Does f/2=f/2?

[highsee3]

I always though f/2=f/2, but I've read several posts about how 35mm sensors "gather more light" and thus 35mm f/4 is really equal to 4/3 f/2 and perhaps 35mm f/4 is equal to APS-C f/2.8.

I understand the DOF is different, that's not my concern.

The thing is, I can't find an example where 35mm (5D or D3 etc.) clearly shows 2 stops better noise (ISO 6400 vs. ISO 1600). I do understand larger sensors have cleaner high ISO, but it's not the same across the board and depends largely on sensor technology and isn't always a huge advantage. (40D vs. 5D)

People have said "but the 35mm sensor gathers 4x as much light as 4/3" meaning that you don't benefit from fast 4/3 or APS-C lens. If this is true, can someone tell me how "fast" the following lenses really are (in %) using the above logic. I'm a Mamiya user and would like to know how fast my Mamiya lenses really are compared to 35mm, APS-C and 4/3.

Olympus 14-35 f/2
Canon EFS 17-55 f/2.8
Canon 24-105 f/4
Mamiya (645) 55-110 f/4.5

Also:

Leica 25mm f/1.4
Sigma APS-C 30mm f/1.4
Canon 50mm f/1.4
Mamiya (645) 80mm f/1.9
Mamiya (6x7) 110 f/2.8

And finally

Olympus 55 f/1.2 (4/3 adapter)
Canon 85 f/1.8 (Aps-c)
Canon 135 f/2 (35mm)
Mamiya 200 f/2.8 APO (645)

thanks

--
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'A camera maker that simply copies others' idea has no right to call itself an original
maker in the first place.' -Mr. Maitani, creator of the OM photographic system.

 

[nickoly]

People have said "but the 35mm sensor gathers 4x as much light as
4/3" meaning that you don't benefit from fast 4/3 or APS-C lens.

Baloney. As you say, f2=f2.

F2 has everything to do with the size of the glass and the focal length - and nothing to to do with the size of the sensor. If you put a view camera lens onto a Leica, there no magical change of properties, only a change of field of view, and there is nothing magic about that either. If they can't see that, don't bother to enlighten them.

They are probably alluding to inherrently cleaner operation at higher ISO, which is true, as you already know.

 

[jfriend00]

f/2 is f/2 no matter what the sensor size is.

But, what you may have been seeing is that if you take two sensors with the same number of pixels and one is an APS size sensor and the other is a 35mm sensor (like say Nikon's new D3 and D300), then the sensor cells on the larger sensor will be around twice as large as the sensor cells on the smaller sensor. All else being equal (like micro-lens designs), that means that for a given exposure, the cells in the larger sensor gather twice as many photons per cell.

The fact that the larger sensor cells are capturing more light photons has two interesting advantages:

1) For a given amount of light, you are getting more signal on a given sensor cell. This can be taken advantage of and makes it appear that the sensor is more sensitive. You see this in cameras when they raise their base ISO from 50 to 100 or in recent cameras from 100 to 200.

2) With more signal for a given amount of light in the scene, you get a better signal to noise ratio. Since noise is largely not proportional to the size of the cell, but signal is, the larger cells capture more signal, but have the same noise as the smaller cells, thus giving them a much better signal-to-noise ratio. This makes the most difference for very low light - either in deep shadows in a normal image or anywhere in the image at high ISO. So, the larger sensor delivers better shadow detail and better high ISO performance.

Both of these benefits are similar in some ways to shooting at a larger aperture thus I could see how some people might say a larger sensor is "like" having a faster lens with a larger aperture. It isn't totally the same because the depth of field and other optical properties are still exactly like the real f/2 aperture, but the light gathering ability will seem to have improved with the large sensor even though the aperture stayed the same size.

Does any of this make sense?
--
John
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[ojohn]

F stops are an absolute ratio of the aperture opening to the focal length, but there can be differences in the amount of light coming through.

