Technically speaking, more signal (light), more noise, and higher SNR. If you normalize two images so the max signal is max white in both, then the apparent noise is lower for the photo with the larger signal, owing to its higher SNR.
What myths? I strive for accuracy and not one of the above mentioned people has ever emailed me about any issues with anything on my web site that I can remember.
Thanks for the more specific links to my reply.
In the audio world, equipment with response from 20Hz to 20Khz records better quality than equipment with 40Hz to 14Khz. And playback of 24bits/192Khz from a hi rez album sounds better than its 16/44.1 low rez brother.
Could you be more specific & explain further? From what I gather :
Indeed. This is why I make every effort (especially in this forum) to say "less noisy" as opposed to "less noise".
For sure. A "less noisy" photo has less "apparent noise" (lower NSR; higher SNR).
The lux is not just the unit of light intensity (really luminous power or emittance): it is the light intensity per unit area (its units are lumens per square-meter, and lumens are luminous energy per unit time). The problem is that having the same amount of light per unit area doesn't mean the same light over different areas. Exposure is the light per unit area accumulated over an exposure time, thus lux-seconds. Equal exposures, then for the same light source place the same amount of light per unit area on the sensor. A larger sensor therefore, such as FF vs mFT with 4 times the area, will receive and record 4 times the light, assuming equal QE (properly used). And, since it has 4 times the total light, it has twice the s/n, considering only shot noise.Could you be more specific & explain further? From what I gather :
The lux (symbol: lx) is the SI unit of illuminance and luminous emittance, measuring luminous flux per unit area. It is equal to one lumen per square metre. In photometry, this is used as a measure of the intensity, as perceived by the human eye, of light that hits or passes through a surface.
How did I use lux wrongly? Would it be more correct if I said the image circle is 0.xx sq meter since it is not a sq meter in size?
Or maybe the pic of the light meter is wrong, perhaps changing to this one is more accurate
So if lux as a unit for light intensity is wrongly used here, what should be the correct unit?
.Now, the amount of light a photo is made from begins with the amount of light that falls on the sensor, where more light falls on the sensor for larger sensor systems for the same exposure (but the same amount of light falls on the sensor for photos of the same scene at the same DOF with the same exposure time).This thread is hoping to address the common misconception regarding sensor size, pixel size & noise from a recent discussion on the m43 forum here. Do spare a few mins & read through what I have to say to determine if it is correct and makes sense. I try to be as concise & simple, so this is going to be long.
My argument is that bigger sensor size (eg. FF) doesn't bring about having more light captured than the smaller sensor. It is the pixel size & its quantum efficiency (QE) that determines the output signal's quality or fidelity (Signal to Noise Ratio). QE is used to describe how well a pixel can convert light photons to electrical charges, the bigger the pixel size, the better the QE, the higher the signal & thus higher SNR.
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Have a good day y'all. Feel free to exchange your pointers.
Still couldn't quite make out what you mean Joseph. More light falls on the sensor for larger sensor systems? From Clarkvision:
The argument for higher sample rates is good. The equivalent in photography is a larger number of pixels. This reduces, or even removes, aliasing and moire. There are not yet any consumer cameras with over-sampling sensors (unless a very low resolution lens, or a pinhole, is used).
I think we all have the bad habit of saying "noise" when we mean "Signal-to-noise ratio".
It might be better to think of it as 4 times the total number of photons.
I do not think this is true. It would be true for a single photodetector taking 4 sequential measurements to find the average light level, but not for an array of detectors where the signal is the differences between them (such as black and white bars on a test chart).
Good. Although this change is only appropriate if, again, we assume equal QE, for, otherwise, a different proportion of those photons will manifest themselves as charges on the sensor, which is what is really relevant.
And now, having established that it has 4 time the number of photons, I guess what you're saying is that an image made from the larger sensor with 4 times the photons, and therefore having Sqrt(4) = 2 times the noise, would not have 4/2 = 2 times the s/n. Is that right?
