"Keith F. Lynch" <kfl@KeithLynch.net> wrote:

If your first thought is that is that the moon would be a burning glass, leaving a trail of destruction during every solar eclipse, think again. In a week I will explain why that wouldn't happen unless someone beats me to it, which I hope they will.

Nobody did, so here goes. In any purely optical system, "etendue" (look it up) can never decrease, due to the second law of thermodynamics. That means any refraction, reflection, diffraction, etc., of all or part of the sun can never be any brighter per unit angular area than the sun itself. As every child chasing ants with a burning glass knows (it's what my generation had instead of video games), to roast an ant you need to hit it with light from multiple directions at once. The lens has to look much larger than the sun from the ant's perspective. This lesson was reinforced a couple decades later when I worked with what was then the world's largest solar furnace, at Kirtland Air Force Base. A medium-sized desert was lined with large mirrors, and anyone at the focal point when it was sunny would have a very bad day. At least one person suffered from moonburn (!) while aligning it all night using the full moon. He should have remembered to wear moonscreen. As seen from Earth's surface, the sun and the moon are almost exactly the same angular size. So if the moon is replaced with a lens that focuses the sun, the result will look exactly the same as the sun, no brighter or dimmer. You'll actually be viewing just one point on the sun, but that won't make a visible difference unless you're using a telescope. That's if the sun is centered exactly behind the moon, as with totality during a total solar eclipse. If the sun is not at all behind the moon, then the moon will be dark (or, with a very low probability, will show some planet for a few seconds). The brightest the sky can get in such a setup is if the edge of the sun is just barely behind the center of the moon. In that case you get two overlapping bright circles. That should be 1.609 times brighter than the sun alone. Not enough to set anything on fire.

How would the refractive index have to vary with depth for the moon to have a focal length equal to the distance to Earth's surface?

Nobody has answered this yet. A closed-form solution is beyond me. I guess I'll write a simulation. I can drop a dimension and use a circle instead of a sphere. And I can drop half the circle, due to symmetry. I'll approximate the smoothly varying refractive index with perhaps a hundred concentric semi-circles, each with a different fixed refractive index, and do ray tracing. That will give me a hundred different interacting knobs to twiddle. Perhaps a hill-climbing algorithm will get it to converge on the optimal solution. First I'll try focusing on a point on the moon's surface rather than Earth's, and see if my program rediscovers the Luneburg lens. Any thoughts on such a program? Thanks. It's hard to believe nobody has solved this already.