Beam spot size for a laser beam from space

Question

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?

EDIT

From this question

$w_z=w_0\sqrt{1+M^2(\frac{z-z_0}{z_R})^2}$

Could anyone provide a reference for this?

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 0 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

Answer 2

If the transmitter on the satellite has an aperture of radius, say 0.1 m, does this mean that $ω_0=0.1$?

No. Exit aperture size does not mean that laser initial waist is such,- it can be smaller as well. As bigger or equal to the aperture beam waists can include diffraction effects from the aperture edges and worsen beam quality.

Exact waist radius at the exit must be given in laser system manufacturer specification or can be measured experimentally by some method as Knife-edge method , etc.

How does the beam spot in this case depend on the M squared parameter?

Since Beam parameter product is $\mathrm {BPP} =M^{2}\cdot {{\lambda }/{\pi }}$, you need to scale Rayleigh length into new value :

$$ \tag 1 z_R \to \frac {z_R}{M^{2}} $$

as Gaussian ideal beam has $M^2 = 1$, all other beams has $M^2 \gt 1$ and so for them laser beam divergence will be higher at the target spot.

BTW, your given reference of non-$TEM_{00}$ modes beam width $w_z$ formula is wrong, since there should be dependence on $M^4$ (not $M^2$ as given in the formula) under square root. To verify this, substitute (1) mapping into standard Gausian $w(z)$ case and see result.