# What is the formula for the power of a laser at distance $x$ (taking attenutation into account)?

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:

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?

• 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. Commented Apr 11, 2023 at 18:56

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 $$x=0$$).