Telescopes and Camera Objectives are both optical systems which image objects from far away to a finite image distance. Although camera objectives are often used for finite object distances, in most calculations and even in optical design, it is assumed that the object is placed in infinity.

However, if you look up the resolution formulas for both optical systems, they are substantially different. For telescopes, it seems that only the diameter of the aperture and the wavelength influence the angular resolution 1. The focal length only governs the magnification, but the resolution does not depend on it.

$\theta = K \cdot \lambda / D, \ K \approx 1.22$

In contrast to that, the resolution limit of photo objectives seem to depend on the focal length 2 3.

$\theta \propto 1 / (\lambda \cdot \text{f-number}) = D / (\lambda \cdot f)$

I really don't get it. Why should the resolution of a telescope be independent of the focal length but for photo objectives it should be different?

Can I make photos of the sky with better resolution if I use a camera objectives with a smaller focal length? I mean, it would be stupid if that was possible. Then you could replace all telescopes with large photo objectives so that you have one more degree of freedom for optimising the resolution. It should be optically possible to build a photo objectives with mirrors, however it seems that nobody does that.

  • 1
    $\begingroup$ Your formula for photo objectives is faulty. And note that diameter is approximately (focal length/f-number). $\endgroup$
    – John Doty
    Dec 23, 2022 at 21:46
  • $\begingroup$ Your equation $\theta\, [\propto 1 / (\lambda \cdot \text{f-number})] = D / (\lambda \cdot f)$ is dimensionally incorrect. $\endgroup$
    – Farcher
    Dec 24, 2022 at 8:59
  • $\begingroup$ @JohnDoty That would make sense. Do you have a reference to a correct equation? With derivation would be best. I guess the photographers just change the aperture diameter with the f-number Setting and then tell everyone that the f-number is the key quantity instead of the aperture diameter. $\endgroup$
    – cakelover
    Dec 24, 2022 at 9:41

1 Answer 1


The equation that you have quoted in your question, $\theta \propto 1 / (\lambda \cdot \text{f-number}) = D / (\lambda \cdot f)$ is dimensionally incorrect because the left-hand side of the equation is not a angle but a reciprocal distance.

In your link 2 - The Airy Disk and Diffraction Limit there is the following equation,

$\xi_{\small{\text{Diffraction Limit}}} = \frac{1}{\left( \text{f} / \# \right) \times \lambda} \times \left( \frac{1000 \large{\unicode[Cambria Math]{x03BC}} \normalsize{\text{m}} }{1\text{mm}} \right)$

where $\xi_{\small{\text{Diffraction Limit}}}$ is not an angle but is the largest number of lines per unit length which can be resolved and in table $2$ the resolution which is called 0% Contrast Limit @ 0.520µm is given a lines per millimetre which for an f-number of $2$ is given as $962$ lines per millimetre.
So for an f-number of $2$ the minimum separation of lines which can be resolved is $1/962000 = 1.04\times 10^{-6}\,\rm m$.

Simplifying your equation $\theta = K \cdot \lambda / D$ with $K=1$ the minimum separation in the focal plane of a lens of focal length $f$ is $\theta \,f = \lambda / D \cdot f = \lambda \cdot \text {f-number} $ which in the numerical example that I chose gives $0.520\times 10^{-6} \cdot 2 = 1.04\times 10^{-6}\,\rm m$.
Thus, the difference is that as a Physicist you set $K=1.22$ and Edmund Optics simplified things by setting $K=1$.


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