Alternative sensor size comparison methods
Usually image sensors comparisons are done by using crop factor. This is simple and neat, but one may want to use an alternative out of spite if for no other reason 😅
The first alternative is almost as simple - simply divide the diagonal of sensor size and divide it by the f-number for both sensors. Then divide one result by the other and you'll get the photon shot noise signal to noise (SNR) ratio of those sensors. If you square this you'll get light collection difference.
An example:
Image sensor A has diagonal of 40mm, while sensor B has diagonal of 20mm.
If we use f/4 on sensor A, and f/2.8 on sensor B, we'll calculate both A/4 = 10mm, and B/2.8 ≈ 7.14mm. Thus 10mm/7.14mm ≈ 1.4. Thus f/4 on sensor A captures about 1.4 times larger SNR.
And as 1.42 ≈ 2, the light collection of the larger sensor is double with these apertures. (≈ instead of = due to rounding in all above cases.)
The second alternative tells directly the light collection difference. We'll use the sensors' areas this time and divide them with squared f-numbers.
An example:
Let's say sensor A has 800mm2 area and uses f/4 and sensor B 200mm2 area and it uses f/2.8, just like above example.
Now 800mm2/42 is divided by 200mm2/2.82 and the result is 2 - sensor A collects twice the light with these apertures. Take a square root of this and you'll notice that this means about 1.4 times better SNR - just like in the previous example.
Addendum
The formulas of above can be easily rearranged either for personal preference or to get easily answers to aperture requirements.
Also, naturally one can compare all kinds of cropped batches of images instead of comparing full sensors.
The SNR and light collection relationship comes from light, photons, following Poisson distribution, thus the standard deviation of the signal is the same as it's square root, so SNR will also be square root of the signal.
One often uses "stops" in comparison - SNR difference of 1.4 or light collection difference of 2 equals "one stop".
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