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

[ectobuilder]

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

I bought an "f2 lens" which is designed only for an APS-C lens.

Most manufactures don't state if the aperture is in Full Frame currency or in the Crop currency.

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

I did a test with a Full Frame lens at f2 (on the same camera) and the Full Frame lens appeared brighter than the APS-C lens at f2 (on the same camera).

 

[J Michael]

f number is independent of the sensor size. It's the ratio of the focal length to the diameter of the aperture. Focal length does not change with sensor size but the lens covers a smaller area of a smaller sensor so there is a narrower angle of view. There is another metric T number which is a measure of the actual transmission of light through a lens. I would expect some variation between lenses at the same f number but none for the same T number.

 

[Samuel H]

As J Michael said, that result is because of different T/ numbers for the same f/ number. Sensor size doesn't play a role at all here.

If you want to get confused, I can tell you that there is an equivalent f/ number in terms of DoF, and you can calculate it by multiplying by the crop factor, just as you do to calculate the equivalent focal length in terms of FoV. So, a 30mm f/2 lens on APS-C will have the same FoV and DoF as a 50mm f/3.2 on full frame (more here). But you don't have to multiply by anything if you are thinking about exposure, it just doesn't have an effect.

 

[Ebrahim Saadawi]

Exposure is the same. It's an f2. An f2 full frame lens on a ff camera will have the same exposure/brightness as an aps-c f2 lens on an aps-c camera (per square area, the ff is more light but hitting new area, it's just larger to the sides)

In terms of field of view and depth of field yes multiply by 1.5x to get the full frame equivalent.

 

[ectobuilder]

Exposure is the same. It's an f2. An f2 full frame lens on a ff camera will have the same exposure/brightness as an aps-c f2 lens on an aps-c camera (per square area, the ff is more light but hitting new area, it's just larger to the sides)

So a m4/3 Voiglander 25mm F0.95 on a GH4 receives more light than a EF 50mm F1.2L on a Canon 5D Mark III?

 

[Joshua_G]

So a m4/3 Voiglander 25mm F0.95 on a GH4 receives more light than a EF 50mm F1.2L on a Canon 5D Mark III?

Yes and no.
Your statement is true only when in both cases the maximum f number possible by each lens is used.
If you'd mount your Canon 50/1.2 lens on your M4/3 camera (via a proper adaptor) and set the lens to f/1.2, both the Canon camera and the M4/3 camera will receive the same amount of light.

The lens' maximum f number isn't necessarily the f number used at any particular shot, it only shows the maximum possible f number of any given lens. On most lenses the user can dial-in f numbers other than the maximum one. In your example, the max f numbers of those 2 lenses are f/0.95 and f/1.2. Both lenses can be set to a different f number, say, f/5.6.

Any lens set to a given f number, say f/5.6, or f/1.4, or f/0.95, or f/16, or f/64, or whatever f number that particular lens is capable of, - will pass through the same amount of light to all cameras, of all sensor sizes (or all film sizes). The amount of light any lens passes to the camera is being determined by the f number used at any particular shot (actually the T number used, see below), regardless of the sensor size (or film size)

The amount of light passing to the camera by the lens, which is determined by the f number used (actually the T number used, see below); combined with the shutter speed; combined with the sensitivity set at the camera (or the sensitivity of the film); - the three of them together determine the exposure. With the same shutter speed and same sensitivity, certain f number used in a shot, by any lens, will give the same exposure to all cameras of all sensor sizes (or all film sizes).

What changes with different sensor sizes (or film sizes) is the DOF (Depth of Field). The exact DOF for any particular shot is derived from, or being determined by:

1. The lens' Focal Length (in your example 25mm or 50mm). When it's a zoom lens, the focal length used in that particular shot.
2. The aperture (f number) used in that particular shot. (Actually it's the T number used, but that gets too complicated here. F numbers are calculated, while T numbers are measured and they take into account light losses inside the lens' elements. When the T numbers aren't known, f numbers are good practical proximity).
3. The camera's sensor size (or film size).

 

[harddrive]

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

I did a test with a Full Frame lens at f2 (on the same camera) and the Full Frame lens appeared brighter than the APS-C lens at f2 (on the same camera).

The strict answer to your question (as you've phrased it, and especially with your capitalising "AMOUNT") is a strong YES!

