# How to calculate losses due to diffraction?

Assume a transmitter with a laser source generating gaussian beams and is transmitted using a telescope with radius of aperture $$a_T$$. The light is collected at the reciever end using a telescope of aperture $$a_R$$. They are seperated by a distance $$z$$.

The power transmitted is equal to $$$$\frac{2}{\pi w_0^2} \int_0^{a_T} 2 \pi r e^{-2 r^2 / w_0^2} d r=1-e^{-2 a_T^2 / w_0^2},$$$$ where $$w_0$$ is the initial beam waist.

Assuming a collimated beam, free space propagation along $$z$$ broadens the beam as,

$$$$\label{diff} w_d(z)=w_0 \sqrt{1+\left(\frac{z}{z_{\mathrm{R}}}\right)^2},$$$$

where $$$$z_{\mathrm{R}}=\frac{\pi w_0^2 n}{\lambda},$$$$ with $$n$$ being the refractive index of the medium and $$\lambda$$ the wavelength of the beam.

The power recieved is equal to $$$$1-e^{-2 a_R^2 / w_d^2},$$$$ The diffraction induced transmittivity is given by the fraction of the output and input powers, $$$$\eta_d(z)=\frac{1-e^{-2 a_R^2 / w_d^2}}{1-e^{-2 a_T^2 / w_0^2}} .$$$$

1. Although simplified, is this the right approach to calculate the losses due to diffraction induced beam broadening?
2. Realistically lasers dont produce perfect gaussian beams. Transmittting telescope could also cause further deviations. How does beam broadening equation vary in such situations? Can such deviations be accounted for using just the $$M^2$$ Would the beam broadening equation be then modified to be, $$w_d(z)=w_0 \sqrt{1+M^2\left(\frac{z}{z_{\mathrm{R}}}\right)^2}?$$ And what would be some realistic values for $$M^2$$ using a transmitting telescope?