"Total Light Gathered" - how is it calculated?

xpatUSA · Nov 8, 2013

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

RussellInCincinnati · Nov 8, 2013

xpatUSA said:
xpatUSA wrote: In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out. How is it calculated and what are units of light thus gathered?

If you have a 50mm F/2 lens, you can say at a glance that the effective light-gathering lens aperture diameter is 25 millimeters, 50 divided by 2.0. And assuming that all lenses have the same transmission efficiency of maybe 90% or so (not too bad an assumption), you can come up with an easy-to-understand light-gathering index simply by squaring the 25 millimeter lens diameter. And then multiplying by the relative sensor area, assuming say that a 24x36mm "full frame" sensor has an area of "1".

So a 50mm F/2 lens imaging to a full frame sensor gathers about (50 divided by two) (then 25 squared) (equals) 600 units of light with a certain normal-ish angle of view. That same lens used with a micro four thirds sensor would gather about 150 units of light (since the final sensor size multiplier is 0.25), with twice-as-narrow angle of view.

This kind of rough calculation helpfully leads quickly to seeing that the micro four thirds camera normal lens, is going to have to be 25mm F/1.0 lens, to "gather the same amount of light" as a full-frame optic. Thus u43 lenses need to be "a lot faster" than their full frame counterparts to get you the same low-light capability.

Great Bustard · Nov 8, 2013

xpatUSA said:
In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

Here we are:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

Mathematically, we can express these four quantities rather simply:

Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)
Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)
Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)
Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400. However, the brightness for the two photos will be the same for a given scene since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

DSPographer · Nov 9, 2013

Great Bustard said:
xpatUSA said:

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

Click to expand...

Here we are:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

Mathematically, we can express these four quantities rather simply:

Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)

Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)

Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)

Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400. However, the brightness for the two photos will be the same for a given scene since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

This doesn't take into account the light lost because of vignetting. The exposure is only correct at the image center, with fast lenses the image boundaries could lose 2 stops or more of light versus the center. So we really should include a loss factor caused by the vignetting of the lens.

xpatUSA · Nov 9, 2013

DSPographer said:
Great Bustard said:

Here we are:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

Mathematically, we can express these four quantities rather simply:

Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)

Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)

Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)

Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400. However, the brightness for the two photos will be the same for a given scene since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

Click to expand...

This doesn't take into account the light lost because of vignetting. The exposure is only correct at the image center, with fast lenses the image boundaries could lose 2 stops or more of light versus the center. So we really should include a loss factor caused by the vignetting of the lens.

If I understand responses so far, the concensus is that "light gathered" is indeed an amount rather than a flux. And the sensor is usually included as an element in the light "gathering" process; thus a FF lens on a FF camera gathers more light than an APS-C lens (i.e. vignetting in extremis) on the same camera. Thus the light gathered would be the integral of the exposure with respect to sensor total area (poorly put, calculus is a foreign language to me). It's like saying the exposure for each pixel all added up gives the light gathered for the whole sensor. That would account for vignetting as opposed to simply multiplying the exposure per unit area by the sensor total area.

Interesting that James ends up with electrons as the unit of gathered light, not photons. Odd, that. And his use of units of electrons per photon for QE is a bit shaky although we all know what he means, I suppose.

GeorgianBay1939 · Nov 9, 2013

xpatUSA said:
DSPographer said:

Great Bustard said:

Here we are:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

Mathematically, we can express these four quantities rather simply:

Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)

Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)

Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)

Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400. However, the brightness for the two photos will be the same for a given scene since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

Click to expand...

This doesn't take into account the light lost because of vignetting. The exposure is only correct at the image center, with fast lenses the image boundaries could lose 2 stops or more of light versus the center. So we really should include a loss factor caused by the vignetting of the lens.

Click to expand...

If I understand responses so far, the concensus is that "light gathered" is indeed an amount rather than a flux.

You've put your finger on a source of much confusion, heat and noise by various debaters here.

It would safer (and wiser IMHO) to use the term "total light gathered" to clearly indicate the integral of exposure wrt area. .... which results in an "amount of light", not a flux or a light density.

xpatUSA said:
And the sensor is usually included as an element in the light "gathering" process; thus a FF lens on a FF camera gathers more light than an APS-C lens (i.e. vignetting in extremis) on the same camera.

The "total light gathered" will depend on the (effective) area of the sensor since it is the integral of exposure over the (effective) area of the sensor. The size of the sensor form the "limits" of the integration.

You could also deal with the vignetting by using the T/, which compensates for loss of light during transmission through the lens. That would also work if you were using averages of exposure over the sensor surface.

xpatUSA said:
Thus the light gathered would be the integral of the exposure with respect to sensor total area (poorly put, calculus is a foreign language to me). It's like saying the exposure for each pixel all added up gives the light gathered for the whole sensor. That would account for vignetting as opposed to simply multiplying the exposure per unit area by the sensor total area.

Sure.

xpatUSA said:
Interesting that James ends up with electrons as the unit of gathered light, not photons. Odd, that.

He is speaking using reality. Since it is pretty difficult to gather up a "bunch of photons"! So he ( I suspect) is using the "useful effect" caused by a "bunch of photons" by saying Total Light collected = photons collected X Quantum efficiency of the sensor to give electrons. Electrical engineers might use signal, opticians might use an area and time integral of some variable listed here ;-) . I don't know, but GB seems to think like a physicist, ie as simply as possible.

xpatUSA said:
And his use of units of electrons per photon for QE is a bit shaky although we all know what he means, I suppose.

