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?

  • $\begingroup$ 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

1 Answer 1


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 $x=0$).


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.