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Micro 4/3
- Thread starter StuartT50
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bobn2
Forum Pro
While this is often true, it doesn't have much relevance to anything.
bobn2
Forum Pro
And in turn, FT was based on the old 110 film format. Kodak decided to make CCD sensors based on their three major film formats, 135, APS-C and 110. The 135 sensors were prohibitively expensive for everyday photography, so manufacturers went for smaller sensors. Kodak and those that followed used the APS-C size and put them in their 135 bodies. Olympus, who had exited the SLR market some years earlier, and had no SLR bodies to use, built a completely new system around the 110 sensor.
Porky89
Leading Member
What he means is that with a full frame camera you have to reduce the aperture by two stops to get the same depth of field as with a M4/3 camera, all else being equal.
Porky89
Leading Member
Why so offensive? Rather rude. Incidentally, can you name a text book that explains why, under typical conditions, most of the noise that comes from a camera is due to photon shot noise? I really would be most grateful if you would be so kind as to do so as I have been quite unable to find one, and yet presumably if it is basic text book stuff they must be readily available.
I am not disputing anything. I am asking for information.
Really? I did not realise digital cameras went that far back - or electronic cameras of any kind for that matter. You learn something new every day, don't you? You don't have any links to articles on the subject, or the names of text books on the subject, do you? That would be most helpful.
Porky89
Leading Member
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Mark S Abeln
Forum Pro
The best sensors these days have an average read noise of about two electrons or lower at best or maybe ten times that much at worst. Of course, older cameras tend to have much higher read noise.
If a sensor has a full well capacity of say 60,000 electrons, the estimated shot noise is no less than the square root of that value, or about 255 electrons, which overwhelms the read noise; even for a well-exposed middle gray tone, the shot noise will be at least 100 electrons on average. The actual values of the shot noise will be higher as most camera sensors have a quantum efficiency considerably less than 100%: they don't absorb 100% of the light, which is a good thing if you want color photography.
Quality terrestrial photography generally tries to keep the image well above the read noise floor.
Right. The read noise floor limits the dynamic range of a camera. But most folks don't operate their cameras in that range: astronomers do of course, and I'll use dark frame subtraction when I'm taking long exposures at night, but the majority of photography is well within the range where shot noise predominates.
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http://therefractedlight.blogspot.com
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StuartT50
Well-known member
Mark S Abeln
Forum Pro
Shot noise - Wikipedia
Loading…
www.clarkvision.com
You can calculate the relative amounts of shot versus read noise from the data found on this site:
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www.sensorgen.info
Single pixel, single bit depth digital cameras have been around for a long time — they are called optical proximity switches — and Albert Einstein got the Nobel Prize in Physics based on an article he wrote on this photoelectric effect way back in 1905. Basically, he interpreted the experimental data as being from the quantum properties of light, which immediately leads to the statistics of shot noise.
Porky89
Leading Member
You say that what I said is completely false and then say exactly the same thing using different words. Why are you being deliberately argumentative? But, more importantly, surely the reason we have a bigger number of grains on the bigger negative is because it is bigger? In fact if you actually do such a comparison in practice (as I have done many times) one finds the number of grains per unit area and the size of the grains is identical. Where does this business of more light to expose them come from? The average light intensity per unit area is exactly the same and the grains are exactly the same size so where is the connection between total quantity of light and grain size. An explanation of your reasoning showing the connection would be most welcome as it seems to be a non-issue.
Really? Try taking a photograph through a telescope of the night sky and then put the lens cap on and make another exposure. If you then examine both frames closely you will find that the noise in both images is identical even though absolutely no photons whatsoever are reaching the sensor during the exposure of the blank frame.
That is why in astrophotography we take blank exposures at the same time as the actual photographs and then subtract them from the image. The noise produced by the camera being constant - even when no photons are striking the sensor at all. Whereas the random fluctuations in the rate of arrival of photons average out during the exposure and are not visible in the final image. Try it for yourself.
Mark S Abeln
Forum Pro
They are typically bigger.
This website gives us some data to consider:
http://www.sensorgen.info
The "Max Saturation capacity" figure is a rough estimate of pixel size, and typically bigger pixels have bigger values. Of course, more megapixels on any sensor will tend to reduce the size of these, but generally speaking, the full frame cameras usually have much higher pixel capacities than compact cameras.
The Quantum Efficiency value incorporates color filtering, so 100% QE would mean that you'd only get black and white images; but QE values also take into account the spacing of pixels: widely spaced pixels will have lower QE than densely packed pixels.
