MFT Thread Number 60,000 ...

Great Bustard wrote:

Basically, the presumption is that when we compare the IQ of two photos, we do so at the same display size. For example, we wouldn't compare a 4x6 inch photo with an 8x10 inch photo.

Indeed, the measure of resolution, lw/ph (line widths per picture height) is the measure of choice for MTF-50 tests, as opposed to lp/mm (line pairs per millimeter on the sensor).

So, when we compare the noise of two different photos, it makes sense to compare area for area rather than pixel for pixel (unless, of course, both photos are made of the same number of pixels).

My preferred unit of measurement is the μphoto (microphoto -- one millionth of a photo). For example, if we were comparing the 16 MP EM5 to the 36 MP D800, we wouldn't compare pixel against pixel, but 16 pixels from the EM5 to 36 pixels from the D800.

The effect of the sensor size is simply that larger sensors usually allow more pixels, and that lenses don't need to be as sharp as lenses for smaller sensors since the final photo is enlarged less.

For example, let's say we were going to make a 24 x 24 inch (610mm x 610mm) print from an EM5 and D800 (the square aspect ratio slightly favors the EM5 due to the more square aspect ratio of the 4:3 sensor as opposed to the more elongated aspect ratio of the D800's 3:2 sensor).

The EM5 sensor measures 13mm x 17.3mm and the D800 sensor measures 24mm x 36mm. So, the enlargement factor for the EM5 photo is 610 mm / 13mm = 47x and the enlargement ratio for the D800 sensor is 610mm / 24mm = 25x. So, any imperfections in the EM5 photo are magnified by approximately a factor of 2 compared to the D800 photo, thus a lens for the D800 need only be half as sharp as a lens for the EM5 to deliver the same detail.

Well, it's a little more complicated, since the D800 has more pixels, so the above would be valid if both sensors had the same pixel count. But the greater pixel count of the D800 extends the advantage.

On the other hand, the smaller pixels of the EM5 have less read noise (the additional noise added by the sensor and supporting hardware) due to their smaller size (since the sensors likely share very similar tech). Thus, as the light gets lower and lower, the read noise becomes more and more prominent. And noise, as we all know, robs detail.

So, as the light lowers, the detail advantage of the D800 over the EM5 lessens, much in the same way that as the DOF deepens, the effects of diffraction softening lessen the advantage, or how motion in the scene reduces the resolution if the shutter speed is not so fast that the subject does not move more than two pixels (for a Bayer CFA) during the course of the exposure.

Anyway, no real mysteries here. First, we compare area for area (on the photo, not on the sensor). Secondly, we need to consider the lens in the context of the enlargement factor. Third, we have to understand that more pixels results in more resolution than fewer pixels. Lastly, we must account for degradation due to noise, diffraction, and motion blur.

Simple, is it not?