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 Finger's lasers was
λ =10^{‑15}*m*. (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* = 0.15*m*.

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

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

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

Holy crap, one part in 10^{14}?
Finger was only 5×10^{14}*m* 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 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, 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 π*r*^{2}, and the Earth's orbital radius is
*r* = 1*au* =1.5×10^{11}*m*. So the area of the Earth's orbit is
π(1.5×10^{11}*m*)^{2}
= 7×10^{22}*m*^{2}.

Earlier we showed Finger's apparent exhaust energy was
5.6×10^{15}*W*.

So exhaust flux, measured at Earth, was *energy* / *area*
= (5.6×10^{15}*W*) / (7×10^{22}*m*^{2})
= 8×10^{‑8}*W*/*m*^{2}.

In other words, at a distance of 3600 *au*, the flux from 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^{‑14}*W/m*^{2}.

So Finger’s exhaust flux was
(8×10^{‑8}*W*/*m*^{2}) / (2.4×10^{‑14}*W/m*^{2})
= 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, just to say *X-ray crabapples*?