The Unexpected Finger’s laser exhaust was detected at a distance of 3600 au. At that range, how bright could it be?

The exhaust would have been pretty bright, if it were perfectly collimated. (That is, if the laser beams never spread out.) But narrow beams would miss the Earth, and hence not be detected, which would be less fun.

In any case, lasers cannot be perfectly collimated, even by perfect lenses, because of stuff I don't understand. At best, the beam will spread out to a circle called the Airy disk.

The smallest possible beam spread (I.e., Airy disk radius / distance) is given by r/d  = 1.22 λ/w, where r/d is beam radius/distance, λ is the laser wavelength, and w is the width of the laser emitter.

So beam spread depends on wavelength. Earlier we showed the wavelength of the Finger's lasers was λ =10^‑15m. (And I bet you wondered why I bothered.)

Beam spread also depends on w, the width of the laser emitter. We will assume the effective width of each laser equalled the diameter of its fuel rods, 6 in  = 0.15m.

Oh wait, I don't have to assume. I am the author.

I DECREE THE EFFECTIVE DIAMETER OF THE FINGER'S LASERS WAS w  = 0.15m.

We can now compute the Finger lasers' best possible collimation (I.e., beam spread divided by distance) which was r/d  = 1.22 λ/w = 1.22×10^‑15m / 0.15m = 10^‑14.

Holy crap, one part in 10¹⁴? The Finger was only 5×10¹⁴m away when she fired. So she could, in theory, have hit the Earth with beams only ten meters wide.

That would be ideal, but just too wacky. I will wimp out, and assume the Finger's laser beams dispersed by 0.06 percent.

I DECREE THAT the Finger'S LASERS DISPERSED 0.06 PERCENT.

So at 3600 au they spread to a diameter of 0.0006 × 3600 au  = 2 au. By lucky coincidence, this happens to be the diameter of Earth’s orbit.

In other words, the Finger's exhaust energy was spread over a circle the size of Earth's orbit.

To simplify, you will assume the energy was spread evenly, even though it was not.

Dividing total energy by total area gives average flux, the energy per square meter.

The area of a circle is πr², and the Earth's orbital radius is r  = 1au = 1.5×10¹¹m. So the area of the Earth's orbit is π(1.5×10¹¹m)² = 7×10²²m².

Click to Embiggen

Earlier we showed the Finger's apparent exhaust energy was 5.6×10¹⁵W.

So exhaust flux, measured at Earth, was energy / area = (5.6×10¹⁵W) / (7×10²²m²) = 8×10^‑8W/m².

In other words, at a distance of 3600 au, the flux from the Finger’s exhaust was 80 nanowatts per square meter.

This is not much, but to a telescope it would be pretty bright. For comparison, the X-ray flux (at Earth) from the Crab Nebula pulsar is 2.4×10^‑14W/m².

So the Finger’s exhaust flux was (8×10^‑8W/m²) / (2.4×10^‑14W/m²) = 3.3 million Crabs.

At this point an astronomer will object that I have not matched wavelengths, so I am comparing gamma-ray oranges to X-ray crabapples. But don't you think it was worth it to say, X-ray crabapples?