Careful with the analogies. You're saying that a 10 second sound clip sounds better tan a 5 second one, but that's the wrong quantity. In sound recording, temporality is an essential aspect of the data record. In still photography, as long as the subject doesn't move you can trade recording time for signal quality, if you're considering only shot noise. Compare an ISO 6400 image to an ISO 100 image. 5 stops more light in the latter, higher acuity and smoother backgrounds as a consequence. Since the shot noise in each pixel is independent, we can apply the same statistics to the spatial array that we would for a series of repeated trials on a single pixel.
I remember quite a lengthy discussion on the Usenet forums between yourself, Emil Martinec, John Sheehy and myself, where, unless I'm remembering wrongly, you flat out refused to countenance the errors that were pointed out to you. Email is not the only means of communication.
exactly, but the discussion was here and based from my side on a page which I also referred to at that time.
yes, like to say that the noise will increase at higher iso, nope , but the signal to noise ratio are different
I am saying that it doesn't. The "total light" theory says that it does, simply because more samples are taken.
And in still photography space (2-dimensional) is the equivalent aspect. A bar test chart for sensors is the same as a square wave signal for audio tests.
That is about ETTR, not sensor size.
A series of exposures on one pixel gives you a more accurate measurement of the DC illuminance on that pixel. A picture made up from these more accurate measurements will have a better SNR.
Likewise, bigger pixels give you lower resolution and worse aliasing. Nothing to do with sensor size, although I suspect that spatial frequency bandwidths (of noise and signal) do have something to do with the better image quality from larger sensors.
Sound particles are called phonons. They are quantised. But sound recording seldom explores very low sound levels (although the quality of room acoustics can be quite hard to record, and so can the reverberations in a piano). Photographers regularly work in very low light levels.
Sound recording is mostly done in the audio equivalent of bright sunshine.
For the same exposure, more light falls on larger sensors in proportion to the ratio of the sensor areas (e.g. 4x as much light falls on a FF sensor as an mFT sensor for the same exposure)..Now, the amount of light a photo is made from begins with the amount of light that falls on the sensor, where more light falls on the sensor for larger sensor systems for the same exposure (but the same amount of light falls on the sensor for photos of the same scene at the same DOF with the same exposure time).This thread is hoping to address the common misconception regarding sensor size, pixel size & noise from a recent discussion on the m43 forum here. Do spare a few mins & read through what I have to say to determine if it is correct and makes sense. I try to be as concise & simple, so this is going to be long.
My argument is that bigger sensor size (eg. FF) doesn't bring about having more light captured than the smaller sensor. It is the pixel size & its quantum efficiency (QE) that determines the output signal's quality or fidelity (Signal to Noise Ratio). QE is used to describe how well a pixel can convert light photons to electrical charges, the bigger the pixel size, the better the QE, the higher the signal & thus higher SNR.
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Have a good day y'all. Feel free to exchange your pointers.
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Still couldn't quite make out what you mean Joseph. More light falls on the sensor for larger sensor systems?
Indeed. The light density (exposure, measured in lux seconds, or, equivalently photons/mm²) is the same for the same scene, relative aperture, lens transmission (t-stop vs f-stop). and exposure time.
The same light density over a larger area results in a greater total amount of light. For example, a bowl with an 8 inch diameter placed in the rain will collect 4x as much water as a bowl with a 4 inch diameter. Likewise, the same exposure on a sensor with 4x the area results in 4x as much light falling on the sensor.
For the same perspective, framing, and aperture (entrance pupil) diameter, the DOF will be the same. For example, 50mm f/2 on mFT and 100mm f/4 on FF both have the same [diagonal] angle of view and aperture diameter (50mm / 2 = 100mm / 4 = 25mm), so if photos of the same scene are taken from the same position (same perspective), the DOFs will be the same.
For example, 50mm f/2 1/100 on mFT and 100mm f/4 1/100 on FF (set the ISO setting to taste for the desired output brightness, or just use Auto ISO).
You can select any exposure time you like. Alternatively, you can let the camera choose the exposure time for you in various AE (auto exposure) modes and/or adjust the exposure time indirectly with the ISO setting.
Let's see if you'll still thank me for the more in-depth explanation:Thanks for the simple explanation. I understood the whole CFA thing.
The one I was thunking about was on the Usenet forums, I very distinctly remember Emil encouraging me to join the discussion. At the time, Roger wasn't posting here.
