I wanted to calculate the beam spot size for a laser beam from a satelite (100 - 36000 Km radius) orbiting the earth. I want to calculate the results to be as accurate as possible to the real world scenario. For example, given the radius of aperture of the transmitting telescope ($0.1$ m) and the wavelength ($1550$ nm), how do I calculate the diffraction at a distance of $40,000$ km ?

Currently I'm assuming a collimated gaussian beam with initial spot size $\omega_0$ the spot size at a distance $z$ is given by,

$$ \omega\left(z\right)= \omega_0 \sqrt{1+\left(\frac{z}{z_R}\right)^2},$$

where $z_R=\pi\omega_0^2n/\lambda$ with $n$ being the refractive index and $\lambda$ the wavelength.

If the transmitter on the satelute has an aperture of radius, say $0.1$ m, does this mean that $\omega_0=0.1$?

I've also read that in practise laser beams are not perfect gaussian, and its deviation from the ideal case is represented using the M squared or the laser beam quality parameter (wikipedia link). How does the beam spot in this case depend on the M squared parameter?


From this question


Could anyone provide a reference for this?


1 Answer 1


You need to clearly define the height of the satellite above the earth. Sometimes the coordinates are such that the altitude is measured from the Earths center.

You also need to be clear about diameters and radii.

I believe the equation you have is the 1/e^2 radius of the intensity of a Gaussian beam. But you should check this and clarify in your question.

Also, is the aperture of 0.1 m a diameter or radius?

All that being said, you should assume that the you've illuminated that aperture with a Gaussian profile beam with a nominally flat phase front. If you equate 𝜔0 with the aperture radius the truncation of the Gaussian beam will modify propagation relative to pure Gaussian propagation. So you'd have to compute the effects of the truncation.

The light from a single mode fiber gives a nearly perfect Gaussian with M^2 ~1.03

HEre's a good reference: https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/lecture21_2.pdf

  • $\begingroup$ From an article I'm reading - "...but the transmitting antenna truncates the beam sent to the ground, causing significant losses if the beam waist is larger than the telescope radius. For a downlink, it is recommended to set the FWHM beam waist to be half of the transmitting telescope diameter (Stutzman and Thiele, 2012)." .Does this imply that I can just set $\omega_0$ as the radius of my transmitting antenna right? $\endgroup$
    – Dotman
    Sep 19, 2023 at 9:08
  • $\begingroup$ Do you know where I can get the typical values of the M squared parameter? $\endgroup$
    – Dotman
    Sep 19, 2023 at 9:11
  • $\begingroup$ If you “overfill” the aperture with the beam by making the beam width larger than the diameter of the aperture then you should assume you have a uniform plane wave. The beam at the Earth will be a Bessel function and you can use the relation that the aperture x divergence angle is equal to the wavelength $\endgroup$
    – JQK
    Sep 19, 2023 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.