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Solution of Spherical Aberration

 
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mawyatt



Joined: 22 Aug 2013
Posts: 2041
Location: Clearwater

PostPosted: Mon Jul 08, 2019 1:05 pm    Post subject: Solution of Spherical Aberration Reply with quote

A former colleague noted this!

Great work by these folks!! That equation looks formidable enough to be not obvious to the casual observer!!

https://petapixel.com/2019/07/05/goodbye-aberration-physicist-solves-2000-year-old-optical-problem/

Best,
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rjlittlefield
Site Admin


Joined: 01 Aug 2006
Posts: 19842
Location: Richland, Washington State, USA

PostPosted: Mon Jul 08, 2019 3:22 pm    Post subject: Reply with quote

I saw this one yesterday. Google News offered it to me, I suppose because it fits my search and click profile.

The new formula is nice work indeed, but at this point I am not getting excited about the practical implications.

As I read the article, the new contribution is that the solution is analytical, which means here that it can be calculated as a finite sequence of arithmetic and taking of roots.

In contrast, the previous method required numeric solution of a pair of first-order differential equations.

While it's true that numeric solution is always an approximation, the result can be made arbitrarily accurate at the cost of crunching more numbers. In particular, the numeric solution can be made more accurate than any fabrication process.

The computational cost for getting that accuracy through numeric approximation is probably higher, perhaps much higher, than doing a direct calculation using the new formula. But for well behaved differential equations like I'd expect to find in this problem, I seriously doubt that the cost of getting enough accuracy has been much of an impediment to the lens designer.

Hence my lack of excitement. The new formula is very nice work, but at this point I don't see it as having much impact on lens quality.

I could always be wrong of course, and that would be a good thing. Maybe the added speed will let designers explore more possibilities than they could do previously.

--Rik
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