[MarkJerling]

This is only the lens part of the equation.
When you capture the images you also add shutter speed and sensor area to the mix.
Longer shutter speed captures more light, a larger sensor capture a larger area of the same illuminance.
The amount of light that images are made of is: illuminance x sensor area x shutter speed.

So with same illuminance and shutter speed FF capture 2.25x the light of APS-C in the images.

No, it's not only "part of the equation", in terms of the OP's question!

Let's look at that again:

My question is:
When using my SMC Pentax 55mm F1.8 @ 1.8 on a APS-C camera does that sensor get approximately the same amount of light as using it on a FF @ the same f:stop?

The answer is clear: The area of the sensor which is exposed to light gets the same amount of light in any format, using the same lens, which explains why a light meter will give the same answer for any number of formats that are able to accept the same lens.

[Fogel70]

Why do you not include all questions the OP had? Why not also in clude this question?

When using my SMC Pentax 55mm F1.8 @ 1.8 on a APS-C camera does that sensor get approximately the same amount of light as using it on a FF @ the same f:stop?

Which is about exposure and sensor size.

[GUB]

Why do you not include all questions the OP had? Why not also in clude this question?

Which is about exposure and sensor size.

It is pretty clear that he and his buddy are talking about exposure. But lets say he was a secret equivalence fan and meant total quantity of light over the image circle. What difference does it make in photography?

[Fogel70]

It is pretty clear that he and his buddy are talking about exposure. But lets say he was a secret equivalence fan and meant total quantity of light over the image circle. What difference does it make in photography?

It means that at the same exposure FF captures images with higher dynamic range, lower noise. better color reproductionand other image paramters. (and possibly also higher resolution)
The whole photgraphic industry seems to have gone full monty becase users are willing to pay a lot extra for this.

Which is a little frustrating for us that are willing to sacrifice a little on IQ for a smaller and more portable APS-C system.

[GUB]

Ummm Dynamic Range parameters are based at the pixel level. Noise is based at the pixel level. Any RGB value issued is at the pixel level. So essentially no - the size of the image circle is irrelevant to them. Try again

[beholder3]

. And each pixel, when using the same lens, whether on APS-C or FF is receiving the same amount of light, that light having passed through the same aperture and the same lens. So, by your argument, every sensor is throwing away light. But that does not alter the amount of light on any given part of the sensor that is exposed to light.

The one question on exposure times has been answered early on. No argueing on that. It does cover the light per area aspect.

Light per area is not what will define the photo results alone. Here total amount of light gathered and magnification are key as well.

With my previous comment you quoted I was just pointing out that a FF lens on aAPSC body smacks a lot of captured photons against the backside innards of the mirror box instead of on the sensor when not using a wide angle converter. All this previously captured light is lost.
No issue if you magnify your output 1.5 times less than then the FF image. If you use same print sizes you will see the differences in technical image quality.

And this happens with wet plate analog film captures compared to tiny crop factor FF just the same.

---------- Post added 23rd Aug 2019 at 09:40 ----------

Ummm Dynamic Range parameters are based at the pixel level. Noise is based at the pixel level. Any RGB value issued is at the pixel level. So essentially no - the size of the image circle is irrelevant to them. Try again

I do expect every person on a photography forum to focus on dynamic range and noise in a complete photo.
So that is where they „are based at“ usually.

[GUB]

No issue if you magnify your output 1.5 times less than then the FF image. If you use same print sizes you will see the differences in technical image quality.

No understanding of optics I see. The apsc lens image circle is a cropped version of the FF circle, not a shrunk one.

I do expect every person on a photography forum to focus on dynamic range and noise in a complete photo.
So that is where they „are based at“ usually.

And sourced from the pixel

[Dartmoor Dave]

The amount of light that images are made of is: illuminance x sensor area x shutter speed.

Yep, that's the crux of the problem right there. So let's amend it to state how a digital camera actually works:

The amount of light that images are made of is: illuminance per pixel x shutter speed.