??? What else? (Efficiency (output/input) is clear and so are the (important) measurables.)

xpatUSA said:
--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

The_Suede · Nov 9, 2013

xpatUSA said:
In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

If you want a really simplified version...

This image is a part of the simplified object field. It's an infinitely large grid/wall of perfect point lights spaced at an even distance from each other. They all have the same intensity. We're aiming a 2x crop camera with a 50mm lens straight at it, and get this:

2x crop sensor - 50mm lens - f/4.0 aperture

Each individual point included in the image will send light to/through the optical system. How much light from each point that will actually pass through the optical system is determined by the aperture area, or more accurately the front pupil area. Nothing else - at least not in the simplified version.

Let's just say/assume that the aperture used was f/4.0, and that the intensity of the point lights in front of the system each contribute 1 (one) photon to the image with the front pupil area you get at 50mm F4.0. Then you have a total of 150 photons on the image plane (10x15 points are included, 1 photon from each). That was the "total light gathered".

Now - with the same camera - change the lens to a 25mm (half focal length), while keeping aperture at f/4.0. What happens? The field of view widens, and you get four times as many light points included in the image plane.

2x crop sensor - 25mm lens - f/4.0 aperture

-But since the front pupil area is now four times smaller (half focal length, same aperture > four times smaller front pupil area) - you still get the same total amount of light passed through!

Four times as many (150x4 = 600) points of light that each contribute four times less (1/4 = 0.25) photons to the image = still a total of (600x0.25) = 150 photons on the sensor / image plane! Same as with the 50mm F4.0 lens.

That's why the "f/stop" system works - it keeps image plane exposure (light per area on the image plane) constant if you keep the f/# constant - no matter what focal length you choose.
Longer focal length > larger front pupil area, but fewer points of light included.
Shorter focal length > smaller front pupil area, but more points of light included.

But what happens if we increase the included angle of view by using a 2x larger image plane/sensor (four times larger area) with the 50mm lens in stead? Well, you still include four times as many light points, but the front pupil area doesn't change. So each point still gives the same amount of light on the image plane.

1x sensor - 50mm lens - f/4.0 aperture

So now you have 600 (20x30) points of light included again - just as with the 2x crop sensor and a 25mm lens - but in this case each point still contributes 1 photon to the sensor/image plane - resulting in a total photon count of 600x1.0 = 600 in stead of 600x0.25 = 150.

...................

Then there are some losses, like cos4, mechanical vignetting, optical reflection/absorption and so on - but that's rather beside the point from a purely "simple system" point of view.

The two basic parts are really:

f/# determines image plane exposure
exposure is "amount of light energy per area unit"

-So you get the light energy amount that your system "gathered" by:

(scene average luminance) / (f-stop)^2 * (sensor area) * (exposure time)

And from that you can get that the system "light gathering ability" when keeping scene average luminance and exposure time constant is:

Light gathering ability = (sensor area) / (f-stop)^2

Since (sensor area) is the square function of crop ratio this can be further simplified by doing a square root on both, and then you get:

Light gathering ability = (crop ratio) / (f-stop)

In the end that means that to collect as many photons per second from a certain scene, you need to change f/# with the same factor as you change the crop ratio.

Great Bustard · Nov 9, 2013

DSPographer said:
Great Bustard said:

xpatUSA said:

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

Click to expand...

Here we are:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

Mathematically, we can express these four quantities rather simply:

Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)

Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)

Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)

Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400. However, the brightness for the two photos will be the same for a given scene since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

Click to expand...

This doesn't take into account the light lost because of vignetting.

Actually, it does. The exposure (the density of the light falling on the sensor) is not uniform for any photo.

DSPographer said:
The exposure is only correct at the image center, with fast lenses the image boundaries could lose 2 stops or more of light versus the center. So we really should include a loss factor caused by the vignetting of the lens.

What you're saying is that, due to vignetting, more light will fall in the middle of the photo than the edges. This is correct. But the formulas above are still correct, if we take the various measures to be averages. That is, average exposure, average scene luminance, etc. Clearly, the greater the vignetting, the lower the average exposure will be for a given scene luminance, f-ratio, and shutter speed, and thus less total light.

Great Bustard · Nov 9, 2013

xpatUSA said:
DSPographer said:

Great Bustard said:

Here we are:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

Mathematically, we can express these four quantities rather simply:

Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)

Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)

Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)

Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400. However, the brightness for the two photos will be the same for a given scene since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

Click to expand...

This doesn't take into account the light lost because of vignetting. The exposure is only correct at the image center, with fast lenses the image boundaries could lose 2 stops or more of light versus the center. So we really should include a loss factor caused by the vignetting of the lens.

Click to expand...

If I understand responses so far, the concensus is that "light gathered" is indeed an amount rather than a flux.

Correct. For example, the total amount of water collected by a cup placed out in the rain would be measured in cm³, not cm.

xpatUSA said:
And the sensor is usually included as an element in the light "gathering" process; thus a FF lens on a FF camera gathers more light than an APS-C lens (i.e. vignetting in extremis) on the same camera.

