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If we have a light source, say a laser or LED, diffracting through a circular aperture of radius $r$, how would we find the radius of the beam at a position $x$ meters from the optical axis? I understand that we will see a diffraction pattern, so we will witness fringes - what I am looking for is the radius of the zero-order / central "fringe".

Would the correct approach be to model the Airy disk function as described here? https://en.wikipedia.org/wiki/Airy_disk

I have written the code below to model the Airy disk, but am having trouble interpreting how $x = ka \sin\theta$ relates to the radius of the beam.

% System Inputs
ApertureRadius = 1e-3;      % Units [m]
Wavelength = 532e-9;        % Units [m]
Theta = 0 : pi/1024 : 2*pi;

% Simulation of Airy Disk
I0 = 1;
K = 2 * pi / Wavelength;
X = K * ApertureRadius * sin(Theta);

I = I0 * (2 * besselj(1, X) ./ X).^2;
plot(X, I);
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  • $\begingroup$ The airy disc uses gives the intensity pattern in terms of bessels functions as written in your program. The first minima occurs at the first zero of the bessel function. That is called the radius of the central disc and is given by the expression for $sin\theta$ as given in one answer below. $\endgroup$
    – Bruce Lee
    Commented Jan 13, 2016 at 6:05
  • $\begingroup$ @BruceLee Okay this makes sense, and the plots using the above program match the answer described below. However, for a larger aperture, I noticed that the first zero of the Bessel function decreases. Wouldn't a larger aperture let in more light, and shouldn't the resultant beam be larger? Perhaps the first minimum is not the correct metric to use? $\endgroup$
    – ArKi
    Commented Jan 13, 2016 at 6:41
  • $\begingroup$ There is one approximation used in the derivation and that is the slit aperture is very small compared to the distance between the slits and screen. So the angle decreasing as slit width increases and vice versa makes perfect sense. $\endgroup$
    – Bruce Lee
    Commented Jan 13, 2016 at 6:48

1 Answer 1

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From the link you shared Far away from the aperture, the angle at which the first minimum occurs, measured from the direction of incoming light, is given by the approximate formula $$\sin\theta\approx 1.22\frac{\lambda}{d}$$ for small $\theta$ $$\theta\approx 1.22\frac{\lambda}{d} $$ where θ is in radians, λ is the wavelength of the light and d is the diameter of the aperture. Now Lets say the radius of the central fringe is $r$ and the distance at which you are measuring it is $x$ then We have $$\tan\theta=\frac{r}{x} $$that is for small $\theta$ $$r\approx x \theta$$

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  • $\begingroup$ Thanks for the explanation! Isn't tan(θ) = r/x? $\endgroup$
    – ArKi
    Commented Jan 13, 2016 at 6:03
  • $\begingroup$ Sorry!!I have corrected it. $\endgroup$ Commented Jan 13, 2016 at 7:06

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