# What is the Fraunhofer diffraction pattern of a circular aperature with a rectangular obstruction?

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.

• 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 $a$...
– JEB
Commented Mar 30, 2020 at 2:40

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.