For a rough example, you could expect an 11 element f/2.8 lens to let less light through than a same focal length f/2.8 3 element lens, just because of loss of transmission through all that extra glass (all else being equal including coatings, and even more unequal if the coatings were different).

For a different rough example, your 50mm seven element lens set at f/8 probably lets through more light than your 18-250 16 element lens set at f/8, just because of all the extra glass in the zoom.

Extreme comparisons between simple vs. complex lenses could show a "1 stop" difference in light transmission using the f/ number alone.

Some professional movie camera lenses are marked in "T" stops. T for "transmission". This was particularly important in the days of outside light meters and narrow latitude and horrendously expensive Technicolor film stocks. "T" was also an homage to Technicolor, maybe. With T stops, setting "T/8" should have the same amount of light between different lenses. (Contrast and flare were corrected in the lab, somewhat.)

Most "f/2.0" lenses were rated "T/2.2" for example, then higher T stops of that lens followed the normal progression of numbers, even through the T stop did not directly relate to the f/ stop.

Not so important now with through the lens metering.

 

[hollis_f]

I always though f/2=f/2, but I've read several posts about how 35mm
sensors "gather more light" and thus 35mm f/4 is really equal to 4/3
f/2 and perhaps 35mm f/4 is equal to APS-C f/2.8.

Utter rot.

Do these people think that, when sunbathing, their legs will tan faster than their arms just because their legs gather more light due to their being larger?

--
Frank Hollis
Mass Spectroscopist in the UK
Can0n 2oD and 4oD

 

[Pedagydusz]

You are correct, there is no area difference. However, in a post above, John (jfriend00) points out a difference: if two sensors are of the same size, and one has more pixels, each photocell of this one gets less light that each photocell in the other sensor, showing in effect less effective sensitivity.

I must say that I made a quick and "unscientific" experiment, pointing two cameras with the same size sensor, the Olympus E-330 (7.4 MPx) and the Olympus E-510 (10.0 MPx) to the same scene and measuring exposure in the same way with the same lens. According to the theory shown in the 1st paragraph, there should be a 10/7.4 better sensitivity for the E-330, but I could not find any difference whatsoever.

Of course, that does not mean a lot, as the cameras sensitivity could be "rigged" to compensate for that effect. But it means that, in practice, not much difference can be expected.
--
Antonio

http://ferrer.smugmug.com/

 

[nickleback]

Utter rot.

How so?

Do these people think that, when sunbathing, their legs will tan
faster than their arms just because their legs gather more light due
to their being larger?

The legs won't tan any faster, but the total amount of light received by the legs is higher, due to the larger area.

--
Seen in a fortune cookie:
Fear is the darkroom where negatives are developed

 

[mike703]

The legs won't tan any faster, but the total amount of light received
by the legs is higher, due to the larger area.

... which is analogous to using a different sized sensor with the same lens. The 'exposure' (light intensity falling on each square inch of skin, or sensor) is the same.

The people mentioned in the original post seem to think that changing the sensor size will alter the intensity of light falling on the sensor. That IS utter rot - like imagining that the sun becomes more intense the more square inches of skin you expose to it.

Best wishes
--
Mike

 

[nickleback]

You are correct, there is no area difference. However, in a post
above, John (jfriend00) points out a difference: if two sensors are
of the same size, and one has more pixels, each photocell of this one
gets less light that each photocell in the other sensor, showing in
effect less effective sensitivity.

You've got the example mixed up. The D3 has a 36x24mm sensor, the D300 has a 23.6x15.8mm sensor. The pixel count is irrelevant. The point is that if you shine f/2 light on both sensors, the intensity of light is the same, but the total amount of light collected by the larger sensor is... larger. So, all things being equal, on a larger sensor you'll get less noise, as you've collected more light.

You are painting a house. You apply the same thickness of paint regardless of house size. Given the above, on a larger house do you use the same or more cans of paint than you would on a smaller house?

