount of light captured by the lens and projected on to the sensor. That does change. A lot.

Sorry, I've been away off-grid...

I've read most things on the topic... and they all seem to convolute things by trying to start from a point of "equivalence" (even that DPreview article, and the example images on pg 4 are particularly poor/misleading IMO).

To really understand it you must start with FOV/scene illuminance. A given area (FOV) can only emit a given amount of light over a given amount of time (it's illuminance)... that cannot be changed, nor can the physics of how light behaves. And that is what we are recording...

Put a 100mm DX lens (if such exists, but it doesn't actually matter) on a 1.5x DX camera and a 150mm FF lens on a FF body. Using the same f(t)-stop on both lenses and from the same distance they will capture the same FOV with the same source illuminance and transmit the same amount of light, and **the exposures will be the same. However, the image circle for the larger sensor is greater than it is for the DX sensor, and that dictates that the light per sq mm is less while the total light (exposure) remains the same (there are more sq mm's recording it).
Everyone seems to get this wrong... exposure (and aperture) is not about "same light per sq mm," as if the light somehow retained the same intensity per each mm as a FOV/scene is spread over a larger image circle... that is impossible. And if a lens could do that, then a lens optimized for a FF sensor would somehow have to increase the total exposure (transmit more light spread over a larger area in order to keep each mm constant), which would necessarily affect it's use/exposure on a crop sensor body. Instead, exposure is about the total light transmitted over the total area/image circle used. I.e. the per in "light per sq mm" means "divided by" and results in the total/overall exposure.
To help understand this we can apply the inverse square law... one way to make an image circle larger is to have a greater distance to the sensor/film plane, and greater distance means less light per sq mm because the image circle is larger. This is in fact what happens with macro/large format photography, it is known as "bellows factor"... in macro photography moving the lens farther away in order to reduce the minimum focus distance increases the size of the image circle and the resulting magnification, and this reduces the amount of light reaching the sensor affecting the exposure (smaller effective aperture). In large format photography it can be either due to actual bellows extension (increased distance/close focusing), or due to using longer FL's for "equivalent/normal" FOV's (spreading a smaller FOV/greater magnification over a larger image circle). The aperture doesn't really have anything to do with the size of the image circle used... it is "constant" only because as you use less of the original/fixed image circle/FOV you are also getting less total light... that is the "same light per square millimeter/less total light" everyone seems to get stuck on, but it has nothing to do with "equivalence" because it does not include the same total scene/FOV/source illuminance.
BTW, bellows factor is the only way to change the size of image circle projected/used while keeping the sensor size constant, but the physics/effect is the same.

From there we have to understand how the inverse square law affects the exposure of a scene/object. If you make a scene/object larger/smaller within the FOV it's exposure (total light) does not change because the light per sq mm changes inversely with size. I.e. if you take a white door and make it larger w/in the FOV it's emitted light per sq mm is reduced, but it occupies more sq mm's and the correct exposure for the door (and the scene overall) remains constant. I.e. the size of an object affects it's "brightness." That's essentially what allows manual exposure to work in a constant light/variable subject/scene situation.

So we have a larger image circle with less light per sq mm -vs- a smaller image circle with more light per sq mm and equivalent total exposures. Next we place the 150mm FF lens on the DX body without changing distance or f(t)-stop. Now the DX sensor is recording a smaller portion of the scene, but it has become "larger" w/in the FOV due to the increased FL. You have less of the scene occupying more of the sensor (what remains is "brighter") and the exposure does not change (assuming even scene illuminance).
And then we can back up 50% of the distance to regain the same/original 100mm FOV which reduces the size of the object(s) w/in the FOV but includes more within it. And again the exposure does not change. And again, that's essentially what allows manual exposure to work in a constant light/variable subject/scene situation. One might think "aha, recording the same FOV from farther w/ DX must mean less light than using the same lens/settings on FF from closer." But it doesn't because the greater distance is offset by the greater concentration/magnification onto the smaller area (same FOV).