But reading some of the comments above then two things are being confused - the AMOUNT of light and the INTENSITY of light. Two different things, albeit related. The terms are *NOT* interchangeable.

If you want the technical terms, "amount" corresponds to luminous flux (measured in lumens) and "intensity" is called "luminous intensity" (measured in candelas).

What would be constant for the same f stop (regardless of sensor size) would be the intensity of light in each case - how bright the images are, if you like. But that's completely different to the amount of light - the latter is a function of both brightness and area.

If it helps, think of 100 watt and 20 watt bulbs. I think everybody will agree that the 100 watt bulb will give out a greater AMOUNT of light? (More photons/second?) But if we use them to illuminate a given scene, then by placing the 20 watt bulb closer, we can make the brightness the same in each case.

 

[arniepix]

f number is independent of the sensor size. It's the ratio of the focal length to the diameter of the aperture. Focal length does not change with sensor size but the lens covers a smaller area of a smaller sensor so there is a narrower angle of view. There is another metric T number which is a measure of the actual transmission of light through a lens. I would expect some variation between lenses at the same f number but none for the same T number.

T stops are equivalent to F stops. The largest difference is that lenses marked in T stops are individually calibrated, where as mass produced lenses marked in F stops are not.

 

[Samuel H]

It's not only the copy-to-copy calibration, it's the fact that they give you that information. Depending on the specifics of the design, some lenses have a T/ that is very close to their f/, others not that close. For example, the original Canon 24-70 f/2.8L USM was T/3.4, but the new II version is T/3. Here, here.

 

[josh-ozirigpro]

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

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

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

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

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

 

[harddrive]

So a m4/3 Voiglander 25mm F0.95 on a GH4 receives more light than a EF 50mm F1.2L on a Canon 5D Mark III?

No. (And assuming we are talking about both lenses set wide open.) For the reasons I gave above, then in the case of the f0.95 lens, the image on the sensor will be BRIGHTER than the f1.2 lens on the 5D, but the AMOUNT of light will be more in the case of the f1.2/5D. Because amount is light depends on image area as well as brightness.

Any lens set to a given f number, say f/5.6, or f/1.4, or f/0.95, or f/16, or f/64, or whatever f number that particular lens is capable of, - will pass through the same amount of light to all cameras, of all sensor sizes (or all film sizes). The amount of light any lens passes to the camera is being determined by the f number used at any particular shot (actually the T number used, see below), regardless of the sensor size (or film size)

The amount of light passing to the camera by the lens, which is determined by the f number used .....

And for the same reasons, again, no. That isn't true. You need to replace "amount of light" with "brightness of the image at the sensor".

If it sounds pedantic, it's not, and the whole subject takes on far more significance with digital sensors and format convertors such as the SpeedBooster. With a very large **ALL ELSE EQUAL** qualifier, then increasing the sensor size will give a higher ISO rating (bigger photosites). But what happens if we compare two cameras, where "A" has 4x the sensor surface area of "B"? All else equal, if the base ISO of B is ISO100, the base ISO of A is likely to be around ISO400. That's why in the digital camera world, then as well as talking about "equivalent focal length" (for equivalent angles of view) the concept of "equivalent aperture" is being increasingly used. The usefulness is that only does it give an equivalence for DOF purposes, but also for *TOTAL* light on the sensor. (Intensity x sensor area.)

Take it further, and it works out that (with the all else equal qualifier) the ONLY factor that really matters in terms of DOF and low light ability is the actual max physical size of the iris and front element - NOT the max f stop. For a given diameter, then if it corresponds to (say) f2 on camera B above, then it will correspond to f4 on camera A. But because the base ISO of A if 2 stops better than B, then we get performance equivalence. The more you think about the "equivalent f stop" concept, the cleverer it is.

 

[josh-ozirigpro]

And this is the reason I love DVX and should've been back sooner!

Thank you for the awesome summary!
Even if to some it may seem a bit pedantic, I think these are the things that we should be ensuring we understand.

 

[M. Gilden]

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

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

 

[Samuel H]

I'm on the same page

But I'll add that a bigger sensor does receive more photons, and that also counts... though not in the terms the OP put the question. At ISO 100, both images will be just as bright, with APSC and FF sensors; but with the FF sensor you can bring the ISO up an extra stop or two without any more noise in your image (at same tech, which is nearly always not the case; but this is the reason you could shoot video at ISO 1600 on the 5D2 with the same noise as the 7D had at ISO 800: more area, bigger pixels, less noise when the output is of the same resolution).