--
Seen in a fortune cookie:
Fear is the darkroom where negatives are developed

 

[Ron Parr]

I always though f/2=f/2, but I've read several posts about how 35mm
sensors "gather more light" and thus 35mm f/4 is really equal to 4/3
f/2 and perhaps 35mm f/4 is equal to APS-C f/2.8.

f/2 is always f/2 (ignoring minor differences between lenses). This mans that, given the same source, the number of photons per unit area will be the same at f/2 no matter what lens you use.

Of course, a larger sensor has more area, so it captures more total light at f/2 than a smaller sensor does.

The thing is, I can't find an example where 35mm (5D or D3 etc.)
clearly shows 2 stops better noise (ISO 6400 vs. ISO 1600). I do
understand larger sensors have cleaner high ISO, but it's not the
same across the board and depends largely on sensor technology and
isn't always a huge advantage. (40D vs. 5D)

I think it would make more sense to compare the 5D with its contemporary, the 20D. Even then, I do agree that the difference is less than you might expect at first glance when you consider that the 5D is taking in about 2.5X as much light at the same f-stop and exposure. A few reasons for this:

• When people do such comparisons, they typically look at the pixel level, which doesn't treat output size as a constant. If you want to measure total noise, and not just noise per pixel, then you would need to consider the noise over a constant sized output area. Note that the 5D has more total pixels than the 20D.

• There may be differences in image processing that partly mask the difference in noise performance. In particular, 20D jpegs can get a little soft and de-saturated at high ISO, while this may be less so for the 5D.

• The total amount of light captured is one of many factors that influences the noise in an image.

• There are some manufacturing differences (mask stitching) required for 24x36mm sensors that may result in slightly lower efficiency per unit area of sensor than for smaller sensors.

People have said "but the 35mm sensor gathers 4x as much light as
4/3" meaning that you don't benefit from fast 4/3 or APS-C lens. If
this is true, can someone tell me how "fast" the following lenses
really are (in %) using the above logic. I'm a Mamiya user and would
like to know how fast my Mamiya lenses really are compared to 35mm,
APS-C and 4/3.

All of your lenses have the same f/stops they have always had.

Assuming the same composition and same f/stop, the total amount of light striking your capture medium will be proportional to the area of the capture medium, and the total amount of light you capture will depend upon the efficiency of your capture medium.

--
Ron Parr
Digital Photography FAQ: http://www.cs.duke.edu/~parr/photography/faq.html
Gallery: http://www.pbase.com/parr/

 

[nickleback]

The legs won't tan any faster, but the total amount of light received
by the legs is higher, due to the larger area.

... which is analogous to using a different sized sensor with the
same lens. The 'exposure' (light intensity falling on each square
inch of skin, or sensor) is the same.

Exactly.

The people mentioned in the original post seem to think that changing
the sensor size will alter the intensity of light falling on the
sensor.

I don't see that mentioned.

That IS utter rot - like imagining that the sun becomes more
intense the more square inches of skin you expose to it.

Let's keep going with arms and legs.

Let's say your legs have twice the area of your arms. If you expose both to the sun for the same amount of time, your legs will collect twice as much light, but your arms and legs will be tanned the same. You agree to that, right?

Now lets expose your legs to half the intensity of light (i.e. close aperture 1 stop) as your arms, but still for the same amount of time. Your arms and legs have collected the same amount of light, right? But your legs won't be tanned as much as your arms, because they received half the intensity of light.

Now lets make your legs twice as sensitive (i.e. double ISO). Maybe use some exfoliant cream to wipe away dead skin cells. And again expose your arms and legs for the same amount of time, with the legs exposed half the intensity of light as your arms. Your ams and legs again have collected the same amount of light. But they both have tanned as much because while the legs have received half the intensity of light, they are twice as sensitive.

Best wishes

Likewise. Hope you didn't get skin cancer from the experiments.

--
Seen in a fortune cookie:
Fear is the darkroom where negatives are developed

 

[mike703]

I'm afraid you've lost me. Sunstroke possibly? It fits with the symptoms: 'There may also be some mental and verbal confusion. Speech may be unintelligible or incoherent...'