For the same exposure, yes.

xpatUSA said:
Thus the light gathered would be the integral of the exposure with respect to sensor total area (poorly put, calculus is a foreign language to me).

Exactly correct.

xpatUSA said:
It's like saying the exposure for each pixel all added up gives the light gathered for the whole sensor.

Again, exactly correct. Replace the integral with a sum. ;-)

xpatUSA said:
That would account for vignetting as opposed to simply multiplying the exposure per unit area by the sensor total area.

As I said to DSPographer, the exposure is not constant across the sensor regardless, so the formulas still hold if you speak in terms of averages (e.g. average exposure, average scene luminance, etc.)

xpatUSA said:
Interesting that James...

That's me, by the way.

xpatUSA said:
...ends up with electrons as the unit of gathered light, not photons.

The Total Light Collected is the Signal that results from the light falling on the sensor, and is thus measured in electrons.

xpatUSA said:
Odd, that. And his use of units of electrons per photon for QE is a bit shaky although we all know what he means, I suppose.

The QE (Quantum Efficiency) is the proportion of photons falling on the sensor that are recorded, and they are recorded by the means of the electrons they release in the silicon. Thus, a QE of unity means that one photon releases one electron. Modern sensors have a QE of around 50% (sensors with the newer BSI tech have QEs in the 75% range), so it takes two photons, on average, to release one electron.

Great Bustard · Nov 9, 2013

The_Suede said:
xpatUSA said:

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

Click to expand...

If you want a really simplified version...

This image is a part of the simplified object field. It's an infinitely large grid/wall of perfect point lights spaced at an even distance from each other. They all have the same intensity. We're aiming a 2x crop camera with a 50mm lens straight at it, and get this:

2x crop sensor - 50mm lens - f/4.0 aperture

Each individual point included in the image will send light to/through the optical system. How much light from each point that will actually pass through the optical system is determined by the aperture area, or more accurately the front pupil area. Nothing else - at least not in the simplified version.

Let's just say/assume that the aperture used was f/4.0, and that the intensity of the point lights in front of the system each contribute 1 (one) photon to the image with the front pupil area you get at 50mm F4.0. Then you have a total of 150 photons on the image plane (10x15 points are included, 1 photon from each). That was the "total light gathered".

Now - with the same camera - change the lens to a 25mm (half focal length), while keeping aperture at f/4.0. What happens? The field of view widens, and you get four times as many light points included in the image plane.

2x crop sensor - 25mm lens - f/4.0 aperture

-But since the front pupil area is now four times smaller (half focal length, same aperture > four times smaller front pupil area) - you still get the same total amount of light passed through!

Four times as many (150x4 = 600) points of light that each contribute four times less (1/4 = 0.25) photons to the image = still a total of (600x0.25) = 150 photons on the sensor / image plane! Same as with the 50mm F4.0 lens.

That's why the "f/stop" system works - it keeps image plane exposure (light per area on the image plane) constant if you keep the f/# constant - no matter what focal length you choose.
Longer focal length > larger front pupil area, but fewer points of light included.
Shorter focal length > smaller front pupil area, but more points of light included.

But what happens if we increase the included angle of view by using a 2x larger image plane/sensor (four times larger area) with the 50mm lens in stead? Well, you still include four times as many light points, but the front pupil area doesn't change. So each point still gives the same amount of light on the image plane.

1x sensor - 50mm lens - f/4.0 aperture

So now you have 600 (20x30) points of light included again - just as with the 2x crop sensor and a 25mm lens - but in this case each point still contributes 1 photon to the sensor/image plane - resulting in a total photon count of 600x1.0 = 600 in stead of 600x0.25 = 150.

...................

Then there are some losses, like cos4, mechanical vignetting, optical reflection/absorption and so on - but that's rather beside the point from a purely "simple system" point of view.

The two basic parts are really:

f/# determines image plane exposure

exposure is "amount of light energy per area unit"

-So you get the light energy amount that your system "gathered" by:

(scene average luminance) / (f-stop)^2 * (sensor area) * (exposure time)

And from that you can get that the system "light gathering ability" when keeping scene average luminance and exposure time constant is:

Light gathering ability = (sensor area) / (f-stop)^2

Since (sensor area) is the square function of crop ratio this can be further simplified by doing a square root on both, and then you get:

Light gathering ability = (crop ratio) / (f-stop)

In the end that means that to collect as many photons per second from a certain scene, you need to change f/# with the same factor as you change the crop ratio.

Now add to it that the photon noise is the square root of the total amount of light collected (signal), and the NSR (Noise-to-Signal Ratio) is inversely proportional to the total amount of light falling on the sensor. For example, quadruple the amount of light falling on the sensor, and you halve the photon noise (there is still read noise to contend with, of course, when the shadows are heavily pushed and/or the ISO setting is very high).

Ian Stuart Forsyth · Nov 9, 2013

Great insight and thank you for the work put into this

xpatUSA · Nov 10, 2013

Great Bustard said:
Great Bustard said:

Interesting that James...

Click to expand...

That's me, by the way.

Great Bustard said:

...ends up with electrons as the unit of gathered light, not photons.

Click to expand...

The Total Light Collected is the Signal that results from the light falling on the sensor, and is thus measured in electrons.