 

[Joshua_G]

No. (And assuming we are talking about both lenses set wide open.) For the reasons I gave above, then in the case of the f0.95 lens, the image on the sensor will be BRIGHTER than the f1.2 lens on the 5D, but the AMOUNT of light will be more in the case of the f1.2/5D. Because amount is light depends on image area as well as brightness.

And for the same reasons, again, no. That isn't true. You need to replace "amount of light" with "brightness of the image at the sensor".

If it sounds pedantic, it's not, and the whole subject takes on far more significance with digital sensors and format convertors such as the SpeedBooster. With a very large **ALL ELSE EQUAL** qualifier, then increasing the sensor size will give a higher ISO rating (bigger photosites). But what happens if we compare two cameras, where "A" has 4x the sensor surface area of "B"? All else equal, if the base ISO of B is ISO100, the base ISO of A is likely to be around ISO400. That's why in the digital camera world, then as well as talking about "equivalent focal length" (for equivalent angles of view) the concept of "equivalent aperture" is being increasingly used. The usefulness is that only does it give an equivalence for DOF purposes, but also for *TOTAL* light on the sensor. (Intensity x sensor area.)

Take it further, and it works out that (with the all else equal qualifier) the ONLY factor that really matters in terms of DOF and low light ability is the actual max physical size of the iris and front element - NOT the max f stop. For a given diameter, then if it corresponds to (say) f2 on camera B above, then it will correspond to f4 on camera A. But because the base ISO of A if 2 stops better than B, then we get performance equivalence. The more you think about the "equivalent f stop" concept, the cleverer it is.

Your above statements are wrong.
In order to enable you and others to get it, let's use film cameras examples.

Take 4 movie cameras, 70mm, 35mm, 16mm and 8 mm - all operating at the same shutter speed, say 1/48. Load each camera with reversal film of the same sensitivity, say 180 ISO. Light a scene and measure the amount of light falling on the subject with hand-held light meter. If outdoors, just measure the amount of light falling on the subject, without any additional lighting. Say the reading is f/5.6 (for 1/48). Set the f stop (or T stop) on the lens of each of those cameras to 5.6 - lo and behold - you'll get same exposure (and correct exposure) on all 4 cameras.

The same goes for stills film cameras.
Take 3 cameras, a view camera of 8x10 inch, a medium format camera of 6x6 cm and a 35mm camera. Set all 3 cameras to same shutter speed, say 1/50. Load each camera with reversal film of the same sensitivity, say 100 ISO. Light a scene and measure the amount of light falling on the subject with hand-held light meter. If outdoors, just measure the amount of light falling on the subject, without any additional lighting. Say the reading is f/8 (for 1/50). Set the f stop on the lens of each of those cameras to f/8 - lo and behold - you'll get same exposure (and correct exposure) on all 3 cameras.

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

The size of the media (film or sensor) is absolutely irrelevant when it comes to exposure.
See: http://en.wikipedia.org/wiki/Exposure_(photography) , http://www.cambridgeincolour.com/tutorials/camera-exposure.htm , http://www.geofflawrence.com/exposure.html , http://digital-photography-school.com/learning-exposure-in-digital-photography/ .

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

The formula for calculating the f number is never changed and the size of the media (film or sensor) isn't a part of that formula.

 

[harddrive]

Your above statements are wrong.
In order to enable you and others to get it, let's use film cameras examples.

No Joshua, your examples are correct in as far as they go, but you're the one who is missing something vital, and it's the **ALL ELSE EQUAL** qualifier that I thought I'd stressed.

What you say about the movie cameras is correct as far as it goes - it's got be if we're assuming the same ISO filmstock. But in that case, I'm sure you'll agree that the quality of the image from the 70mm frame will be vastly better than from the 8mm frame? That's no accident, and it's possible because there is a lot more TOTAL light to make use of. Yes, the INTENSITY is the same, the brightness of the image if you prefer, that's why with the same film stock it's the same f stop in each case. But in the case of the 70mm image, it's the same INTENSITY spread over a much bigger area, so a far greater QUANTITY of light. A lot more photons in total, if you like.