On each pixel, a photodiode converts photons into electrons. Then (on each pixel on a CMOS sensor) readout electronics measure the accumulated charge. Then an analog-to-digital converter turns that measurement into a digital value. A bigger surface area of the photodetector per pixel will convert more photons into electrons, so that the charge that we actually want to measure will be easier to distinguish from the sensor's inevitable background noise. The sensor's dynamic range will be higher.

Now there's two ways to get that bigger surface area per photodetector. You can have a smaller sensor with fewer megapixels but a bigger surface area per pixel, or you can have a bigger sensor with more megapixels and also a bigger surface area per pixel. But whatever balance you decide to strike between megapixels and individual photodetector surface area, the signal-to-noise ratio is determined at the pixel level. The only thing that's affected by the amount of light hitting the overall surface area of the sensor is exposure, and all but the most lunatic fringe of equivalentists accept that exposure is the same on different sensor formats.

So can we please be clear about this once and for all: a lens that collects more light has no direct impact on signal-to-noise ratio. A lens with a wider maximum aperture means that you can get away with using lower ISO in lower light, and indeed that does have an impact on the signal-to-noise ratio. But only because of the lower ISO, which means less amplification at the pixel level.

I notice that magnification has also been mentioned, so let's go back and look at those pixels again. Remember that the accumulated charge per pixel is measured and converted to a digital value? The analog-to-digital converter doesn't include the physical size of the pixel, and certainly not the physical size of the sensor, in the data it records. It doesn't need to. All it needs is a digital value that records the measured charge for each individual pixel of whatever resolution in megapixels the sensor happens to have.

And now, finally, you want to look at the photo on your monitor. Those digital values per pixel from the sensor are downsampled to your monitor resolution and displayed as a luminance level (and of course a colour) per pixel on the monitor. The physical size of the sensor that took the photo, and even the physical size of the monitor, are irrelevant. All that matters is the pixel resolution. If you had a 24 megapixel APS-C sensor and a 24 megapixel FF sensor with the same signal-to-noise ratio per pixel, the resulting images would be indistinguishable in terms of noise.

The only time you are ever magnifying or enlarging a digital photo is if you upsample it to a higher viewing resolution than the original sensor resolution.

The thing is, there's an actual way that cameras work, and it's well understood, and it isn't open to debate. It's just the way they work.

[beholder3]

So can we please be clear about this once and for all: a lens that collects more light has no direct impact on signal-to-noise ratio.

That is wrong.

If you plug a 50mm f1.4 lens with apsc image circle on a apsc sensor camera you will get SNR A. Here both the image circle and the sensor size are limiting bottlenecks.
If you use a FF image circle 50mm f1.4 lens which gathers more light in total ( not per unit of area) and add a wide angle converter to it you get SNR B. And B will be one stop better than A.
To be clear: SNR is based on the noise in the whole image. Both lenses shot at F1.4. Same shutter times.

---------- Post added 23rd Aug 2019 at 11:11 ----------

The only time you are ever magnifying or enlarging a digital photo is if you upsample it to a higher viewing resolution than the original sensor resolution.
)

The Apsc sensor is 2;25 times smaller than a FF sensor. And you print to what size?

There is a huge magnification going on with every digital photo. Or do you watch your apsc photos on a screen sized 16x24 mm?

[pschlute]

So with same illuminance and shutter speed FF capture 2.25x the light of APS-C in the images.

...and presumably by this logic you can prove that a view camera captures a humongous amount of light more than an aps-c sensor ?

Doesn't alter the fact that the "exposure" is exactly the same (shutter speed/f stop/ISO)

[GUB]

That is wrong.

If you plug a 50mm f1.4 lens with apsc image circle on a apsc sensor camera you will get SNR A. Here both the image circle and the sensor size are limiting bottlenecks.
If you use a FF image circle 50mm f1.4 lens which gathers more light in total ( not per unit of area) and add a wide angle converter to it you get SNR B. And B will be one stop better than A.
To be clear: SNR is based on the noise in the whole image. Both lenses shot at F1.4. Same shutter times.