All of this is easily tested with a FF camera that has "DX mode" and a constant aperture zoom lens. And all of it is how light/sensors/exposures actually work.
And none of this takes into consideration "DOF equivalence" which would dictate using a larger aperture for the DX sensor. That would increase the amount of light reaching the DX sensor, allowing the use of a lower ISO which offsets the smaller pixel efficiency/sensitivity deficit.

**Complexities arise when you change bodies due to REI/SOS variances/WB handling/flange distances(image circle spread)/sensor characteristics, and when changing lenses due to f vs t stop variances. All of these are typically minor on their own but can combine to create apparent/obvious differences.

 

[HoppyUK]

<snip>
To really understand it you must start with FOV/scene illuminance. A given area (FOV) can only emit a given amount of light over a given amount of time (it's illuminance)... that cannot be changed, nor can the physics of how light behaves. And that is what we are recording...
<snip>

^^^This is where you are going wrong.

Light radiated from the subject may not change, but the amount of light collected, and in turn delivered to the sensor, changes hugely - according to the diameter of the aperture and focal length.

 

[Deleted member 49549]

Sorry, I've been away off-grid...

I've read most things on the topic... and they all seem to convolute things by trying to start from a point of "equivalence" (even that DPreview article, and the example images on pg 4 are particularly poor/misleading IMO).

To really understand it you must start with FOV/scene illuminance. A given area (FOV) can only emit a given amount of light over a given amount of time (it's illuminance)... that cannot be changed, nor can the physics of how light behaves. And that is what we are recording...

Put a 100mm DX lens (if such exists, but it doesn't actually matter) on a 1.5x DX camera and a 150mm FF lens on a FF body. Using the same f(t)-stop on both lenses and from the same distance they will capture the same FOV with the same source illuminance and transmit the same amount of light, and **the exposures will be the same. However, the image circle for the larger sensor is greater than it is for the DX sensor, and that dictates that the light per sq mm is less while the total light (exposure) remains the same (there are more sq mm's recording it).
Everyone seems to get this wrong... exposure (and aperture) is not about "same light per sq mm," as if the light somehow retained the same intensity per each mm as a FOV/scene is spread over a larger image circle... that is impossible. And if a lens could do that, then a lens optimized for a FF sensor would somehow have to increase the total exposure (transmit more light spread over a larger area in order to keep each mm constant), which would necessarily affect it's use/exposure on a crop sensor body. Instead, exposure is about the total light transmitted over the total area/image circle used. I.e. the per in "light per sq mm" means "divided by" and results in the total/overall exposure.
To help understand this we can apply the inverse square law... one way to make an image circle larger is to have a greater distance to the sensor/film plane, and greater distance means less light per sq mm because the image circle is larger. This is in fact what happens with macro/large format photography, it is known as "bellows factor"... in macro photography moving the lens farther away in order to reduce the minimum focus distance increases the size of the image circle and the resulting magnification, and this reduces the amount of light reaching the sensor affecting the exposure (smaller effective aperture). In large format photography it can be either due to actual bellows extension (increased distance/close focusing), or due to using longer FL's for "equivalent/normal" FOV's (spreading a smaller FOV/greater magnification over a larger image circle). The aperture doesn't really have anything to do with the size of the image circle used... it is "constant" only because as you use less of the original/fixed image circle/FOV you are also getting less total light... that is the "same light per square millimeter/less total light" everyone seems to get stuck on, but it has nothing to do with "equivalence" because it does not include the same total scene/FOV/source illuminance.
BTW, bellows factor is the only way to change the size of image circle projected/used while keeping the sensor size constant, but the physics/effect is the same.