With my engineering background, I can't agree with that. If it said:

"The Total Light Collected is [proportional] to the Signal that results from the light falling on the sensor, [which] is measured in electrons"; that would be more precise.

Point being that the units of light in the context of this thread are photons, not electrons.

Great Bustard said:
Great Bustard said:

Odd, that. And his use of units of electrons per photon for QE is a bit shaky although we all know what he means, I suppose.

Click to expand...

The QE (Quantum Efficiency) is the proportion of photons falling on the sensor that are recorded, and they are recorded by the means of the electrons they release in the silicon. Thus, a QE of unity means that one photon releases one electron. Modern sensors have a QE of around 50% (sensors with the newer BSI tech have QEs in the 75% range), so it takes two photons, on average, to release one electron.

Again, the Engineer in me rebels: Efficiency is dimensionless, it has no units other than relative ones like 'per cent'. You can not say 'photons per electron' because that creates a non-dimensionless unit which, by definition, can not be called 'efficiency'. Normally, the term for such as photons per electron would include the word 'specific', for example, 'specific power' for a gas turbine, e.g. bhp/lb/hr or W/l/s (if you must).

Of course we all know that electrons (>1) per photon is not possible in a camera image sensor and that a fractional photon is impossible anywhere.

--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

crames · Nov 10, 2013

The_Suede said:
xpatUSA said:

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

Click to expand...

If you want a really simplified version...

This image is a part of the simplified object field. It's an infinitely large grid/wall of perfect point lights spaced at an even distance from each other. They all have the same intensity. We're aiming a 2x crop camera with a 50mm lens straight at it, and get this:

2x crop sensor - 50mm lens - f/4.0 aperture

Each individual point included in the image will send light to/through the optical system. How much light from each point that will actually pass through the optical system is determined by the aperture area, or more accurately the front pupil area. Nothing else - at least not in the simplified version.

Let's just say/assume that the aperture used was f/4.0, and that the intensity of the point lights in front of the system each contribute 1 (one) photon to the image with the front pupil area you get at 50mm F4.0. Then you have a total of 150 photons on the image plane (10x15 points are included, 1 photon from each). That was the "total light gathered".

Now - with the same camera - change the lens to a 25mm (half focal length), while keeping aperture at f/4.0. What happens? The field of view widens, and you get four times as many light points included in the image plane.

But only in this artificial case which is like a flat field. Real world, when you zoom out you get something entirely different added to the periphery.

The_Suede said:
2x crop sensor - 25mm lens - f/4.0 aperture

-But since the front pupil area is now four times smaller (half focal length, same aperture > four times smaller front pupil area) - you still get the same total amount of light passed through!

You've just shown that on a 2x crop sensor, 25mm @ f/4 can give the same total light as 50mm @ f/4! Only because you have added additional light sources in the 25mm shot that were not in the original 50mm shot.

The_Suede said:
Four times as many (150x4 = 600) points of light that each contribute four times less (1/4 = 0.25) photons to the image = still a total of (600x0.25) = 150 photons on the sensor / image plane! Same as with the 50mm F4.0 lens.

If you just look at the central 150 points of light that were in the 50mm shot, you have the same number of points, each contributing 1/4 photons, but in 1/4 the area, giving the same photons/area (exposure), but 1/4 the "total light."

The_Suede said:
That's why the "f/stop" system works - it keeps image plane exposure (light per area on the image plane) constant if you keep the f/# constant - no matter what focal length you choose.
Longer focal length > larger front pupil area, but fewer points of light included.
Shorter focal length > smaller front pupil area, but more points of light included.

You should only be comparing the same scene elements. What more or fewer "points of light" are included in the field of view are completely arbitrary.

The_Suede said:
But what happens if we increase the included angle of view by using a 2x larger image plane/sensor (four times larger area) with the 50mm lens in stead? Well, you still include four times as many light points, but the front pupil area doesn't change. So each point still gives the same amount of light on the image plane.

1x sensor - 50mm lens - f/4.0 aperture

So now you have 600 (20x30) points of light included again - just as with the 2x crop sensor and a 25mm lens - but in this case each point still contributes 1 photon to the sensor/image plane - resulting in a total photon count of 600x1.0 = 600 in stead of 600x0.25 = 150.

...................

Then there are some losses, like cos4, mechanical vignetting, optical reflection/absorption and so on - but that's rather beside the point from a purely "simple system" point of view.

The two basic parts are really:

f/# determines image plane exposure

exposure is "amount of light energy per area unit"

-So you get the light energy amount that your system "gathered" by:

(scene average luminance) / (f-stop)^2 * (sensor area) * (exposure time)

And from that you can get that the system "light gathering ability" when keeping scene average luminance and exposure time constant is:

Light gathering ability = (sensor area) / (f-stop)^2

Since (sensor area) is the square function of crop ratio this can be further simplified by doing a square root on both, and then you get:

Light gathering ability = (crop ratio) / (f-stop)

Shouldn't that be sqrt(Light gathering ability) = (crop ratio) / (f-stop) ?

The_Suede said:
In the end that means that to collect as many photons per second from a certain scene, you need to change f/# with the same factor as you change the crop ratio.

xpatUSA · Nov 10, 2013

xpatUSA said:
Again, the Engineer in me rebels: Efficiency is dimensionless, it has no units other than relative ones like 'per cent'. You can not say 'photons per electron' because that creates a non-dimensionless unit which, by definition, can not be called 'efficiency'. Normally, the term for such as photons per electron would include the word 'specific', for example, 'specific power' for a gas turbine, e.g. bhp/lb/hr or W/l/s (if you must).

