Deriving circular aperture diffraction

[timetraveller123]

Homework Statement

i don't understand where the 1.22 number comes from like in
$sin \theta = 1.22 \lambda /D$
so i tried to derive but to not much help

Homework Equations

electric field at distance R due to slit of length D is
$E = \frac{\epsilon_l D}{R} sinc(\frac{kD}{2}sin \theta)sin(wt - kR)$
so i thought of breaking the circle into series of strips with theta being zero so sinc being 1

The Attempt at a Solution

$\epsilon_ l = \epsilon_a dx\\ D= \sqrt{R^2 - x^2}\\ R = R+x sin\theta\\ dE = \frac{\epsilon_A dx 2 \sqrt{R^2 - x^2}}{R}sin(wt - kR + kxsin\theta)\\ E= \frac{2 \epsilon_a}{R} \int _{-R}^{R} \sqrt {R^2 - x^2}sin(wt -kR + kxsin\theta)dx\\ E = \frac{2 \epsilon_a}{R} sin(wt - kR) \int_{-R}^{R} \sqrt {R^2 - x^2} cos(kx sin \theta)dx\\$
the other term drops out as it is odd
i tried to put this into integral calculator but it didnt give an answer so help where does the 1.22 come from[/B]

 

[Charles Link]

The Airy disc mathematics gets rather advanced. Here is a previous homework post that you might find useful: https://www.physicsforums.com/threa...n-using-bessel-functions.916241/#post-5775639

 

[timetraveller123]

oh wow so is the solution to this integral also bessel function?

if so would you be kind enough to provide the solution (as in the thread you mentioned the op seems to be doing in polar coordinates so the integrals are slightly different) then i might think about how the 1.22 comes about(i don't know bessel functions)

 

[Charles Link]

On this one I think you can take $\frac{\pi r d }{\lambda z} =3.8317$ (3.8317 is the first zero of $J_1(x)$). that means $\frac{r}{z} \approx \sin(\theta)=(\frac{3.8317}{\pi}) \frac{\lambda}{d}=1.22 \frac{\lambda}{d}$. $\\$ Note: I googled the $J_1$ Bessel function. I'm not real familiar with Bessel functions either.

 

[timetraveller123]

oh i see thanks for the help i will try to learn a bit about bessel functions