From there we have to understand how the inverse square law affects the exposure of a scene/object. If you make a scene/object larger/smaller within the FOV it's exposure (total light) does not change because the light per sq mm changes inversely with size. I.e. if you take a white door and make it larger w/in the FOV it's emitted light per sq mm is reduced, but it occupies more sq mm's and the correct exposure for the door (and the scene overall) remains constant. I.e. the size of an object affects it's "brightness." That's essentially what allows manual exposure to work in a constant light/variable subject/scene situation.

So we have a larger image circle with less light per sq mm -vs- a smaller image circle with more light per sq mm and equivalent total exposures. Next we place the 150mm FF lens on the DX body without changing distance or f(t)-stop. Now the DX sensor is recording a smaller portion of the scene, but it has become "larger" w/in the FOV due to the increased FL. You have less of the scene occupying more of the sensor (what remains is "brighter") and the exposure does not change (assuming even scene illuminance).
And then we can back up 50% of the distance to regain the same/original 100mm FOV which reduces the size of the object(s) w/in the FOV but includes more within it. And again the exposure does not change. And again, that's essentially what allows manual exposure to work in a constant light/variable subject/scene situation. One might think "aha, recording the same FOV from farther w/ DX must mean less light than using the same lens/settings on FF from closer." But it doesn't because the greater distance is offset by the greater concentration/magnification onto the smaller area (same FOV).

All of this is easily tested with a FF camera that has "DX mode" and a constant aperture zoom lens. And all of it is how light/sensors/exposures actually work.
And none of this takes into consideration "DOF equivalence" which would dictate using a larger aperture for the DX sensor. That would increase the amount of light reaching the DX sensor, allowing the use of a lower ISO which offsets the smaller pixel efficiency/sensitivity deficit.

**Complexities arise when you change bodies due to REI/SOS variances/WB handling/flange distances(image circle spread)/sensor characteristics, and when changing lenses due to f vs t stop variances. All of these are typically minor on their own but can combine to create apparent/obvious differences.

At f2, the 100mm lens will have an aperture of 50mm, the 150mm lens an aperture of 75mm. The light intensity is the same on the sensor.

Neither does it matter that you are using a dx lens on the dx camera. A 100mm fx will function the same.

 

[HoppyUK]

At f2, the 100mm lens will have an aperture of 50mm, the 150mm lens an aperture of 75mm. The light intensity is the same on the sensor.

Neither does it matter that you are using a dx lens on the dx camera. A 100mm fx will function the same.

Yes. And crucially, if you increase the size of the aperture, brightness on the sensor is increased.

Bigger aperture = more light. Simple as. The way Steven is explaining things, changing the aperture would have no influence on exposure.

 

[LCPete]

^^^This is where you are going wrong.

Light radiated from the subject may not change, but the amount of light collected, and in turn delivered to the sensor, changes hugely - according to the diameter of the aperture and focal length.

Forgive my ignorance I've never understood the physics behind it if I use F2.8 on a 100 mm lens and F 2.8 on a 300 mm lens is the amount of light collected the same? will I have to use the same shutter speed and iso to correctly expose the shot ?

 

[StephenM]

It's the same shutter speed and iso for both. The f numbers are devised to ensure that the exposure is constant for a constant f number, regardless of the focal length.

 

[petersmart]

Forgive my ignorance I've never understood the physics behind it if I use F2.8 on a 100 mm lens and F 2.8 on a 300 mm lens is the amount of light collected the same? will I have to use the same shutter speed and iso to correctly expose the shot ?

Yes the light collected is the same but the diameter on the 300mm lens will be 3x the diameter of the 100mm lens.

The reason is THE INVERSE SQUARE LAW.

The 100mm lens will cast a circle of light with a certain luminosity but the 300mm lens will be further away from the sensor so the circle which it casts will be 3 times as large - with an area 9 times the area of the 100mm lens.

If the diameter of the 2 lenses were the same then the light TRANSMITTED through the 300mm lens would be the same as the 100mm lens but AT THE SENSOR would be reduced to 1/9 of the light from the 100mm lens because the same amount of light would be spread across a much larger area.

So the f no guarantees that no matter what the focal length of a lens the light reaching the sensor will always be the same.

It is the same reason that light from a distant star does not have the same intensity as our sun even though the distant star may be bigger and brighter than our sun.
.

 

[sk66]

^^^This is where you are going wrong.

Light radiated from the subject may not change, but the amount of light collected, and in turn delivered to the sensor, changes hugely - according to the diameter of the aperture and focal length.

Not really...
Aperture diameter is relative to focal length. In truth f-stop is not a measurement of light but it relates to t-stop which is. And the transmittance of a lens (t-stop) is not really a measurement of light reaching the sensor (but it does determine that). Transmittance is a measurement of total light received vs total light projected. i.e. it takes into account the FOV/source and size of the image circle. In fact, the instrument for determining/calibrating t-stops is an "integrating sphere" and doesn't involve "a camera" as such. Any given t/f-stop means that a lens receiving X amount of light will project Y amount of light... i.e. the total exposure. This is why t-stop, and by association f-stop, are lens characteristics independent of any sensor/image circle consideration.

Edit to add: OR, that's exactly right... the amount of light transmitted is "only" dependent of FL(recorded FOV/source) and aperture, it is not sensor size related... and spreading the recorded/source FOV over a larger sensor area means less light per sq mm.

A lot of the explanations of equivalence (including the DPR one linked) state that "a larger sensor has more area for each section of the scene." And that is true... it's exactly the same as saying the source area is spread over a larger area (image circle). I.e less light/mm but more mm overall. It is not some kind of "benefit" as they make it out to be.

The way Steven is explaining things, changing the aperture would have no influence on exposure.

How did you get that from anything I wrote? The only thing I said (in a round about way) is that any given aperture (t-stop) generates the same total exposure for a given source illuminance (captured FOV).

This is probably the best explanation of things that I have seen... be prepared for a lot of technical geekery (it can be kind of confusing)...

 

[LCPete]

Yes the light collected is the same but the diameter on the 300mm lens will be 3x the diameter of the 100mm lens.

The reason is THE INVERSE SQUARE LAW.

The 100mm lens will cast a circle of light with a certain luminosity but the 300mm lens will be further away from the sensor so the circle which it casts will be 3 times as large - with an area 9 times the area of the 100mm lens.

If the diameter of the 2 lenses were the same then the light TRANSMITTED through the 300mm lens would be the same as the 100mm lens but AT THE SENSOR would be reduced to 1/9 of the light from the 100mm lens because the same amount of light would be spread across a much larger area.

So the f no guarantees that no matter what the focal length of a lens the light reaching the sensor will always be the same.

It is the same reason that light from a distant star does not have the same intensity as our sun even though the distant star may be bigger and brighter than our sun.
.

Ahh I see thanks for explaining it I had worked out just by experience how the depth of field varies with distance and focal length but never understood that part of it the amount of light on the sensor and the exposure thanks

 

[Deleted member 49549]

@sk66.

The deduction from your 'spreading the light over a larger sensor' statement would be that the sunny 16 rule is sensor size dependant. It isn't.

 

[Deleted member 49549]

Not really...
Aperture diameter is relative to focal length. In truth f-stop is not a measurement of light but it relates to t-stop which is. And the transmittance of a lens (t-stop) is not really a measurement of light reaching the sensor (but it does determine that). Transmittance is a measurement of total light received vs total light projected. i.e. it takes into account the FOV/source and size of the image circle. In fact, the instrument for determining/calibrating t-stops is an "integrating sphere" and doesn't involve "a camera" as such. Any given t/f-stop means that a lens receiving X amount of light will project Y amount of light... i.e. the total exposure. This is why t-stop, and by association f-stop, are lens characteristics independent of any sensor/image circle consideration.

Edit to add: OR, that's exactly right... the amount of light transmitted is "only" dependent of FL(recorded FOV/source) and aperture, it is not sensor size related... and spreading the recorded/source FOV over a larger sensor area means less light per sq mm.

