Wow, that video is spectacular too!
For what I did no calculation was needed; the effect is fairly independent of the size of the object casting the shadow, or the distance behind it that you project it to. I passed the laser through a magnifying glass so that the beam would be expanded (past the focal point) to a few times the size of the coin after a few meters, and then placed the coin in the beam and projected the shadow. The spot begins to become visible in the middle of the shadow not too far beyond the coin -- a few meters. You'd be surprised how quickly the spot can form afterward, since the light can actually spread perpendicularly to its direction of travel!
Every point on the wavefront essentially acts as a source of new spherical wave fronts. It is important that the surface of the object casting the shadow be very smooth. A coin with ridges, like the US quarter or dime, does not work. A spherical object like a marble will work (it's the profile seen by the incoming wave fronts that matters). I thought as well about whether the effect could occur in eclipses. For the Earth and Moon, the biggest problem is that the lunar surface is too rough. We can use the equations from the wikipedia page to check.Assuming parallel rays of light (which is fairly true of the Sunlight), Arago's spot will form if:
1)
[math]L = \frac{d^2}{\lambda l} \geq 12)
[math]\Delta r < \sqrt{r^2 + \lambda l}For Moon casting the shadow, eq 1) imposes the condition that the observer must be less than a few thousand light years away. No problem.
Eq 2) imposes the condition that, for Earth being about 360,000km away during total eclipse, then to form a spot on it the Moon's roughness must be less than about 50 micrometers. Absurdly smooth... this is the real reason why we don't see the effect in solar eclipses, or anywhere else in the solar system. (And a good thing, too, or else it would greatly interfere with viewing the Sun's corona).
For comparison with the coin (diameter ~ 1cm, spot projection distance ~4m), the maximum size for roughness is about 0.1mm. Very reasonable for a coin.
What would it take to view a natural Arago/Fresnel/Poisson spot from a celestial body? Perhaps it happens in wide binary systems containing a cool neutron star (although the cooling time of a neutron star might be an issue). Neutron stars have a radius of about 10km and, using the
maximum terrain height formula, can probably have surface features of up to about 1mm high. Assuming the partner star is sufficiently far away to act as a point source, how far must the viewer be from the neutron star to see the spot in the shadow? Solve eq 2) for
[math]l and plug and chug, and it's 40,000km. Again reasonable, and could work with a sun-like star ~100AU away.
One last experiment I tried was to use a non-circular coin. I tried the Canadian 1-dollar
Loonie. It does not produce Arago's spot. However, because edge is still smooth and regular (11-sided), it does produce a really neat interference pattern: an 11-sided star, although only 9 points are visible here due to obscuration.