I been trying to find the way I can calculate the telescope lens diameter needed to see a star by the distance of the star or the planet from the earth. Is there any mathematical relation we usually use to build our telescopes lens?

Example : Proxima Centauri b is about 4.24 light years away from us. We need a telescope that will capture its atmosphere. How big would the lens of that telescope needed will be?

  • $\begingroup$ A telescope the size of a galaxy in order to observe a star system at a tiny fraction of the galaxy's diameter: you'd be building a frigging huge microscope, but not a telescope. $\endgroup$ Jan 19, 2023 at 13:35
  • $\begingroup$ joke aside: can you share some of your own thoughts on how to approach this. There are common formula to derive the size of an optic in order to resolve a certain angle. Calculate the angle you want to resolve, then apply that formula for the aperture of the optics. en.wikipedia.org/wiki/Angular_resolution $\endgroup$ Jan 19, 2023 at 13:38
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    $\begingroup$ well , i know but i just need the mathematic relation i will play with it $\endgroup$
    – xone-a
    Jan 19, 2023 at 13:38
  • $\begingroup$ I'm not sure whether you need a telescope as big as galaxies, a telescope the size of earth would be sufficient. Because an telescope the size of Earth known as EHT can capture something more than 2000 lightyears away $\endgroup$
    – user47732
    Jan 19, 2023 at 15:28

1 Answer 1


The relation for angular resolution (angle $\theta$), the wavelength $\lambda$ and the telescope opening $D$ is simple, given by the diffraction formula for the Airy disk:

$$ \sin \theta \approx \theta = 1.22 \frac{\lambda}{D}$$

The angle $\theta$ can be calculated from the triangle from observer to object-of-interest (distance $d$ and its extend $r$: $\theta \approx r/d$.

For a distance of $d = 4.24 \mathrm{LY} = 4.01\cdot 10^{16}$m, you get for 1000km-size features an angle of $\theta = \frac{1,000,000\mathrm{m}}{4.01\cdot 10^{16}\mathrm{m}} = 2.5\cdot 10^{-11}$, and thus $$ D = 1.22\cdot \frac{\lambda}{\theta} = 1.22\frac{500\cdot 10^{-9}}{2.5\cdot 10^{-11}} = 25000\mathrm{m} = 25\mathrm{km}$$

Yet you don't need to resolve the atmosphere or planet at all in order to detect and analyse it: you can do spectroscopy and compare the host star's spectrum to the planet's spectrum. The difference will be the influence of the planet's atmosphere. This is already being done.

  • $\begingroup$ i did some things and find out the D diameter of the leng will be 3 meters :) :( $\endgroup$
    – xone-a
    Jan 19, 2023 at 19:29
  • $\begingroup$ in second attempt i have fixed the theta the equation gave 'D is 54 km.' $\endgroup$
    – xone-a
    Jan 19, 2023 at 19:42
  • $\begingroup$ am pretty sure thats not true at all haha but anyways i think i will need some help $\endgroup$
    – xone-a
    Jan 19, 2023 at 20:13
  • $\begingroup$ @xone-a "i did some things" perhaps you could edit those things that you did into the question... But a 54km optical telescope seems reasonable for resolving an exoplanet (ie utterly beyond current capacity) so that might be right. $\endgroup$
    – James K
    Jan 19, 2023 at 20:49
  • $\begingroup$ For a 1000km resolution, and visible light at a 500nm wavelength, my result is a diameter of 250km. $\endgroup$ Jan 20, 2023 at 9:16

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