A lot of the explanations of equivalence (including the DPR one linked) state that "a larger sensor has more area for each section of the scene." And that is true... it's exactly the same as saying the source area is spread over a larger area (image circle). I.e less light/mm but more mm overall. It is not some kind of "benefit" as they make it out to be.



How did you get that from anything I wrote? The only thing I said (in a round about way) is that any given aperture (t-stop) generates the same total exposure for a given source illuminance (captured FOV).

This is probably the best explanation of things that I have seen... be prepared for a lot of technical geekery (it can be kind of confusing)...

A quote from your link:

The only factors in the exposure are the scene luminance, f-ratio, shutter speed, and transmissivity of the lens (note that neither sensor size nor ISO are factors in exposure).

 

[HoppyUK]

A quote from your link:

The only factors in the exposure are the scene luminance, f-ratio, shutter speed, and transmissivity of the lens (note that neither sensor size nor ISO are factors in exposure).

Perhaps we can all agree on that

Just to avoid confusion though, 'exposure' in this context does not mean 'correct exposure' in terms of the final image - it simply refers to the amount of light/photons falling on the sensor on a per sq mm basis. In practical terms, correct exposure requires a certain amount of light/photons and a certain amount of signal gain applied electronically (amplification, or ISO) for the final result.

 

[sk66]

@sk66.

The deduction from your 'spreading the light over a larger sensor' statement would be that the sunny 16 rule is sensor size dependant. It isn't.

No, it's exactly the reason why it does work...

Ponder this question:
If a larger sensor does receive more light for a given recorded FOV, what happens to it when I normalize that larger/brighter area to the same size output/display? Do we just discard it?

I know my statement(s) are somewhat "controversial," and to that extent I started a discussion the other day in the "Science/Technology" section on DPR where the engineers/designers/uber geeks hang out.

And, I agree with the quoted excerpt.
But there is a common misunderstanding as to what lens transmissivity actually means. It has zero to do with a sensor or how the transmitted light is utilized after the fact (i.e. it does not maintain "density" per se). It is a measure of total light/light loss (FOV/illuminance received vs total output) and is in fact "a ratio." And any lens that transmits more/less of the light received for a given setting has a different transmissivity ratio (t-stop).
Transmissivity actually takes into account (measures losses due to) FL (distance) and f-ratio (entrance pupil/apparent aperture more precisely) . That leaves the only factors in exposure as being transmissivity, scene luminance (total light w/in the FOV), and SS (f-ratio and transmissivity/t-ratio are both included because they may or may not be the same).

 

[Deleted member 49549]

I think I need to clarify a couple of your statements. Can we start with

• "a larger sensor receives less light per sq mm for a given exposure."

I think to best understand the error one of us is making, we need to get back to a simple model. My understanding of your proposition is that you are saying that as a sensor gets larger, all else being equal, it doesn’t receive more light.
Is my understanding correct?

If so, lets try a thought experiment with a piece of old fashioned film. Lets start with a typical 35mm frame size of 36x24mm. In the camera, it receives an amount of light that evenly exposes to a calibrated value. If i then cut this frame in half, all else being equal, it will receive half the light, and the film will be exposed in exactly the same manner as the larger frame. (lets ignore any reciprocity problems and assume the film has a perfectly linear response).

Your proposition means that the film will need a varying amount of light dependant on its area. I don’t think it does...

 

[LCPete]

It's the same shutter speed and iso for both. The f numbers are devised to ensure that the exposure is constant for a constant f number, regardless of the focal length.

Thanks Stephen sorry i missed your post amongst all the others

 

[sk66]

I think I need to clarify a couple of your statements. Can we start with

• "a larger sensor receives less light per sq mm for a given exposure."

I think to best understand the error one of us is making, we need to get back to a simple model. My understanding of your proposition is that you are saying that as a sensor gets larger, all else being equal, it doesn’t receive more light.
Is my understanding correct?

