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Suppose we have a circular aperture of radius 3$\lambda_0$ and we place a vertical rectangle of width $\lambda$ over the center of the aperture (as shown in the picture). What will the Fraunhofer diffraction pattern be in this case?

I understand that to solve this problem, one will have to take the convolution of a circular aperture's diffraction with the inverse of a single slit's diffraction, but I'm having some difficulty getting through the calculation as I'm not entirely confident. Here's what I have so far:

The transmission function for a circular aperture is the step function and the for a slit, it is the rectangular function. So, the total transmission function should be: $\tau = 1+\Pi(3 \lambda_0)-rect(x/(2*3 \lambda_0))$. Then, we'll want to take the Fourier transform of this, which should yield: T = $\delta + \frac{\lambda_0 J_1(\frac{2*3\lambda_0 \pi sin(\theta)}{\lambda_0})}{sin(\theta)} - sinc(\frac{\pi 2*3 \lambda_0 sin(\theta)}{\lambda_0})$ (where I've converted the x in the rect function to polar coordinates). The issue here, is that I'm certain there should be some radial dependence (given that a circular aperture's diffraction pattern has a radial dependence.

If I were to proceed, I would take $T$ and somehow use it to find the diffraction pattern (i.e., the intensity), though I'm also unclear of the details of that.enter image description here

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  • $\begingroup$ Why a convolution? Is 2 slits of width $a$ separated by $2a$ the convolution of a slit of width $3a$ and a a (negative) slit of width $a$... $\endgroup$
    – JEB
    Commented Mar 30, 2020 at 2:40

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The aperture is a circular aperture multiplied by a function which is 1 everywhere, except at $-\lambda_0/2< x < \lambda_0/2$, where it has a value of zero. No addition is involved.

The Fourier transform is then the convolution of the 2D Fourier transforms of these two functions.

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