Great questions! Answering them actually leads to some pretty profound insights into how quantum physics works.
To answer these questions, let's generalize the double slit experiment in a couple of ways. First, let's imagine a triple slit experiment. Just like before, the photon will go through all of the slits, and each of them will interfere with each other to form the final pattern on the screen, except now the pattern will come from the superposition of three different point sources instead of two. This will work the same for a quadruple slit experiment, quintuple slit experiment, etc.
Now take this to the extreme--imagine we make a slit through every point along our barrier. We now have an infinite number of slits, and thus our barrier has completely disappeared. So you can actually think of empty space as being a barrier with an infinite number of slits in it--the photon will travel through every point along our imaginary barrier, and these infinite numbers of photon paths will all interfere with each other to form the final image you see on the screen.
Now let's go back to the double slit, and imagine another generalization--adding another layer of slits behind the first one. I'll call them 1A, 2A, 1B, and 2B, where A is the first layer of slits, and B is the second. Now there are four possible paths the photon can go through: (1A,1B), (1A,2B), (2A,1B), and (2A,2B). These four paths will interfere with each other just like the original two paths did.
Now let's start adding detectors. Imagine adding a detector at all four slits. When the photon goes through the apparatus, it will trip one of the detectors at layer A, continue on to trip one at layer B, and then continue on to the screen. So there are four possible outcomes in the experiment: the detector at 1A and the detector at 1B can go off, 1A and 2B, 2A and 1B, or 2A and 2B--each corresponding to one of the possible paths listed above. Because we've made a measurement of which path the photon took, it doesn't interfere with itself anymore, so the wavefunction collapses and we just see a distribution corresponding to the two point sources of the second layer of slits.
Next, let's get a little more complicated. Imagine putting a detector at 1B and 2B only. Now there are two possible outcomes from the experiment: either 1B fires, or 2B does. However, there are still four paths through the apparatus. This means that if the detector at 1B fires, the photon could have gone through 1A and then 1B, or it could have gone through 2A and then 1B--we can't tell which. Because of this, these two paths will interfere with each other just like before. Similarly, the two paths that go through 2B will interfere with each other.
The end result of this is that by the time the photon gets to the screen, we've collapsed the wavefunction again (because the second round of slits had detectors), so we will see two noninterfering point sources on the plate, but the intensity of each of those point sources will be determined by the interference pattern that the first layer of slits produced. You could imagine putting the plate at layer B, getting the interference pattern just like we always do, and then turning around and shining light of that intensity onto the real plate behind layer B.
Finally, let's combine the two ideas I've presented. Imagine layer A has two slits just like before, and layer B has an infinite number of slits (i.e. it is just empty space), but one of those slits has a detector at it. Now there are two possibilities: either the detector fires, or it doesn't. If it does, we know there are two things that could have happened--either it went through slit 1A or 2A. So we will see a point source on the screen whose intensity is determined by the interference of the two slits at the point of the detector.
On the other hand, if the detector didn't fire, then we know it did not go through the detector slit, but we don't know which of the other (infinite number of) slits it did go through. Thus, we will have the interference of all possible paths that go through 1A and then some point on the second layer, interfering with all possible paths that go through 2A and then some point on the second layer. This will basically add up to the original interference pattern created by just the two slits. So what you will ultimately see on the screen in the scenario you have proposed is the original interference pattern plus a point source from the detector's position, whose intensity is equal to what you would have gotten at the point of the detector if you had put the screen in its plane.
This answer got really long, so I hope I got the point across ok. Let me know if you'd like clarification on any part of it, and I'd be happy to go into a little more detail. I'm still learning some of this too, so I'm not absolutely positive about it, but I'm 99% sure that what I've just described is accurate. If it's not, hopefully somebody else can correct me.