If so, lets try a thought experiment with a piece of old fashioned film. Lets start with a typical 35mm frame size of 36x24mm. In the camera, it receives an amount of light that evenly exposes to a calibrated value. If i then cut this frame in half, all else being equal, it will receive half the light, and the film will be exposed in exactly the same manner as the larger frame. (lets ignore any reciprocity problems and assume the film has a perfectly linear response).

Your proposition means that the film will need a varying amount of light dependant on its area. I don’t think it does...

You are correct... given "all else being equal" also means the same source area (lens FOV/t-ratio/total light). That is essentially "cropping/crop factor" and the smaller area does indeed get less light.

 

[Deleted member 49549]

You are correct... given "all else being equal" also means the same source area (lens FOV/t-ratio/total light). That is essentially "cropping/crop factor" and the smaller area does indeed get less light.

So what is the cause of the larger sensor receiving less light per area?

 

[HoppyUK]

You are correct... given "all else being equal" also means the same source area (lens FOV/t-ratio/total light). That is essentially "cropping/crop factor" and the smaller area does indeed get less light.

In which case, we are all in agreement and the confusion all along has just been semantics! I alluded to that earlier, based mainly on my belief that Steven generally knows what he's talking about

It's a same problem I often find with complicated stuff - it's very hard to explain clearly and easily. Like one of the links Steven posted - badly written in pseudo-scientific language that can be as impenetrable as legal documents on occasion, even apparently contradictory in places

 

[sk66]

So what is the cause of the larger sensor receiving less light per area?

When you have the same source area (different FOV, same t-ratio/total light). If you spread a given source area (total light) over a larger area it has a lower flux density (same brightness).

Most think that using a longer lens with the same t-ratio means the longer lens is letting more light through. That is not correct. The same t-ratio means it is letting thru the same amount of total light received. The "more light" due to a larger entrance pupil is only to compensate for the greater light loss thru the longer optical path. A lens does not, and can not increase the amount of light emanating from a source area. And if a lens does transmit more light for a given setting (aperture) it has a different transmissivity (t-ratio) rating.

The common explanation is that a constant t-ratio maintains flux density (photons/area) but it does not. It maintains a constant brightness, which is not the same thing. But it is in fact "brightness/relative brightness" that we are photographing due to variable flux densities.

 

[HoppyUK]

So what is the cause of the larger sensor receiving less light per area?

It doesn't. Not in your example.

My understanding would be that a larger sensor can collect the same amount of total light/photons at a lower brightness level than a smaller sensor, simply due to greater surface area. Eg double the sensor area and halve the brightness (one stop less) results in the same total light/photon capture.

This is one aspect of the basic equivalence explained in the DPReview article I linked earlier. It means, for example, that an APS-C sensor at ISO100 has the same noise and dynamic range performance as a full-frame sensor at ISO200 (very roughly).

 

[HoppyUK]

When you have the same source area (different FOV, same t-ratio/total light). If you spread a given source area (total light) over a larger area it has a lower flux density (same brightness).

Most think that using a longer lens with the same t-ratio means the longer lens is letting more light through. That is not correct. The same t-ratio means it is letting thru the same amount of total light received. The "more light" due to a larger entrance pupil is only to compensate for the greater light loss thru the longer optical path. A lens does not, and can not increase the amount of light emanating from a source area. And if a lens does transmit more light for a given setting (aperture) it has a different transmissivity (t-ratio) rating.

The common explanation is that a constant t-ratio maintains flux density (photons/area) but it does not. It maintains a constant brightness, which is not the same thing. But it is in fact "brightness/relative brightness" that we are photographing due to variable flux densities.

What does that mean in plain English in three sentences?

 

[sk66]

My understanding would be that a larger sensor can collect the same amount of total light/photons at a lower brightness level than a smaller sensor, simply due to greater surface area. Eg double the sensor area and halve the brightness (one stop less) results in the same total light/photon capture.