Whatever: f/2 = f/2, as everybody seems to agree, which is reassuring.
--
Mike

 

[nickleback]

I'm afraid you've lost me. Sunstroke possibly?

Perhaps. If you are confused, read it again, I made it quite simple. But don't dismiss what you can't understand as "utter rot".

Whatever: f/2 = f/2, as everybody seems to agree, which is reassuring.

f/2 = f/2, and a larger sensor will receive more total light than a smaller sensor at the same aperture.

--
Seen in a fortune cookie:
Fear is the darkroom where negatives are developed

 

[Haider]

From what I understand if you have acertain size aperture and allow light through it you will get a certain volume of light/photons. These are collected by a the sensor or bucket. The smaller the sensor/bucket the more it will be filled therfore the higher the signal compared to the noise. Therefore a higher signal compared to noise value (aka signal to noise ratio) on the smaller sensor, than a larger one for the given aperture size and shutter speed.

 

[jfriend00]

From what I understand if you have acertain size aperture and allow
light through it you will get a certain volume of light/photons.
These are collected by a the sensor or bucket. The smaller the
sensor/bucket the more it will be filled therfore the higher the
signal compared to the noise. Therefore a higher signal compared to
noise value (aka signal to noise ratio) on the smaller sensor, than a
larger one for the given aperture size and shutter speed.

You have it backwards. A larger sensor behind the same lens collects more total photons and thus has a higher signal to noise ratio (assuming other sensor characteristics are the same) which means lower visible noise in the resulting image.

--
John
Popular: http://jfriend.smugmug.com/popular
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[nickleback]

From what I understand if you have acertain size aperture and allow
light through it you will get a certain volume of light/photons.

Yes. Just to clarify, I assume you mean apparent aperture size (i.e. "x" mm, not aperture ratio like f/2). If you are talking aperture ratio, then nothing else you wrote makes much sense, as with different sensor sizes you'll be using different focal lengths, which at the same aperture ratio the longer lens will have a larger apparent aperture.

These are collected by a the sensor or bucket.

OK.

The smaller the
sensor/bucket the more it will be filled therfore the higher the
signal compared to the noise.

Not quite. Total light is the largest component of signal to noise. Total light is the same.

--
Seen in a fortune cookie:
Fear is the darkroom where negatives are developed

 

[Haider]

yes aperture size, not the f number.

From what I understand if you have acertain size aperture and allow
light through it you will get a certain volume of light/photons.

Yes. Just to clarify, I assume you mean apparent aperture size (i.e.
"x" mm, not aperture ratio like f/2). If you are talking aperture
ratio, then nothing else you wrote makes much sense, as with
different sensor sizes you'll be using different focal lengths, which
at the same aperture ratio the longer lens will have a larger
apparent aperture.

These are collected by a the sensor or bucket.

OK.

The smaller the
sensor/bucket the more it will be filled therfore the higher the
signal compared to the noise.

Not quite. Total light is the largest component of signal to noise.
Total light is the same.

If you point your camera at bright white light with the shutter open long enough and with a big enough aperture hole, the ensuing photo will be just bright white with no noise because the bucket (sensor) has filled with good photons and there is no room for bad noise photons. This is in layman's terms. The less the bucket is full the more room for bad noise photons. Take a photo that is heavily underexposed you will notice much more noise than an over heavily exposed photo. Or just take alook at your shadows in photos they have more noise where as your burnt highlights don't have noise. That why small sensor (with the same MP) equals smaller photosites (buckets), equals easier to fill (therefore less room for noise), equals lower dynamic range (a draw-back, well you can't have it all).

 

[Haider]

Think in photosite size, look at my explanation below...

From what I understand if you have acertain size aperture and allow
light through it you will get a certain volume of light/photons.
These are collected by a the sensor or bucket. The smaller the
sensor/bucket the more it will be filled therfore the higher the
signal compared to the noise. Therefore a higher signal compared to
noise value (aka signal to noise ratio) on the smaller sensor, than a
larger one for the given aperture size and shutter speed.