Oops, dyslexia rules, KO: it should read 'electrons per photon', duh

DSPographer · Nov 10, 2013

Great Bustard said:
What you're saying is that, due to vignetting, more light will fall in the middle of the photo than the edges. This is correct. But the formulas above are still correct, if we take the various measures to be averages. That is, average exposure, average scene luminance, etc. Clearly, the greater the vignetting, the lower the average exposure will be for a given scene luminance, f-ratio, and shutter speed, and thus less total light.

Yes, the formula is correct using average luminance. We just need to be careful because it is not conventional to list f-stops or even t-stops of a lens based on the average light over the entire sensor. The usual ratings are only correct at the image center. So, when we compare the total light gathered by an f/1.4 lens set to f/1.4 and f/2.8 the ratio is not 4: because the lens will have less vignetting when stopped down.

The_Suede · Nov 10, 2013

crames said:
The_Suede said:

xpatUSA said:

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

Click to expand...

If you want a really simplified version...

This image is a part of the simplified object field. It's an infinitely large grid/wall of perfect point lights spaced at an even distance from each other. They all have the same intensity. We're aiming a 2x crop camera with a 50mm lens straight at it, and get this:

2x crop sensor - 50mm lens - f/4.0 aperture

Each individual point included in the image will send light to/through the optical system. How much light from each point that will actually pass through the optical system is determined by the aperture area, or more accurately the front pupil area. Nothing else - at least not in the simplified version.

Let's just say/assume that the aperture used was f/4.0, and that the intensity of the point lights in front of the system each contribute 1 (one) photon to the image with the front pupil area you get at 50mm F4.0. Then you have a total of 150 photons on the image plane (10x15 points are included, 1 photon from each). That was the "total light gathered".

Now - with the same camera - change the lens to a 25mm (half focal length), while keeping aperture at f/4.0. What happens? The field of view widens, and you get four times as many light points included in the image plane.

Click to expand...

But only in this artificial case which is like a flat field. Real world, when you zoom out you get something entirely different added to the periphery.

Also, I could be in the center of the sun. Or the dots could be green. So? This is a uniform flatfield, discretized. It's a model, not a landscape scenario.

crames said:
crames said:

2x crop sensor - 25mm lens - f/4.0 aperture

-But since the front pupil area is now four times smaller (half focal length, same aperture > four times smaller front pupil area) - you still get the same total amount of light passed through!

Click to expand...

You've just shown that on a 2x crop sensor, 25mm @ f/4 can give the same total light as 50mm @ f/4! Only because you have added additional light sources in the 25mm shot that were not in the original 50mm shot.

Yes, and that is also what happens if you have a uniform gray target. Except then you have limes infinity points giving off 1/limes infinity of light. Which is harder to convey in a simplified form.

crames said:
crames said:

Four times as many (150x4 = 600) points of light that each contribute four times less (1/4 = 0.25) photons to the image = still a total of (600x0.25) = 150 photons on the sensor / image plane! Same as with the 50mm F4.0 lens.

Click to expand...

If you just look at the central 150 points of light that were in the 50mm shot, you have the same number of points, each contributing 1/4 photons, but in 1/4 the area, giving the same photons/area (exposure), but 1/4 the "total light."

Ummm.... Yes?

crames said:
crames said:

That's why the "f/stop" system works - it keeps image plane exposure (light per area on the image plane) constant if you keep the f/# constant - no matter what focal length you choose.
Longer focal length > larger front pupil area, but fewer points of light included.
Shorter focal length > smaller front pupil area, but more points of light included.

Click to expand...

You should only be comparing the same scene elements. What more or fewer "points of light" are included in the field of view are completely arbitrary.

Who are you to tell me what I should compare? I'm sorry I didn't use polar bears, if that's your preference. Suggestions?

Once again, this is a model for a flat field. Should you wish to get picky about it, compare the 2x crop 25F4.0 shot with the 1x crop 50F4.0, and voila - there you have identical targets, identical target coverage, and 4x more light energy per second through the system with one of teh systems - AND the reason why. What the questions was originally asking for. Difference in light gathering ability, and "why" in a slightly more formalized manner.

crames said:
crames said:

But what happens if we increase the included angle of view by using a 2x larger image plane/sensor (four times larger area) with the 50mm lens in stead? Well, you still include four times as many light points, but the front pupil area doesn't change. So each point still gives the same amount of light on the image plane.

1x sensor - 50mm lens - f/4.0 aperture

So now you have 600 (20x30) points of light included again - just as with the 2x crop sensor and a 25mm lens - but in this case each point still contributes 1 photon to the sensor/image plane - resulting in a total photon count of 600x1.0 = 600 in stead of 600x0.25 = 150.

...................

Then there are some losses, like cos4, mechanical vignetting, optical reflection/absorption and so on - but that's rather beside the point from a purely "simple system" point of view.