That is the common understanding and I contend that it is "wrong" when (and only when) both sensors are receiving the same total light (source FOV/transmittance) and are recording the same scene.

Exposure is simply a "photon count" and sensor size/pixel size is irrelevant as such. If I take the same scene and spread it over a larger area (with a lens of the same transmissivity)it has lower photon density but the same total light and brightness (the ISL) resulting in the same exposure. If the larger sensor has the same MP count the individual pixels will be proportionally larger, and as a result each pixel will count (receive) the same number of photons giving the same per pixel exposure (as the same scene on a smaller sensor).

The *advantages larger sensors have is (generally) larger pixels which (generally) equates to a lower noise floor (easier to count photons), (generally) greater fill factor, a larger FWC, and less susceptibility to shot noise in low light.
(*in terms of light)

 

[woof woof]

It doesn't. Not in your example.
My understanding would be that a larger sensor can collect the same amount of total light/photons at a lower brightness level than a smaller sensor, simply due to greater surface area. Eg double the sensor area and halve the brightness (one stop less) results in the same total light/photon capture.

I was an electronics engineer not an optical one, but that will never stop me sticking my ore in

The way I've always looked at it is that if you take a picture with either film or a digital sensor and proceed to cut it in half both halves will retain the same image quality and any loss in image quality will be due to any additional magnification to produce the final image from the smaller film/sensor? Yes? No?

To me cutting a film or sensor in half is no different to cutting a print in half, or folding it in half or putting a sheet of paper over half of it or your display screen. None of these things should affect the image quality and any loss in image quality should be due only to the effect of enlarging the picture back to its original size.

FF cameras generally give better image quality than crop cameras and I can see that this is true for me when I compare my FF A7 to my MFT cameras. But. All things are not equal and the only real way for me to compare quality between the two systems is to not do it as the cameras are just too different. However, cropping an A7 image confirms my view of life, the universe and everything and in my world when I crop an A7 image the only way I can then degrade the image quality is to enlarge it.

So, if my MFT cameras had the exact same innards as my A7 they'd give the exact same image quality even if the sensor was cut in half and any reduction in image quality would be due to the additional enlargement that the smaller picture needed.

As an electronics engineer, now retired, what I will say is that if some people who have never done so were to look at a signal on an oscilloscope they'd (probably) be amazed at the difference a length of track or even the shape of that track can make. To me the design, components, materials used and layout of electronic chips and circuits (including the size of the pixels and where the A/D converter is etc) is the key and something that many people possibly don't give sufficient weight to and it's these things that to me make the difference.

 

[sk66]

The way I've always looked at it is that if you take a picture with either film or a digital sensor and proceed to cut it in half both halves will retain the same image quality and any loss in image quality will be due to any additional magnification to produce the final image from the smaller film/sensor? Yes? No?

Yes, basically.
The main difference in sensor size in terms of light is that it is easier to achieve a higher QE (quantum efficiency) with a larger sensor with any given level of technology. Combine that with the inherent greater contrast/sharpness of larger sensors, and the reduced amount of enlargement required, they (almost) always produce a higher total IQ.

 

[woof woof]

Yes, basically.
The main difference in sensor size in terms of light is that it is easier to achieve a higher QE (quantum efficiency) with a larger sensor with any given level of technology. Combine that with the inherent greater contrast/sharpness of larger sensors, and the reduced amount of enlargement required, they (almost) always produce a higher total IQ.

Yup,

I think that when using the smaller systems it helps to use good lenses specifically designed for that smaller system as this can go some way to mitigating the inherent quality disadvantage their smaller size gives them. Luckily MFT is quite well served with some excellent lenses but I think that this is one area in which Canon/Nikon could do better as they have (arguably) less well developed APS-C lens line ups and some users will be using lenses not specifically designed to get the best out of the smaller system.