<p>The relation for <a href="https://en.wikipedia.org/wiki/Optical_resolution" rel="nofollow noreferrer">angular resolution</a> (angle <span class="math-container">$\theta$</span>), the wavelength <span class="math-container">$\lambda$</span> and the telescope opening <span class="math-container">$D$</span> is simple, given by the diffraction formula for the Airy disk:</p>
<p><span class="math-container">$$ \sin \theta \approx \theta = 1.22 \frac{\lambda}{D}$$</span></p>
<p>The angle <span class="math-container">$\theta$</span> can be calculated from the triangle from observer to object-of-interest (distance <span class="math-container">$d$</span> and its extend <span class="math-container">$r$</span>: <span class="math-container">$\theta \approx r/d$</span>.</p>
<p>For a distance of <span class="math-container">$d = 4.24 \mathrm{LY} = 4.01\cdot 10^{16}$</span>m, you get for 1000km-size features an angle of <span class="math-container">$\theta = \frac{1,000,000\mathrm{m}}{4.01\cdot 10^{16}\mathrm{m}} = 2.5\cdot 10^{-11}$</span>, and thus
<span class="math-container">$$ D = 1.22\cdot \frac{\lambda}{\theta} = 1.22\frac{500\cdot 10^{-9}}{2.5\cdot 10^{-11}} = 25000\mathrm{m} = 25\mathrm{km}$$</span></p>
<p>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 <a href="http://spiff.rit.edu/classes/resceu/lectures/spectra/spectra.html" rel="nofollow noreferrer">being done</a>.</p>

CC BY-SA 4.0
2023-01-19T19:29:44.163Z
i did some things and find out the D diameter of the leng will be 3 meters :) :(
xone-a
CC BY-SA 4.0
2023-01-19T19:42:58.483Z
in second attempt i have fixed the theta  the equation  gave 'D is 54 km.'
xone-a
CC BY-SA 4.0
2023-01-19T20:13:15.540Z
am pretty sure thats not true at all haha but anyways i think i will need some help
xone-a
CC BY-SA 4.0
2023-01-19T20:49:27.907Z
@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.
James K
CC BY-SA 4.0
2023-01-20T09:16:26.643Z
For a 1000km resolution, and visible light at a 500nm wavelength, my result is a diameter of 250km.
Ralf Kleberhoff
CC BY-SA 4.0
2023-01-20T09:41:15.653Z
Previously I definitely mixed up km and m at some stage. But I get 25km (see edit above)
planetmaker
CC BY-SA 4.0
2023-01-19T15:08:26.777Z
2023-01-20T09:40:30.243Z
2023-01-20T09:40:30.243Z
planetmaker
<p>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?</p>
<p>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?</p>

CC BY-SA 4.0
2023-01-19T13:35:22.047Z
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.
planetmaker
CC BY-SA 4.0
2023-01-19T13:38:11.307Z
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. https://en.wikipedia.org/wiki/Angular_resolution
planetmaker
CC BY-SA 4.0
2023-01-19T13:38:51.740Z
well , i know but i just need the mathematic relation i will play with it
xone-a
<galaxy><telescope><light><mathematics><space-probe>
What is the diameter of a telescope lens that will capture the surface of Proxima Centauri b?
xone-a
# What is the diameter of a telescope lens that will capture the surface of Proxima Centauri b?

(Asked by: username_0 on 2023-01-19)

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?

**Tags:** galaxy, telescope, light, mathematics, space-probe
> **Comments:**
> **username_1**: 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.  
> **username_1**: 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. https://en.wikipedia.org/wiki/Angular_resolution  
> **username_0**: well , i know but i just need the mathematic relation i will play with it  

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## 1 answers

### Answer by username_1 on 2023-01-19. Score: 1

The relation for [angular resolution](https://en.wikipedia.org/wiki/Optical_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](http://spiff.rit.edu/classes/resceu/lectures/spectra/spectra.html).

> **Comments:**
> **username_0**: i did some things and find out the D diameter of the leng will be 3 meters :) :(  
> **username_0**: in second attempt i have fixed the theta  the equation  gave 'D is 54 km.'  
> **username_0**: am pretty sure thats not true at all haha but anyways i think i will need some help  
> **username_2**: @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.  
> **username_3**: For a 1000km resolution, and visible light at a 500nm wavelength, my result is a diameter of 250km.  
> **username_1**: Previously I definitely mixed up km and m at some stage. But I get 25km (see edit above)  

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