Micro four thirds - does the crop factor apply to MFT lenses as well?

We are speaking on a signal/noise ration in an image, aren't we? The image is not created by a single, two or even by three pixels. This is trivial, isn't it? To form an image we need what is called a frequency band that is created by a huge number of pixels. The wider the band the more detailed image we can get. Different content of the image belongs to different space frequency range inside the whole band. A particular "signal" is always occupies a part of the whole spectral band or even can be located at a single frequency.  The wider the band the more noise power we have inside this band. If the noise is homogeneously distributed over the frequency range then decreasing the frequency band two times results in double decreasing of the noise power, or we have a square root of 2 the decrease in its average spectral amplitude (the decrease of noise power by 2 results in increase of signal/noise by a factor of sqrt(2) for a single-frequency signal). And this is just a law for any physical system.  So it remains valid for both photonic noise and dark noise if the last is characterized by a homogeneous spectral distribution.

The lower frequency band results in a better signal/noise ratio and higher dynamic range. However, the payment for this is lost of fine details in the image.

What is the frequency band in the case we are speaking on?

It is defined by several basic elements.

i) The first element that has a finite frequency band is a lens. In the ideal case its highest frequency edge is defined by an inverse value of a size of the diffraction spot it provides for the plane waves of the visible light spectrum. The lowest frequency edge of the band is equal to an inverse value of the size of the whole image (limited by a sensor size). For example, a bad lens that has aberrations limited resolution can decrease a frequency band and improve a signal/noise ratio and the dynamic range.

ii) The discrete nature of a sensor results in a periodically repeated spectrum of the signal (image). For this reason so-called AA-filters are used for cutting-off just a single spectrum from the repeated series in order to escape artifacts in the image.  In an ideal case the highest spectral edge of an AA-filter is limited by an inverse value of the pixel pitch.

Thus the band width of the sensor is just the difference: (1/pixel_pitch -1/sensor_size). In the case of the pixel pitch is much less than the sensor size (that is the typical case we speak on), the sensor band width is just defined by the value 1/pixel_pitch.

The pixel pitch of m43 16M sensor is two times less than the pixel pitch of 16M 35mm sensor. Thus in this case the frequency band of the m43 sensor is two times wider for each of the two independent directions (height and width). This fact results in that the power of noise is two times higher for each of the independent directions in case of m4/3 sensor compared to 35mm sensor. The final result is that signal/noise is just 2 times (one stop) better (of course, herein we speak explicitly on the influence of the geometry of the sensor, neglecting the lens and reading circuit contributions).

iii) the electronic circuit should be capable to provide fast reading the signals from individual pixels. The faster the reading (now we are in time domain instead of the space domain) the wider frequency band of the electric circuit is necessary. The "payment" for the faster reading is, of course, an additional noise.

A bit more on the time domain. If we think on an accumulation of photons in a single pixel then to get more photons we need more time. The longer time duration means the lower frequency band.

I have pointed only some basic thins. The detailed picture is, of course, more complicate, but I hope that these my short comments will be useful for those who really wants to know a bit more on what is going on.

SP.