We have been through all this before - add a speed booster to a 50 1.4 and you have a 35 f1 --- go back to post 98 and start reading. And if you have used "same shutter times" the speedbooster + lens will be overexposed by a stop.

[GUB]

The Apsc sensor is 2;25 times smaller than a FF sensor. And you print to what size?

There is a huge magnification going on with every digital photo. Or do you watch your apsc photos on a screen sized 16x24 mm?

It would be quite cool to have a monitor at 7372:4924 resolution. But I don't think we are there yet even with large screen tvs are we?

[GUB]

Yep, that's the crux of the problem right there. So let's amend it to state how a digital camera actually works:

The amount of light that images are made of is: illuminance per pixel x shutter speed.

On each pixel, a photodiode converts photons into electrons. Then (on each pixel on a CMOS sensor) readout electronics measure the accumulated charge. Then an analog-to-digital converter turns that measurement into a digital value. A bigger surface area of the photodetector per pixel will convert more photons into electrons, so that the charge that we actually want to measure will be easier to distinguish from the sensor's inevitable background noise. The sensor's dynamic range will be higher.

Now there's two ways to get that bigger surface area per photodetector. You can have a smaller sensor with fewer megapixels but a bigger surface area per pixel, or you can have a bigger sensor with more megapixels and also a bigger surface area per pixel. But whatever balance you decide to strike between megapixels and individual photodetector surface area, the signal-to-noise ratio is determined at the pixel level. The only thing that's affected by the amount of light hitting the overall surface area of the sensor is exposure, and all but the most lunatic fringe of equivalentists accept that exposure is the same on different sensor formats.

So can we please be clear about this once and for all: a lens that collects more light has no direct impact on signal-to-noise ratio. A lens with a wider maximum aperture means that you can get away with using lower ISO in lower light, and indeed that does have an impact on the signal-to-noise ratio. But only because of the lower ISO, which means less amplification at the pixel level.

I notice that magnification has also been mentioned, so let's go back and look at those pixels again. Remember that the accumulated charge per pixel is measured and converted to a digital value? The analog-to-digital converter doesn't include the physical size of the pixel, and certainly not the physical size of the sensor, in the data it records. It doesn't need to. All it needs is a digital value that records the measured charge for each individual pixel of whatever resolution in megapixels the sensor happens to have.

And now, finally, you want to look at the photo on your monitor. Those digital values per pixel from the sensor are downsampled to your monitor resolution and displayed as a luminance level (and of course a colour) per pixel on the monitor. The physical size of the sensor that took the photo, and even the physical size of the monitor, are irrelevant. All that matters is the pixel resolution. If you had a 24 megapixel APS-C sensor and a 24 megapixel FF sensor with the same signal-to-noise ratio per pixel, the resulting images would be indistinguishable in terms of noise.

The only time you are ever magnifying or enlarging a digital photo is if you upsample it to a higher viewing resolution than the original sensor resolution.

The thing is, there's an actual way that cameras work, and it's well understood, and it isn't open to debate. It's just the way they work.

Too many words for them Dave. They are not up to absorbing information. Dreams they can do.

[beholder3]

We have been through all this before - add a speed booster to a 50 1.4 and you have a 35 f1 --- go back to post 98 and start reading. And if you have used "same shutter times" the speedbooster + lens will be overexposed by a stop.

Exactly, the combo now behaves like a 35 f1 and guess what? this does collect more light than the apsc 50 1.4 alone. And it is able to do this solely because the base FF 50 1.4 had a larger image circle collecting more light.

The image will not be overexposed if you chose lower ISO. Anyhow by stating this you admit that the lens combo has captured more light. Which was all I explained.

[GUB]

Anyhow by stating this you admit that the lens combo has captured more light

But the combo is now a 35mm f1 - why can't you grasp that you have literally put a different lens on with consequently different result. You have not had time to go back and read that we have already covered this. So stop talking nonsense and read up.