Finding the lens diameter of a telescope by magnitude 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?
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.