You have it backwards. A larger sensor behind the same lens collects
more total photons and thus has a higher signal to noise ratio
(assuming other sensor characteristics are the same) which means
lower visible noise in the resulting image.

--
John
Popular: http://jfriend.smugmug.com/popular
Portfolio: http://jfriend.smugmug.com/portfolio

 

[jfriend00]

If you point your camera at bright white light with the shutter open
long enough and with a big enough aperture hole, the ensuing photo
will be just bright white with no noise because the bucket (sensor)
has filled with good photons and there is no room for bad noise
photons. This is in layman's terms. The less the bucket is full the
more room for bad noise photons. Take a photo that is heavily
underexposed you will notice much more noise than an over heavily
exposed photo. Or just take alook at your shadows in photos they have
more noise where as your burnt highlights don't have noise. That why
small sensor (with the same MP) equals smaller photosites (buckets),
equals easier to fill (therefore less room for noise), equals lower
dynamic range (a draw-back, well you can't have it all).

I'm sorry, but you don't have the right understanding of how noise happens in sensors.

Let's compare two theoretical sensor cells. Each is from a 12MP sensor, one is from an APS-sized sensor (like the Nikon D300) and one is from a 35mm sensor (like the Nikon D3). For the purpose of focusing on just the size difference, let's assume that all else in the sensor design is identical. For most of the things that contribute to noise, for the purposes of this comparison, we can assume that a given sensor cell will have X amount of noise in a given circumstance. The value X has nothing to do with the size of the cell so both the D300 sized cell and the D3 size cell have the same absolute amount of noise in them.

Now expose the two cells to an image with the same focal length lens, the same aperture setting and the same shutter speed. The larger cell collects more photons because it has a larger area. The number of photos is proportional to the signal that this cell produces. So, let's say the amount of photons collected by the smaller D300 sized cell is N. Since the D3 sized sensor cells are 2.3x as large the area as the D300 sized sensor cells, that would mean that the D3 sized cell collects around 2.3N the number of photons.

So, you can see that the absolute value of the noise is approximately the same in the two examples, but the signal is 2.3 times higher for the larger sensor cell. If you then want to look at the signal-to-noise ratio (which is what determines how visible the noise is), you'd see that the larger sensor cell as a 2.3 times higher signal-to-noise ratio.

The signal-to-noise ratio has absolutely nothing to do with how "full" the bucket is. BTW, in the above example (assuming identical sensors that just differ in size), the two sensors buckets are equally full. It only has to do with the absolute value of the signal compared to the absolute value of the noise. Holding everything else constant, raising the signal will raise the signal to noise ratio and reduce the visible effect of noise. But, the only thing that matters is the signal value, not whether the bucket is 1/3 full or 2/3 full.

The reason you see more noise in shadows or underexposed images or high ISO images is because the signal value (the number of photons collected) in those circumstances is substantially lower. Underexpose by 1-stop and you have only half the number of photons. Raise the ISO from 100 to 400 and expose by the meter and you only have 1/4 the number of photons which makes the signal to noise ratio only 1/4 what it was. That makes noise a lot more visible.
--
John
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[Pedagydusz]

Haider, I'm afraid you are getting it backwards, as others have pointed out. The reason is that you are using a bad "bucket" analogy.

The system does not behave like a faucet, or a hose, pointed at a bucket that collects all the water. Think of it as a steady, and very uniform, rain or drizzle. The photosites are small buckets. You put them out to collect the rain for a short, well determined, time interval. The smaller the bucket, the less water it collects. If, in that time interval, there is a rainfall of 5 cub. cm / square cm, and the bucket section is 5 square cm, it fills with half the water (in volume or weight) than it would if it were 10 sq. cm in section.
If you think in terms of photons instead of water, the result is similar.

That is in terms of light collected. From this, it is easy to deduce that smaller photosites will show more noise, in equal circumstances.
--
Antonio

http://ferrer.smugmug.com/