The discussion revolves around the concept of diffraction-limited spot size in optics, particularly in relation to telescope mirrors and the Airy Disk. Participants explore theoretical aspects, practical implications, and mathematical formulations associated with diffraction in optical systems.
Participants express differing views on the interpretation of mathematical formulations and the implications of energy conservation in diffraction theory. There is no consensus on the correctness of specific claims or the appropriate framing of aperture shapes.
Some discussions involve unresolved mathematical steps and assumptions regarding aperture shapes, which may affect the interpretations of energy conservation and diffraction patterns.
@Mech_Engineer I'm glad you liked the article. The Airy disc I believe is what you get for the shape and intensity of the focused diffraction spot if you assume a circular aperture. The result that I have in the article above results from assuming a square aperture. It is a somewhat specialized topic so that it hasn't received a tremendous number of views, but I'm so glad you found it interesting. :) :)Mech_Engineer said:
My argument is meant to show the conservation of energy in a qualitative sense and also to show that the peak intensity is proportional to the square of the area of the aperture. Very precise energy conservation could be shown in both cylindrical and x-y geometry. The calculations I did, (and it was quite a number of years ago=I simply summarized the result here), shows an energy conservation in x-y geometry where you have a product rule ## I(x,y)=I_o i(x) i(y) ## with a rectangular or square aperture.fizzy said:
Quantitatively, it worked out correctly in x-y coordinates. If you want to compute the diffraction integrals in cylindrical coordinates, I'm sure you would get complete energy conservation, and you would find my on-axis intensity value to be correct. Feel free to post the calculation in cylindrical coordinates, but I did it using x and y independently and I was satisfied with the result.fizzy said:
When I get a chance I will demonstrate the precision of energy conservation of the diffraction pattern in one dimension using the sinc^2 function.(I'm busy at the moment.) (It integrates to be precisely the result that I used in the article. ) Meanwhile, I do have considerable experience working problems in diffraction theory, so that I'm not pretending to be someone that I am not. ## \\ ## Editing... ## \\ ## For a slit of width ## b ## , ## I(\theta)=I_o \frac{sin^2(\frac{\pi \, b \, sin(\theta)}{\lambda})}{(\frac{\pi \, b \, sin(\theta)}{\lambda})^2} ##. It can readily be shown that with ## sin(\theta)=\theta=\frac{x}{f} ##, (small angle approximation with a focused image), and with a narrow pattern (wide slit) that ## \int I(\theta) \, d \theta =I_o (\frac{\lambda}{b}) ##. (The calculation uses the substitution ## u=(\frac{\pi b}{\lambda} )x ## and uses the result that ## \int\limits_{-\infty}^{+\infty} (\frac{sin^2 u}{u^2}) \, du=\pi ##. The limits on the integral are justified (instead of stopping at ## \theta=\pi/2 ## , etc.) because the beam pattern is narrow.) With this kind of precision in a one dimensional calculation, it would seem unnecessary to try to solve a complex diffraction integral in cylindrical coordinates. Notice that this makes the effective width of the pattern ## \Delta \theta_{eff}=\frac{\lambda}{b} ## The result is exactly what is needed. It can be readily extended to two dimensions. ## \\ ## I think it is justifiable to call ## \Delta x=f (\Delta \theta_{eff}) ## the focused spot size diameter for the purposes of the article. Otherwise I could call it the "effective width of the focused spot".fizzy said:
The assumption is made that the aperture of the incident beam onto the paraboidal focusing mirror is square. It will produce a focused spot that has effective area ## A_{eff}=\frac{ \lambda^2 f^2}{b^2} ##. Mathematically it works out just as it should. (In the article, for a beam diameter "b", I simply said the spot size was "approximately" ## \Delta x=\frac{\lambda f}{b} ##.) The computation can be done quite precisely, as I have done in posts #10 and #11. ## \\ ## Additional comment: I believe doing this same calculation using a cylindrical geometry would be quite difficult, but the calculation with the rectangular geometry is rather straightforward. ## \\ ## Edit: 12-31-18 Here is a homework thread that covers the problem of the Airy disc created by a circular aperture: https://www.physicsforums.com/threa...n-using-bessel-functions.916241/#post-5775639fizzy said: