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We have a telescope that can see stars with about +14 magnitude. how to find its lens (or mirror) diameter? (I mean the $D$ in formula $\theta = 1.22 \lambda / D $)

I don't know whether I can assume that the temperature of star is equal to temperature of the sun or not. but if we think that they are equal we get this:

$m_{\mathrm{sun}} - m_{\mathrm{star}} = -2.5 \log_{10} (\frac{b_{\mathrm{sun}}}{b_{\mathrm{star}}}) = -2.5 \log (\frac{\sigma t^4 \times 4 \pi R_{\mathrm{sun}}^2 /(4 \pi r^2) }{\sigma t^4 \times 4\pi R_{\mathrm{star}}^2 / (4\pi r_{\mathrm{star}}^2)})$ if we solve this with what we know about magnitude of sun (-26.83 or something close to it between -25 to -27) and another things like radii we get:

($R/r = 3.16605 \times 10^{-11} \mathrm{rad} = \tan \theta/2 = \theta/2$ and from that and formula $\theta = 1.22 \lambda / D $ and $\lambda = 550 nm$ we get something like D = $10596.8 \mathrm{m}$. which is very big for a telescope. what's the problem? how to solve this? Is this answer correct?

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    $\begingroup$ Something looks very wrong here.... It looks like you have found the size of a mirror needed to resolve a 14 magnitude, sun-like star as a disc. You don't need to resolve a star as a disc to "see" it. $\endgroup$
    – James K
    Commented Apr 9, 2017 at 20:22
  • $\begingroup$ @JamesK I think it's OK - the $\sigma T^4$ times $4 \pi R^2$ is the amount of light, and it's divided by $4 \pi r^2$ the total solid angle, to get the ratio of light received. Then it's... whoa - oh, I see what you mean. Ya the derivation takes a left turn and diverges into an Airy disk. $\endgroup$
    – uhoh
    Commented Apr 10, 2017 at 5:56

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The unaided eye can typically see mag 6 objects. With your telescope, you can see an additional 8 magnitudes. This requires a factor of $100^{8/5}$ of additional light gathering power (since 5 magnitudes equals 100 times the brightness). I.e. about 1580 times larger aperture than the pupil of the eye. Assuming the pupil has a diameter of 7mm, then the diameter of your telescope would have to be $7\times 1580^{1/2}$, which is about 280 mm.

This is fairly close I think - perhaps a bit on the high side but not much.

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    $\begingroup$ Good clear argument. Look at cruxis.com/scope/limitingmagnitude.htm which calculates limiting magnitudes for various telescopes. It agrees with this analysis. What you are missing is the effect of the eyepiece, though the magnification is not as significant for stars as for diffuse objects. $\endgroup$
    – James K
    Commented Apr 9, 2017 at 20:32
  • $\begingroup$ Good start, but since the dark-adapted eye is more sensitive than the non-adapted eye by quite a lot, you should edit to indicate whether you're quoting adapted sensitivities. $\endgroup$
    – Carl Witthoft
    Commented Apr 10, 2017 at 16:30

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