The argument can then be turned around. What if we choose the film stock to get equivalent image quality? We may find that to be at all acceptable, we want Kodachrome at ISO25 for acceptability. Move to 16mm, and we can maybe get away with (say) ISO100? Move to 35mm, and we may find that ISO400, even 800 rivals our ISO25 8mm image? So assuming the same scene illumination, this then means we need to set our 8mm camera to f2, our 16mm camera to f4, and our 35mm camera to f8-11 - *FOR COMPARABLE IMAGE QUALITY*.

[Whilst the numbers above are arbitray to illustrate the point, the principle is far from theoretical. As a boy my father used a 5x4 plate camera, and the lenses tended to have small (by todays standards) f stops, and obviously long focal lengths. But the size of the negative - the small amount it would need to be enlarged for a 10x8 print - meant he could routinely use fairly high ISO film speeds without excessive grain. Come a move to 35mm Pentax and it meant the necessity of using much slower film speeds to keep the grain acceptable - but that was compensated for by lenses with much faster f stops.]

I accept that the releationship is much less clear with film than with digital, and I did state "the whole subject takes on far more significance with digital sensors ..." and "That's why in the digital camera world, ....... the concept of "equivalent aperture" is being increasingly used.". Why? Because we come back to that old **ALL ELSE EQUAL** chestnut. To make a sensible comparison, we have to assume the same basic sensor technology and system resolution just as starters. So in which case then not only is a full frame 35mm frame (roughly) twice the size of APS-C, but - assuming the same resolution - the photosites will be twice the size as well. Consequently, it's base ISO rating will be about one stop faster.

So in your example about stills cameras, the valid comparison comes when you load them with DIFFERENT ISO film stocks, the larger the sensor, the faster the stock you use. Because that is what actually happens in the digital world.

Joshua - I'm not disagreeing with any of the examples you give as far as they go. They may well be all anybody needs to know from an operational point of view. But you are drawing the wrong conclusions because they are incomplete, and by that I mean you're neglecting facts (assuming same ISO film stock in each case) such as the quality of image in the bigger sensor cases is far better than for small sensor. And that's not chance, it's due to the far greater QUANTITY of light being used to form those images - think of it as the "never something for nothing rule". Remember also that for the same angle of view, an f2 lens for 70mm must have a far bigger diameter front element/iris than for 8mm. So hardly surprising it let's a greater quantity of overall light in - but spread over a bigger area, so the brightness at any one point is the same as for an f2 lens and the 8mm image.

If you want an analogy, then think of the f stop as the voltage. If you want the power used by an appliance, you also need to know the current it's taking. In this case, if you want to know the total amount of light, you need to know f stop and sensor area.

The Speed Booster optics changes the effective focal length, thus increases the f number. See: http://www.metabones.com/products/?c=speed-booster .

But that's not all it does - it changes the coverage area of the lens as well, and that is critical to the full understanding. Effectively it's concentrating the light into a smaller area - so it's a BRIGHTER image though the same total amount of light.

 

[Joshua_G]

…
What you say about the movie cameras is correct as far as it goes - it's got be if we're assuming the same ISO filmstock.

As far as exposure goes, it applies just the same to different digital cameras being set to the same sensitivity (ISO), having different frame size.

The correct exposure is being determined by 3 factors:
1
The amount of light that falls on the subject.
2
The sensitivity of the media (the film stock or the sensitivity set in a digital camera).
3
The f number, or T number used.

The formula of the exposure never changes.
The media size (film or sensor) isn't included in the exposure formula.

But in that case, I'm sure you'll agree that the quality of the image from the 70mm frame will be vastly better than from the 8mm frame?

The question that started this thread was about exposure, not about image quality.
It is correct that when *EVERYTHING ELSE BEING EQUAL*, the larger the media is (film area or sensor size), the better the image quality will be.
However exposure doesn't change when the media size is being changed.
A certain lens set to specific f number, will give the very same exposure (which is the amount of light reaching the media) on all cameras it will be attached to, cameras of all sizes of media.

That's no accident, and it's possible because there is a lot more TOTAL light to make use of.

Your above speculation is wrong.
The image quality is better the larger the media is because of other physical factors that have nothing to do with the amount of light.

The amount of light reaching the media is being determined by: (1) the amount of light that falls on the subject, (2) the sensitivity of the media, (3) The f number, or T number used.

The media size (film or sensor) isn't included in the exposure formula. The exposure, along the shutter speed, determines the amount of light reaching the media. The media size has no part here.

