I have been trying to wrap my mind around how the resolving power (in terms of the modulation transfer function, or MTF) of individual elements in an optical system affect the overall resolving power of the system, and I've come across the following formula (which I believe is an approximation) in a few places, namely this document released by Kodak on the subject of film resolution

EQ1: 1/R = 1/r [film] + 1/r [camera lens] + 1/r [enlarging lens] + 1/r [printing paper]

This formula, or some variation of it, crops up a lot in literature about camera system resolution, but I'm having a hard time tracking down what the theory behind it is. Additionally, in my own efforts to measure the resolving power of imaging systems with resolution test targets, I am getting results that seem absurd if this formula is accurate.

Just for example, I have been testing my camera (A7r IV) with a macro lens on a 1951 resolution test target, and I am able to resolve somewhere between Element 4 and 5 in Group 6, which would put the system resolution somewhere in 90-100lp/mm range. I'm going to assume the lower end of that range for the sake of argument, and just focus on the resolving power at the center, at a single point rather than considering MTF as a curve. I also know that calculating resolving power using these charts is pretty subjective, but I am just trying to get a general range.

Regardless, assuming 90lp/mm, this equates to 4572DPI. My camera's sensor captures an image with pixel dimensions of 9504 x 6336, and it's a full frame sensor so let's just assume the smaller dimension is roughly equivalent to the sensor DPI. I'm ignoring the effect of the Bayer array here as well. If I plug these numbers into the system resolving power formula adapted for the elements in my system:

1/r[sensor] + 1/r[lens] = 1/r[system]

1/6336 + 1/x = 1/4572

I get a totally absurd lens MTF of 16422, which is well over 300lp/mm and likely impossible to achieve in a consumer lens, particularly not a macro lens optimized for 1:1 reproduction on a flat field.

So I guess my question here is, am I making a math mistake, or is this formula just nonsense? If the formula is nonsense, is there a better way to calculate the resolving power of an optical system from its components, or at least a shorthand for approximating?


1 Answer 1


I do not know the specific definition of your resolution but if it is just another word for SNR (signal to noise ratio) or, even better, NSR (the noise to signal ratio) then the logic is as follows. You are given a signal power $S$ (mean square signal average) and assume that to the signal is added various independent noise terms whose mean square average, variance, is $N_1,N_2,..$. Each noise term corresponds to an individual SNR of $\rho_k=\frac{S}{N_k}$. For independent (or uncorrelated) noise terms the total noise power (mean square average) is the sum of the individual variances: $N=N_1+N_2+..$. The variances add up to the total variance resulting in the overall $NSR = \frac{\sum_k N_k}{S} = \sum_k \frac{1}{\rho_k}$.

  • $\begingroup$ Ah interesting. I'm trying to wrap my head around how that related to the resolving power in terms of the modulation transfer function. Basically it the ability to distinguish lines pairs apart (by contrast) as the lines increase in frequency. I am using the simplest version of this approach, where I photograph a chart with lines at frequencies that correspond to certain number of line pair per mm, which can be translated to DPI resolution. If I can see the separate lines in the resulting photograph, then the lens can resolve at least that much detail. $\endgroup$
    – flimsy
    Commented Apr 27, 2023 at 20:28

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