Re: Question for you optical gurus...

acl wrote:
On May 8, 12:13 am, "Nicholas O. Lindan" <s...@xxxxxxx> wrote:
<achilleaslazari...@xxxxxxxxxxx> wrote

No, diffraction will not produce circles around the image of the lamp.
Like this, maybe?

http://www.matter.org.uk/tem/diffraction_at_aperture.htm

Won't look near as pretty: the lamp not being a
point source, and having a pretty messy spectrum,
the thing will be a fuzzy dot. It won't be a set of
rays.

The rest of this isn't worth responding to...

It's not worth responding to, eh? Did I somehow insult you, are you
always like this, or did you just decide that I just don't know what I
am talking about?

Anyway, the link you posted to shows diffraction through a circular
aperture. The diffraction pattern is related to the fourier transform
of the aperture shape (rather, of the transfer function, but we take
that to be 1 inside the aperture and 0 outside). The fourier transform
of eg a square is going to have four "rays" coming out (of course
they're not rays, they're modulated with a period related to the size
of the aperture). It's trivial to check that the fourier transform of
a shape with n-fold rotational symmetry will also be n-fold
rotationally symmetric, but working out more details needs explicit
calculation.

Since I suppose you'll dismiss what I wrote again, look, for example,
here
http://www.kw.igs.net/~jackord/df/d1.html
you can see some java applets to calculate the diffraction pattern
through circular, rectangular and triangular apertures (go to where it
says "rectangular and triangular apertures"). But then, if you don't
believe what I say without bothering to investigate, why believe
anything you find on the internet?

Anyway. I can go into as much detail about this as you want, in fact.
Somehow, however, I doubt you'll now google more carefully and say
"oops you're right, sorry for the tone of the response" :)

ACL has the correct explanation in this thread.

Roger
.

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