# Double slit diffraction integral?

In our physics class, the professor went over a so-called "diffraction integral" for single slit diffraction:

$$E = E_0\cos(kx - kct)\qquad\text{then we ignore time dependence}:$$ $$E = E_0\cos(kx)$$ $$\Delta E = E_0 \cos(k\Delta r)$$ $$\Delta E = E_0 \cos(k\sin(\theta)z)$$ $$E = \frac{1}{D} \int_{-D/2}^{D/2} E_0 \cos(k\sin(\theta)z) dz$$ $$E = E_0 \frac{2\sin(k\sin(\theta)D/2)}{k\sin(\theta)D}$$ $$E = \text{sinc}(k\sin(\theta) \frac{D}{2})$$

Can we set up and evaluate such an integral for double slit diffraction? I tried doing so, but I got an answer that doesn't match the commonly accepted equation for double slit diffraction. That commonly accepted equation is $$E = \text{sinc}(k\sin(\theta) \frac{D}{2}) \cos(k\sin(\theta) \frac{D}{2})$$

Any help would be much appreciated. For reference, I've already seen and read the post: Double slit diffraction derivation, and did not understand very much because it uses physics that we haven't been taught yet, like Fraunhofer diffraction.

• I have done this in this post of mine. It may be helpful. Commented Dec 13, 2023 at 9:48
• @MaximalIdeal exactly what I was looking for. Thank you!! Commented Dec 13, 2023 at 10:24