Why doesn't the logic behind destructive points in single slit diffraction apply to double slits?

Question

I was learning about single slit diffraction, and how if you have a slit such that the path difference between the closest and farthest ray is a natural number m * the wavelength, you will have destructive interference because you can pair rays together that interfere destructively (I was watching these two Khan Academy videos: https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/single-slit-interference and https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/more-on-single-slit-interference).

However, this leaves me wondering how you derive the equation for double slits using this logic. Every derivation I have found seems to treat the light coming from each slit as a single ray - and thus, if the path difference between the two rays is an natural number m * lambda, you get constructive interference (even the Khan Academy video does it this way: https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/youngs-double-slit-part-2). But to me, this seems to ignore the destructive interference that occurs from the infinite parallel rays coming from each slit (if you use the same logic that was used for single slits).

Is there some way to use the same logic as in the single slit proof to show that m corresponds to points of constructive interference in the double slit setup, not of destructive interference?

Answer 1

It can be confusing that the formula for single slit dark spots has the same form as double slit bright spots, but the equations come from different places.

For a double slit, the path difference is $dsin\theta$ (where $d$ is the distance between slits). If that path difference is an integer multiple of a wavelength, $dsin\theta = m\lambda$, then constructive interference will occur.

For a single slit, if you have a ray from the top half of the slit and it's mirror image in the bottom half of the slit the path difference is $\frac{a}{2}sin\theta$ (where $a$ is the slit width). If this difference is a half integer multiple of the wavelength, $\frac{a}{2}sin\theta = \frac{m}{2}\lambda \implies asin\theta = m\lambda$, then destructive interference will occur.