So, is it true that a Four Thirds lens needs to be about twice as ‘sharp’ as its Full Frame counterpart in order to be able to display an image of spatial resolution equivalent to the larger format’s?
It is, because of the simple geometry I will describe in this article. In fact with a few provisos one can generalize and say that lenses from any smaller format need to be ‘sharper’ by the ratio of their sensor linear sizes in order to produce the same linear resolution on same-sized final images.
This is one of the reasons why Ansel Adams shot 4×5 and 8×10 – and I would too, were it not for logistical and pecuniary concerns.
‘Sharpness’ is in quotes above because in this post we are referring to the objectively measurable, quantitative, linear spatial resolution IQ of the systems obtained solely from unprocessed raw data – not the subjective, qualitative, perceived sharpness from processed images. See this article if you are interested in understanding the difference.
Imagine that we are evaluating equally sized images displayed side by side, in this case to determine which is ‘sharper’. We don’t know what format produced what image. All we know is that they all show the exact same scene content, a tall tree filling the frame, captured from the exact same spot. This is how the 8×12″ displayed photographs appear to our eye:
From where we are standing we can see the branches clearly. The tree fills the frame, which means that it filled the sensing medium that took it. Not knowing the format that took what image we can imagine how the tree might have looked on various sensing media at the time of capture:
The figure above shows correct relative sensing sizes, the length of their diagonals noted on top. Note how much harder it gets to count branches as the format gets smaller. This is telling of the much harder job that a smaller format lens has to accomplish in order to provide the same spatial resolution as larger ones in the final displayed image.
How much harder? Let’s assume that MTF50 resolution on the displayed 8×12 images is the same, about 6 line pairs per mm. The diagonal of 8×12″ is 366.4mm, so if the detail came from a FF format camera (43.2mm sensor diagonal) that same detail would have been 50.9 lp/mm on its sensor (6 x 366.4 / 43.2). Performing the same projection for the other systems produces the following table:
You can see in blue the ‘sharpness’ required by each format’s imaging system in order to provide a final 8×12″ image of 6 lp/mm spatial resolution. The values are linear with sensor size because of simple geometry as shown below:
So yes, lenses from a smaller format need to be ‘sharper’ by the ratio of their sensor linear sizes in order to produce the same linear resolution on same-sized final images. mFT lenses need to be twice as ‘sharp’ as FF ones for instance, 102 lp/mm vs 51 lp/mm respectively, so that same-sized final images will look equally ‘sharp’ to the viewer.
Are they twice as sharp? It depends on your working f-number but typically not if looking at prime lenses of equivalent price, the subject of a future post.
PS Also note that the issue goes away if we express linear resolution as a function of picture height, as discussed earlier:
mFT’s value is a bit different than the others because it has a slightly different aspect ratio, so it suffers the ‘landscape convention’. If we expressed the resolution in units of line pairs per picture diagonal they would all read the exact same, 2198 lp/pd.