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While this is inspired by science fiction, if I have a hypothetical laser weapon that needs to meet a very high energy requirement of energies of the order $ 1 \times 10^{22} $ Joules, what formula would describe the power of the laser as a function of distance with attenutation and beam divergence taken into account? This is so I can determine the extra power that the laser may require on top of $ 1 \times 10^{22} $ Joules to account for any loss in power so it can still be used as a hypothetical defensive laser weapon for distances limited to our Solar System. As Power = Energy/Time, if the hypothetical laser is active for 1 second, it requires a power of $ 1 \times 10^{22} $ Watts but I want to know how this is attenuated over large distances.

What I have done is used Beer-Lambert's attenutaion law:

$I = I_0 e^{-ax} $

where $a$ is the absorption coefficient (which I believe is high for the atmosphere but low for space) and $x$ is the distance travelled. Then using the following

$ I = P/A$

where $A$ is the area of the beam, I got

$P(x) = I_0 e^{-ax} A(x) = I_0 e^{-ax} 4 \pi x^{2}$

where $P(x)$ and $A(x)$ are the power and area of the laser at distance travelled $x$. However, I'm not sure this is the correct approach. If I ignore the constants $I_0$ and $a$ and plot $e^{-x} 4 \pi x^{2}$, I get the following plot:

enter image description here

Focusing on the positive $x$ axis, where $x$ is distance, this does show an attenuation like expected (where power decreases as distance increases) apart from the region where the $x^2$ dominates. But I'm still not sure if this is a good description of the power of a laser being fired into space?

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  • $\begingroup$ At that power, air isn't a passive medium. I suspect that the beam will convert the air into a super-hot opaque plasma, and push most of it out of the beam. But I don't know how to do a detailed calculation for that. Maybe put your lasers in orbit, or on the Moon. $\endgroup$
    – PM 2Ring
    Commented Apr 11, 2023 at 18:56

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The Beer-Lambert law, in the form you stated it, applies either to confined beams (that aren't diverging due to, for example, being confined in a waveguide), or to plane waves, which aren't diverging because they already extend evenly through all space.

If you want to apply this law to a diverging beam, you should express it in terms of the total beam power rather than in terms of intensity:

$$P(x)=P_0e^{-\alpha x}$$

For a diverging beam in the far field (beyond any point of convergence due to an output optics of the source, etc), the intensity will fall off as $1/x^2$ due to the divergence.

Combining the effects of absorption and divergence, you'll have a peak intensity (typically at the beam center, but the source optics could modify this) of

$$I(x) = \frac{I_0 e^{-\alpha x}}{x^2}$$

Where $I_0$ is a proportionality factor you could determine from a measurement somewhere convenient in the far field (not at $x=0$).

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