I realize that the people here must be sick of answering stupid questions about the double slit experiment, and different ways of setting it up, but if I may, I have a question about what you would expect to see if the double slit experiment were set up in a slightly different manner, and why. First, I am assuming that in the classic double slit experiment we need only put a detector at one slit in order to destroy the interference pattern. But this leaves me with a question. Is it the fact that we then know which slit the particle went through that destroys the interference pattern, or is it the fact that we interacted with the probability wave that destroyed the interference pattern? In other words, could we destroy the probability wave without knowing which slit the particle went through. So my question is, what if, instead of setting up the double slit experiment in the way that I usually see it portrayed, with something that looks like a laser pointed at a wall with two slits in it, but instead we set it up with a light positioned in the center of a box with a pair of slits on each of the six opposing walls, and a screen positioned behind each pair of slits. I assume that absent any detectors we would expect to see an interference pattern on all six of the screens. But what if we then put a detector on just one of the twelve slits? Would that one detector be sufficient to destroy the interference pattern on all six of the screens? Or would it only destroy the interference pattern on the screen positioned behind the wall with the detector? What would you expect to see with such a setup, and why? Thanks to everyone for considering my question.