# What is the diameter of a telescope lens that will capture the surface of Proxima Centauri b?

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?

• 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. Jan 19 at 13:35
• 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 Jan 19 at 13:38
• well , i know but i just need the mathematic relation i will play with it Jan 19 at 13:38
• 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
– user47732
Jan 19 at 15:28

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.

• i did some things and find out the D diameter of the leng will be 3 meters :) :( Jan 19 at 19:29
• in second attempt i have fixed the theta the equation gave 'D is 54 km.' Jan 19 at 19:42
• am pretty sure thats not true at all haha but anyways i think i will need some help Jan 19 at 20:13
• @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. Jan 19 at 20:49
• For a 1000km resolution, and visible light at a 500nm wavelength, my result is a diameter of 250km. Jan 20 at 9:16