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In the reseach paper " A laser link from lunar surface..." , chapter 2.1.3.:

When employing, for example, 20 cm diameter apertures at the transmitter telescopes and considering a carrier wavelength of λ = 1,064 nm, a full width at halfmaximum (FWHM)-divergence angle of approximately θdiv = 3.25 µrad is achieved. This divergence of θdiv = 3.25 µrad will produce spot sizes of about 1.2 km diameter on Earth-ground.

I am trying to follow the calculation and can't get the result.

With the given wavelength and diameter I can compute : $$ \frac{\lambda}{d} = \frac{1064\,\text{nm}}{ 20\,\text{cm}} = 53.2\cdot 10^{-7} $$ The distance I can compute using sin-law: $$ L = \frac{ D}{\sin (\theta)} = \frac{20\,\text{cm}}{\sin (3.25\,\text{mrad})} $$

Footprint =$ \frac{\lambda}{d} L$

At the end, I didnt get the same result.

How was it computed?

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    $\begingroup$ The link distance L does not need to be calculated as it is given in table 2 with 354.000km. The footprint, in general, is λL/d but usually gets multiplied by a factor based on the beam shape. In this case, they would have used a factor of ≈ 0,64. This answer from a similar question might help you how to calculate the factor. $\endgroup$
    – GittingGud
    Commented Mar 21, 2022 at 11:44
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    $\begingroup$ the text you cite says µrad, you use mrad. That's a factor of 1000 off. $\endgroup$ Commented Mar 21, 2022 at 17:44

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