Crop Factor, Low Light and Aperture with m4/3 lenses? Part 2

Started 6 months ago | Discussions thread
Anders W
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Here is how it actually works
In reply to Truthiness, 6 months ago

Truthiness wrote:

knickerhawk wrote:

We're talking about cropped sensors, not cropped images.

It does not matter how you crop the image!

If you need to stretch a 1cm image to 100cm output size, the aberrations of the lens will be more visible than if you stretch a 2cm image to 100cm output size. This is obvious so it is a bit odd how someone gets the motivation to argue it.

Apparently, you overlooked some of the earlier replies in the thread, for example this one:

http://www.dpreview.com/forums/post/53899308

To reiterate, as Cossairt et al. put it in the paper I link to in that post:

"In conventional lens design, resolution is limited by the spot size of the lens. For a lens with aberrations, spot size increases linearly with the scale of the lens."

In other words, for two lenses with identical designs but scaled in all relevant dimensions so as to create image circles of different size (e.g., a 50/1.8 for FF scaled downwards so as to make a 25/1.8 for MFT), the aberration-limited resolution on a per-image basis will be exactly the same at the same f-stop.

Consequently, the claim of yours above is incorrect. What it overlooks is the elementary fact that the loss of resolution due to aberrations in the 1 cm image will be proportionally smaller than in the 2 cm image for two lenses with exactly the same design but made to cover differently sized image circles.

However, the diffraction-limited as opposed to aberration-limited spot size does not scale in the manner described by Cossairt et al. Rather the diffraction-limited spot size remains the same at the same f-stop. Consequently, the smaller format has a disadvantage as far as diffraction is concerned.

The general implications of these two facts (the one regarding aberrations and the one regarding diffraction) can be illustrated graphically by means of the stylized graphs below (borrowed from a post where I demonstrated the same thing to bobn2 four months ago here).

The first graph shows how the two lenses (identical except for being scaled so as to yield image circles of different size) would generally compare at the same f-stop. At very wide apertures there would be very little difference since optical aberrations is the dominant factor in determining sharpness levels and diffraction hardly matters at all. The sharpness level will therefore be pretty much the same for MFT and FF alike. As we stop down, optical aberrations are reduced and diffraction starts to play a bigger role. Since FF has the benefit of less diffraction at any given f-stop, it will gradually get further and further ahead as diffraction becomes a more and more decisive factor and once we get to f/16, there will be a rather significant difference between the two systems.

View: original size

If, instead we compare at FF-equivalent f-stops (those f-stops where the level of diffraction, the DoF, and the total amount of light that falls on the sensor at the same shutter speed are the same), so that f/1 on MFT is compared with f/2 on FF, f/2 on MFT with f/4 on FF, and so on, it will be the other way around. At very wide apertures, the advantage of FF over MFT will be major rather than minor due to the optical aberrations at f/1 (MFT) being far more serious than those at f/2 (FF). As we stop down, the role of optical aberrations is again reduced and that of diffraction increased. Since diffraction at equivalent apertures is the same, the result is that the FF advantage is reduced to eventually reach a trifling level.

But it's your own logic. YOU were the one who started this discussion by noting that greater magnification comes with an unavoidable resolution penalty.

I state(ed) the following, nothing more:

  1. With a perfect sensor and a lens with aberrations we will have a limited resolution which can be measured in line pairs per millimiter on the image plane.
  2. What the resolution of the output image is depends on the crop factor and can be measured in line pairs by image height.

Do you agree or disagree?

The problem is that you only applied that penalty from an interim point in the magnification (i.e. the sensor plane) to the output display plane.

Huh! The lens draws an image with finite resolution. The more you have to enlarge it, the more the aberrations will show.

You didn't account for the "penalty" incurred between the lens and the sensor (i.e., the focal length).

Huh!

A Carl Zeiss Jena 35mm f/2.4 Flektogon has 35mm focal length regardless fo the crop factor.

With ideal perfect sensors a cropped sensor will turn a lower resolution output image on perfect output device than what a full frame sensor would do because you need to enlarge the image more, thus enlarge the aberrations more too (after all, there just part of the image).

If you change the lens or the optical configuration, that is your bussiness, but has nothing to do with the point. For example by using the focal reducer you're improving the optical configuration by reducing the aberrations (for example 1µm aberration on the image with a naked lens would reduce to 0.5µm aberration with a perfect 2x focal reducer). High quality focal reducer is certainly a better option in most cases than a conventional adapter.

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