The two basic parts are really:

f/# determines image plane exposure

exposure is "amount of light energy per area unit"

-So you get the light energy amount that your system "gathered" by:

(scene average luminance) / (f-stop)^2 * (sensor area) * (exposure time)

And from that you can get that the system "light gathering ability" when keeping scene average luminance and exposure time constant is:

Light gathering ability = (sensor area) / (f-stop)^2

Since (sensor area) is the square function of crop ratio this can be further simplified by doing a square root on both, and then you get:

Light gathering ability = (crop ratio) / (f-stop)

Click to expand...

Shouldn't that be sqrt(Light gathering ability) = (crop ratio) / (f-stop) ?

Yes, maybe, depends on in what metric you "want" the unknown unit "light gathering ability" to be. Stops (log2)? Lin? Technical, like in radiometric in stead of photometric? Then you sub log10 in stead. I don't know - your choice.

Since it's a derived metric, you could try and go for some standardization. There are organizations handling this should you want to take it further.

crames said:
crames said:

In the end that means that to collect as many photons per second from a certain scene, you need to change f/# with the same factor as you change the crop ratio.

Click to expand...

Jack Hogan · Nov 10, 2013

xpatUSA said:
In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

Here is my attempt, and when we talk about light gathered I like to think about an actual number of photons, so here it goes:

A light source provides light to the scene. The light is reflected (in Lambertian fashion we assume) by it and heads for the lens. Photographers call this light Luminance. The portion of this light that falls within the Field of View of the camera/lens (theta/theta') and makes it through the lens (transmittance, vignetting) arrives at the sensing medium plane, where the sensor sits in DSCs.

The voyage of light from source to sensor

The light goes through the various filters on top of the sensor and hits the sensing material, mostly silicon. While silicon is exposed to the light it typically produces a certain number of electrons per unit area every second (according to its effective QE). These electrons are captured, scaled (by camera gain, set by in-camera 'ISO') and counted by the sensor and downstream electronics (amplifiers, ADC), with the resulting values written to the Raw file.

The formulas are in the pictures above and as you can see are a function of unit area, so the larger the sensing area the more the photons (err light) collected. One can calculate fairly accurate values of photons/e-/ADU generated by knowing the effective area of a photosite and the parameters above). Typical values for T is 0.9, V 0.95 (and correct the falloff in pp if needed), Eph 3.42 e-19, Effective QE 15%, Camera gain at ISO 100 5 e-/ADU.

As far as untis are concerned it depends what you are interested in. If you are interested in what's happening on top of the sensor, then I would suggest photons (as in: this lighting, scene, and field of view produce this total number of photons per micron^2 during exposure after having gone through this lens - for example about 13000 photons per micron squared).

If you are interested in the number of electrons captured by the sensor, then it probably makes sense to look at the output of an individual photosite of a given area and you need to correct for the photons lost in the sensor layers above silicon (reflection, microlenses, IR, UV, CFA etc.) and for the conversion rate from photon to electrons in the silicon: this is all accounted for by a catch-all factor called Effective QE (For example effective QE of 15%, in which case for the situation above and a 5.9 micron photosite you would have 13000 photons/micron^2 * 0.15 e-/photon * 5.9^2 micron^2/photosite = 68000 e-/photosite)

If you are interested in the values of your Raw file, then you need to know your camera's gain when it took the capture. If 5 e-/ADU at ISO 100 in the example above, that would mean a Raw value of 68000 e- / 5 e-/ADU = 13576.

It all follows from the formulas above.

Cheers,

Jack

crames · Nov 10, 2013

The_Suede said:
crames said:

The_Suede said:

xpatUSA said:

In discussions like "FF vs. m43" or "this lens vs that" or "f-number equivalence" or "crop factor", ad nauseam, it is often said that something "gathers more light" but without giving engineering units or any indication how it was figured out.

How is it calculated and what are units of light thus gathered?

--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

Click to expand...

If you want a really simplified version...

This image is a part of the simplified object field. It's an infinitely large grid/wall of perfect point lights spaced at an even distance from each other. They all have the same intensity. We're aiming a 2x crop camera with a 50mm lens straight at it, and get this:

2x crop sensor - 50mm lens - f/4.0 aperture

Each individual point included in the image will send light to/through the optical system. How much light from each point that will actually pass through the optical system is determined by the aperture area, or more accurately the front pupil area. Nothing else - at least not in the simplified version.

Let's just say/assume that the aperture used was f/4.0, and that the intensity of the point lights in front of the system each contribute 1 (one) photon to the image with the front pupil area you get at 50mm F4.0. Then you have a total of 150 photons on the image plane (10x15 points are included, 1 photon from each). That was the "total light gathered".

Now - with the same camera - change the lens to a 25mm (half focal length), while keeping aperture at f/4.0. What happens? The field of view widens, and you get four times as many light points included in the image plane.

Click to expand...

But only in this artificial case which is like a flat field. Real world, when you zoom out you get something entirely different added to the periphery.

Click to expand...

Also, I could be in the center of the sun. Or the dots could be green. So? This is a uniform flatfield, discretized. It's a model, not a landscape scenario.

I'm sorry I'm missing your point. Would you mind explaining how this model relates taking photos of real subjects? In effect you are comparing different scenes in the two 2x images.

The_Suede said:
The_Suede said:

The_Suede said:

2x crop sensor - 25mm lens - f/4.0 aperture

-But since the front pupil area is now four times smaller (half focal length, same aperture > four times smaller front pupil area) - you still get the same total amount of light passed through!