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http://therefractedlight.blogspot.com
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StuartT50
Well-known member
bobn2
Forum Pro
What you said was: "The total quantity of light the negative receives plays no part and is just an interesting, but irrelevant, correlation." (emboldened above). That is, in fact completely false, as I illustrated.
Well, duh, as the saying goes.
How do you expose more grains without having more light? To expose a grain means incidence of some number of photons (the number depending on the efficiency of the emulsion). The photons don't get re-used. Their energy is dissipated freeing photoelectrons. So, to expose more grains you need more photons. If you didn't have more photons, you wouldn't expose any more grains.
tautology
Because we are not talking about light per unit area, we are talking about the total light, to expose the total number of grains. Light per unit area times area equals amount of light.
It was perfectly well explained the first time round. I've tried to explain it even more simply above. Frankly, if you can't work it out now, you're probably never going to understand. It seems a simple concept. Same light per unit area, bigger area gives more light.
Really.
The night sky is not at all a typical subject.
Actually, that is a different reason. Noise, proper noise, from shot to shot will always be different, just because its random. Subtracting that will not reduce noise. What you're removing with your dark frame subtraction is 'pattern noise', caused by the sensor and its readout having systematic differences between lines and columns.
As above, the night sky is a highly atypical subject. It's all deep shadow. It's not surprising that what you see is shadow noise. Now here is an experiment for you. Take a picture of the night sky like you would one of a normal subject. Take your camera. Point it at the night sky. Set the ISO to the lowest your camera has, take a meter reading using averaged or centre weighted. Now take a photo at using the exposure settings given by the meter. Look at the result. Tell me how noisy the night sky is and whether the electronic noise is the dominant noise.
I re-iterate this. If you cannot understand that amount per unit area times area gives total amount, I suspect you're never ever going to understand a 'scientific explanation'.
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Bob.
DARK IN HERE, ISN'T IT?
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Terry Breedlove
Senior Member
I shoot MFT because the sensor is smaller than FF. This gives me smaller lenses and a total smaller package. It allows me to use a wider aperture for faster shutter speeds yet retain a good depth of field for accurate focus at the same time.
bobn2
Forum Pro
No it doesn't give you a 'wider aperture'. In fact it gives you a 'narrower' one, If you have your mFT fitted with a 25mm f/1.4 lens, the aperture is 17.9mm wide. If you have a FF camera fitted with a 50mm f/1.4 lens (that gives the same angle of view) you have an aperture that is 35.8mm wide. That's twice as wide. It's collecting four times the light, so for the same image quality, you could be setting a shutter speed 1/4 that of the mFT camera.
There is no dodging depth of field. Light gathering and depth of field go together, if you want low light performance you get narrow depth of field. Making the sensor smaller does not magically produce light.
By your logic a phone camera with a 1/5" sensor is going to be miles better than either mFT or FF. Clearly it isn't.
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Bob.
DARK IN HERE, ISN'T IT?
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Terry Breedlove
Senior Member
You don't understand that the aperture is a ratio RATIO. f 2.8 on ff is f 2.8 on MFT is f2.8 on a large format camera concerning exposer. It is a RATIO. Damn people get with it already. Now because it is a smaller sensor the image quality will not be as good as a larger sensor. Guess what we already understand that but are willing to make the trade off for the smaller system. Just as the old 35mm was not as good as a 4x5 camera. However when calculating exposer it is all the same since the sensor area is larger on FF than it is on MFT the aperture needs to be larger because it needs more light to get the same exposer since it has a larger area to spread the light on. The exposure is the SAME because it is a RATIO designed right into the lens. Get it now deal with and go away with that stupid mouthy mentally handicapped garble.
GodSpeaks
Forum Pro
Actually, I think the digital matches or beats 35mm film threshold came at around 10MP.
With FF sensors now at 36, 42 and 50MP, we are definately well into medium format film territory.
Leonard Migliore
Forum Pro
I think aperture is a dimension, like 10mm or 15mm. It's the diameter of the entrance pupil. Now, f/stop is a ratio; some call it relative aperture. And, as you note, it works good for normalizing exposure between lenses of different focal lengths.
Now, in typical photographic parlance, we would say "I used an aperture of f/4" and everybody knows you mean relative aperture. But this thread, at this point, is a somewhat technical discussion and it's important to use accurate terminology.
So I find it somewhat unproductive for you to devolve into insults because your less-then-accurate statement has been taken at face value.
mostlyboringphotog
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mostlyboringphotog
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Cool info - thanks!
Just curious though as 110 instamatic was so much smaller... Also, the image quality was nothing to write home about. :-(
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