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Assuming the SNR is shot limited. Would it be the square root of 2 because of the shutter speed?

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2 Answers 2

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Yes, cutting the amount of light in half cuts the number of collected electrons in half. Under the usual assumption of a Gaussian error the fractional standard deviation is multiplied by the square root of 2, so the signal to noise ratio is divided by the square root of 2.

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  • \$\begingroup\$ You should consider the electronic noise. If going from ISO 100 to 200 it will probably be unnoticeable, but working at high ISOs, the electronic noise is probably much higher than the "light" noise. \$\endgroup\$ Apr 22, 2022 at 5:02
  • \$\begingroup\$ Quite the opposite. The less light there is, noise due to the randomness of light, what we call Poisson distribution or "shot" noise, is a higher percentage of the total noise. \$\endgroup\$
    – Michael C
    Apr 22, 2022 at 9:10
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The primary cause of noise in modern cameras is shot noise, which follows a Poisson distribution... that is, the noise component is equal to the sqrt of the signal collected (photoelectrons).

E.G. if the original SS collected 10,000PE the noise component would be 100 for an SNR of 100:1. Then at 2 x SS you collected 5,000PE the noise component would be 71.7 for an SNR of 70:1. The longer shutter duration collects both more signal and more noise, but it also results in a better SNR.

Read (conversion) noise is also variable in that it is also more prominent when the ADC is provided with a lower quality/noisier signal.

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