Click to expand...

You've just shown that on a 2x crop sensor, 25mm @ f/4 can give the same total light as 50mm @ f/4! Only because you have added additional light sources in the 25mm shot that were not in the original 50mm shot.

Click to expand...

Yes, and that is also what happens if you have a uniform gray target. Except then you have limes infinity points giving off 1/limes infinity of light. Which is harder to convey in a simplified form.

Again, the relevance to real photos?

The_Suede said:
The_Suede said:

The_Suede said:

Four times as many (150x4 = 600) points of light that each contribute four times less (1/4 = 0.25) photons to the image = still a total of (600x0.25) = 150 photons on the sensor / image plane! Same as with the 50mm F4.0 lens.

Click to expand...

If you just look at the central 150 points of light that were in the 50mm shot, you have the same number of points, each contributing 1/4 photons, but in 1/4 the area, giving the same photons/area (exposure), but 1/4 the "total light."

Click to expand...

Ummm.... Yes?

The_Suede said:

The_Suede said:

That's why the "f/stop" system works - it keeps image plane exposure (light per area on the image plane) constant if you keep the f/# constant - no matter what focal length you choose.
Longer focal length > larger front pupil area, but fewer points of light included.
Shorter focal length > smaller front pupil area, but more points of light included.

Click to expand...

You should only be comparing the same scene elements. What more or fewer "points of light" are included in the field of view are completely arbitrary.

Click to expand...

Who are you to tell me what I should compare? I'm sorry I didn't use polar bears, if that's your preference. Suggestions?

No need to get snippy. Feel free to carry on all you like, with models that apply to featureless white walls.

The_Suede said:
Once again, this is a model for a flat field. Should you wish to get picky about it, compare the 2x crop 25F4.0 shot with the 1x crop 50F4.0, and voila - there you have identical targets, identical target coverage, and 4x more light energy per second through the system with one of teh systems - AND the reason why. What the questions was originally asking for. Difference in light gathering ability, and "why" in a slightly more formalized manner.

Yes, that comparison makes sense and has general applicability in photography. I didn't comment about it.

I'm questioning the point of comparing the two 2x images.

The_Suede said:
The_Suede said:

The_Suede said:

But what happens if we increase the included angle of view by using a 2x larger image plane/sensor (four times larger area) with the 50mm lens in stead? Well, you still include four times as many light points, but the front pupil area doesn't change. So each point still gives the same amount of light on the image plane.

1x sensor - 50mm lens - f/4.0 aperture

So now you have 600 (20x30) points of light included again - just as with the 2x crop sensor and a 25mm lens - but in this case each point still contributes 1 photon to the sensor/image plane - resulting in a total photon count of 600x1.0 = 600 in stead of 600x0.25 = 150.

...................

Then there are some losses, like cos4, mechanical vignetting, optical reflection/absorption and so on - but that's rather beside the point from a purely "simple system" point of view.

The two basic parts are really:

f/# determines image plane exposure

exposure is "amount of light energy per area unit"

-So you get the light energy amount that your system "gathered" by:

(scene average luminance) / (f-stop)^2 * (sensor area) * (exposure time)

And from that you can get that the system "light gathering ability" when keeping scene average luminance and exposure time constant is:

Light gathering ability = (sensor area) / (f-stop)^2

Since (sensor area) is the square function of crop ratio this can be further simplified by doing a square root on both, and then you get:

Light gathering ability = (crop ratio) / (f-stop)

Click to expand...

Shouldn't that be sqrt(Light gathering ability) = (crop ratio) / (f-stop) ?

Click to expand...

Yes, maybe, depends on in what metric you "want" the unknown unit "light gathering ability" to be. Stops (log2)? Lin? Technical, like in radiometric in stead of photometric? Then you sub log10 in stead. I don't know - your choice.

Since it's a derived metric, you could try and go for some standardization. There are organizations handling this should you want to take it further.

If in stops or log10 it would be a different equation, such as log(crop ratio) - log(f-stop), no? Anyway, I only mentioned it because usually it's necessary to take a square root on both sides to maintain an equality, but I guess it doesn't matter here since it's not a real equation.

The_Suede said:
The_Suede said:

The_Suede said:

In the end that means that to collect as many photons per second from a certain scene, you need to change f/# with the same factor as you change the crop ratio.

Click to expand...

Click to expand...

Great Bustard · Nov 10, 2013

xpatUSA said:
Great Bustard said:

xpatUSA said:

Interesting that James...

Click to expand...

That's me, by the way.

xpatUSA said:

...ends up with electrons as the unit of gathered light, not photons.

Click to expand...

The Total Light Collected is the Signal that results from the light falling on the sensor, and is thus measured in electrons.

Click to expand...

With my engineering background, I can't agree with that. If it said:

"The Total Light Collected is [proportional] to the Signal that results from the light falling on the sensor, [which] is measured in electrons"; that would be more precise.

You are correct.

xpatUSA said:
Point being that the units of light in the context of this thread are photons, not electrons.

But the number of photons striking the sensor are measured by the current they produce. So, in a very real sense, the number of photons falling on the sensor is proportional to the current (the number of electrons released).

xpatUSA said:
xpatUSA said:

xpatUSA said:

Odd, that. And his use of units of electrons per photon for QE is a bit shaky although we all know what he means, I suppose.