Yes, the INTENSITY is the same, the brightness of the image if you prefer, that's why with the same film stock it's the same f stop in each case. But in the case of the 70mm image, it's the same INTENSITY spread over a much bigger area, so a far greater QUANTITY of light. A lot more photons in total, if you like.

You are confusing yourself and others.
The amount of light reaching the media has nothing to do with the media size.

The argument can then be turned around. What if we choose the film stock to get equivalent image quality? We may find that to be at all acceptable, we want Kodachrome at ISO25 for acceptability. Move to 16mm, and we can maybe get away with (say) ISO100? Move to 35mm, and we may find that ISO400, even 800 rivals our ISO25 8mm image? So assuming the same scene illumination, this then means we need to set our 8mm camera to f2, our 16mm camera to f4, and our 35mm camera to f8-11 - *FOR COMPARABLE IMAGE QUALITY*.

Again, image quality does change with the size of the media.
Exposure and the amount of light reaching the media has nothing to do with the size of the media.

…
So in your example about stills cameras, the valid comparison comes when you load them with DIFFERENT ISO film stocks, the larger the sensor, the faster the stock you use.

You are completely wrong here.
Larger the film size enables us to use faster film stock, however it has nothing to do with the amount of light reaching the film. It is because of the film's grain. The larger the film is, lesser degree of magnification is used to cover certain screen area. The lesser the magnification is, the film's grain is less apparent to the spectator's sight.

…

Repetitive arguments are snipped.
I feel it's enough for me to relate once to each argument.

If you want an analogy, then think of the f stop as the voltage. If you want the power used by an appliance, you also need to know the current it's taking. In this case, if you want to know the total amount of light, you need to know f stop and sensor area.

Since I'm an electronic engineer, I can safely say that what you wrote above is nonsense and completely not analogues to exposure and amount of light reaching the media.

But that's not all it does - it changes the coverage area of the lens as well, and that is critical to the full understanding. Effectively it's concentrating the light into a smaller area - so it's a BRIGHTER image though the same total amount of light.

The change in coverage area has no effect on exposure, or amount of light reaching the media.
What affects the effective f number is the decrease in the effective focal length.

 

[harddrive]

The formula of the exposure never changes.
The media size (film or sensor) isn't included in the exposure formula.

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

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

The question that started this thread was about exposure, not about image quality.

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

And if you start to bring sensor size into the equation it becomes a three way tradeoff (**WITH ALL ELSE EQUAL** - "yawn", I'm getting tired of saying that..... ) as a bigger sensor will have a higher ISO for the same quality. And let's remind ourselves what the original question was:

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

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

It is correct that when *EVERYTHING ELSE BEING EQUAL*, the larger the media is (film area or sensor size), the better the image quality will be.
However exposure doesn't change when the media size is being changed.

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

A certain lens set to specific f number, will give the very same exposure (which is the amount of light reaching the media) on all cameras it will be attached to, cameras of all sizes of media.

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

If you want an analogy, then think of the f stop as the voltage. If you want the power used by an appliance, you also need to know the current it's taking. In this case, if you want to know the total amount of light, you need to know f stop and sensor area.

Since I'm an electronic engineer, I can safely say that what you wrote above is nonsense and completely not analogues to exposure and amount of light reaching the media.

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

But that's not all it does - it changes the coverage area of the lens as well, and that is critical to the full understanding. Effectively it's concentrating the light into a smaller area - so it's a BRIGHTER image though the same total amount of light.

The change in coverage area has no effect on exposure, or amount of light reaching the media.
What affects the effective f number is the decrease in the effective focal length.

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

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

 

[abreu-canedo]

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

However, the OP was not assuming a speed booster.

The intensity of light hitting each digital pixel of a APS-C sensor, a S35mm sensor and a 35mm(FF) sensor are the same at f2 by using the same lens on all of them, given the same shutter speed on all occasions.

It's true that if you're using a 24mm lens that was meant to cover a 35mm sensor, then the 35mm sensor would be receiving more light as an entire sensor because it catches more of the light coming in than a smaller, say APS-C sized sensor in it's entirety with the same lens. However, with the same lens, without the optical conversion of a speed booster, the intensity of light hitting each digital pixel will be the same...equaling to the same exposure on all of them.

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

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

Now, I'm not an engineer, but in 15 years of film and digital experience, I have found this to be reliable info.

However, I'm always open to learning something new, and this thread is intriguing!

 

[harddrive]

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

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

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

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