Total Light Theory continued.

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Total Light Theory continued.
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Continued from this thread.

In particular, I wish to address two replies to the following post:

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Dimitris Servis wrote:

"But the single greatest factor in the noisiness of a photo is how much light the photo is made from"

Yes.

1. How do you define the noisiness of a photo?

In terms of luminance noise (as opposed to chroma noise), the noise is the standard deviation of the recorded signal from the mean signal, where the mean signal is taken to be the "true" signal.

For example, if you take measurements of 90, 105, 97, 110, and 98 electrons, the mean is 100 electrons and the standard deviation (noise) is 7.7 electrons, resulting in a relative noise of 7.7 / 100 = 7.7%.

In addition, noise has both a magnitude (demonstrated above) and a frequency. For example, let's consider two photos of the same scene, one photo made with 4x as many pixels as the other and with the assumption that the electronic noise (the noise from the sensor and supporting hardware) is insignificant compared to the photon noise (the noise from the light itself) or that the electronic noise from the two sensors is essentially the same. The noise of the photo with 4x the number of pixels will have a noise frequency that is twice as high. But, while the individual pixels will be more noisy (have a greater relative deviation), will the photo itself be more noisy? The answer is no, no it will not -- the photo made with fewer pixels will be more blurry.

2. What is the mechanism that connects the definition of noise in (1) with how much light the photo is made from under the assumption that it is captured by an array of nxn independent pixels connected to off-chip amplifiers?

If we have two photos of the same scene displayed at the same size, then, with the same assumptions about electronic noise discussed in the previous paragraph, the photo made with more light will be less noisy.

Some thought experiments:

1. Use a D800 once with a 50 1.4 and once a 35 1.8. How do the two images compare with respect to your noise metric? How do they compare to a D7000+35 1.8?

2. How does a theoretical 4/3 6,549 x 4,912 image compare to the 7,360 x 4,912 D800 image relative to your noise metric?

3. As a frequent user of a D750 and an Om-d Em5 mk ii how are the variables available to me before taking a picture affected by your metric? How does your metric manifest itself in my A3+ prints?

How about a demonstration? Here's the Canon 6D at ISO 6400 and Olympus EM5 at ISO 1600, thus both photos made with the same total amount of light (additionally, both sensors have similar electronic noise levels). All but identical with regards to noise.

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Reply to Dimitris' response:

I don't have images with electrons. Sorry.

Well, Dimitris, you've expressed the crux of your misunderstanding. Light is composed of photons. Those photons release electrons from the silicon in the sensor which the camera records. The counts of those photoelectrons are the very basis of the information that is needed to create the photo.

Reply to Jack Hogan:

He is not interested in equivalence, shutter speed is different and Exposure is the same.

I assume you mean the example I linked. The exposures are not the same for the two photos -- the linked example shows the same noise in two photos made with the same total amount of light.

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