Click to expand...

The QE (Quantum Efficiency) is the proportion of photons falling on the sensor that are recorded, and they are recorded by the means of the electrons they release in the silicon. Thus, a QE of unity means that one photon releases one electron. Modern sensors have a QE of around 50% (sensors with the newer BSI tech have QEs in the 75% range), so it takes two photons, on average, to release one electron.

Click to expand...

Again, the Engineer in me rebels: Efficiency is dimensionless, it has no units other than relative ones like 'per cent'. You can not say 'photons per electron' because that creates a non-dimensionless unit which, by definition, can not be called 'efficiency'. Normally, the term for such as photons per electron would include the word 'specific', for example, 'specific power' for a gas turbine, e.g. bhp/lb/hr or W/l/s (if you must).

Fair enough. By QE (Quantum Efficiency), we mean PPRE (Probability of a Photon Releasing an Electron or Proportion of Photons that Release Electrons).

xpatUSA said:
Of course we all know that electrons (>1) per photon is not possible in a camera image sensor...

Sure -- this simply means that the max possible QE is 100%.

xpatUSA said:
...and that a fractional photon is impossible anywhere.

Just as a fractional electron is impossible anywhere. A photon either releases an electron or it doesn't, but does so with a particular probability, and neither the expectation value of that probability nor the standard deviation need to take on integral values.

GeorgianBay1939 · Nov 10, 2013

xpatUSA said:
Great Bustard said:

xpatUSA said:

Interesting that James...

Click to expand...

That's me, by the way.

xpatUSA said:

...ends up with electrons as the unit of gathered light, not photons.

Click to expand...

The Total Light Collected is the Signal that results from the light falling on the sensor, and is thus measured in electrons.

Click to expand...

With my engineering background, I can't agree with that. If it said:

"The Total Light Collected is [proportional] to the Signal that results from the light falling on the sensor, [which] is measured in electrons"; that would be more precise.

Point being that the units of light in the context of this thread are photons, not electrons.

Although I am reluctant ever to spring to GB's defence, since he doesn't need it, your concern is dealt with here: http://www.dpreview.com/forums/post/52495152

xpatUSA said:
xpatUSA said:

xpatUSA said:

Interesting that James ends up with electrons as the unit of gathered light, not photons. Odd, that.

xpatUSA said:

He is speaking using reality. Since it is pretty difficult to gather up a "bunch of photons"! So he ( I suspect) is using the "useful effect" caused by a "bunch of photons" by saying Total Light collected = photons collected X Quantum efficiency of the sensor to give electrons. Electrical engineers might use signal, opticians might use an area and time integral of some variable listed here . I don't know, but GB seems to think like a physicist, ie as simply as possible.

Click to expand...

Odd, that. And his use of units of electrons per photon for QE is a bit shaky although we all know what he means, I suppose.

Click to expand...

The QE (Quantum Efficiency) is the proportion of photons falling on the sensor that are recorded, and they are recorded by the means of the electrons they release in the silicon. Thus, a QE of unity means that one photon releases one electron. Modern sensors have a QE of around 50% (sensors with the newer BSI tech have QEs in the 75% range), so it takes two photons, on average, to release one electron.

Click to expand...

Again, the Engineer in me rebels: Efficiency is dimensionless, it has no units other than relative ones like 'per cent'. You can not say 'photons per electron' because that creates a non-dimensionless unit which, by definition, can not be called 'efficiency'. Normally, the term for such as photons per electron would include the word 'specific', for example, 'specific power' for a gas turbine, e.g. bhp/lb/hr or W/l/s (if you must).

And the engineer in me rebels at the statement that "Efficiency is dimensionless".

It can be, and is often manipulated to be so, (using "specific" as an example)... thus avoiding conversion factors when changing methods of measurement. Another example, in the aeronautical field, a measure of efficiency of a wing is Lift/Drag a convenient ratio of two forces and susceptible to dimensionless study using Mach, Reynolds, etc numbers. However another important efficiency for an aircraft is AMPG AirMiles/Gallon, a composite efficiency measure of both airframe and powerplants (normally qualified by payload, altitude, etc)

Beside that, the QE that GB used above IS DIMENSIONESS, as it is the ratio of two numbers, (Ne, the number of electrons) to (Np, the number of photons). QE is (Ne)/(Np) .... dimensionless ratio of two numbers. No units of energy, momentum or charge, thank goodness.

Fortunately GB did not force us intothe intricacies !

And (in Canada) we used to (acceptably) compare vehicle efficiency in MPG (miles per gallon) ..... output/input using commonly used measures. In the last few years some official was convinced by some marketer to change that to litres/100km .... RIDICULOUS! That is a good place for engineers to rise up as one to protest the inversion of the ratio .... to reveal input/output .... the reciprocal of efficiency. End of Rant!!!

xpatUSA said:
Of course we all know that electrons (>1) per photon is not possible in a camera image sensor and that a fractional photon is impossible anywhere.

So far!

xpatUSA said:
--
"For every opinion, there is an equal and opposite"
Ted
SD9, SD10, GH1 and a lens or two

Lots of fun, eh?

Tom

(Dreary rainy November day today, not nice to be out collecting photons to liberate electrons from my sensors).

"Total Light Gathered" - how is it calculated?

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