What if you turn the photon detector on after the paired photon has already gone through both slits? Id imagine the answer will be that even if you turn the photon detector on after the paired photon has already hit the screen, the screen will show no interference pattern because the universe sensed the future and adjusted itself to avoid a paradox. I imagine that the photon pair foreshadowed  that a photon detector would be turned on and thus adjusted its its path to only go through one slit. 
Edit:
Arpad thats not what i was asking. Im asking in a situation set up like a quantum eraser. A photon goes through 2 silts then hits a crystal that splits it into 2 identical entangled photons. One photon goes to a screen where its interference pattern (or lack of one) shows up. The second photon travels a very long distance to a photon detector which will determine which slit the #2 photon went through AFTER the #1 photon has already hit the screen. Make sense? My instinct is that the universe as always, will conspire against the experimenter and the interference pattern wont show on photon #1’s screen if the photon detector that detects which way photon #2 went is turned on, EVEN IF that photon detector isnt switched on until AFTER photon #1 has already hit the screen. 
 A: Your title question and your question in the body is different. You are not specifically saying where the detector (and what type of detector it is) is positioned.
Now in the body your the question you are asking what would happen if you would turn on the detector after the photon has already hit the screen and interacted with it (left a dot). In this case, since the detector did not interact with the photon, there will be an interference pattern.
Now in the title you are asking what would happen if you would turn on the detector after the photon has passed through the slits. There are two cases here:


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*the detector is on one of the slits, in this case there will be an interference pattern, since the photon does not interact with the detector, because the photon has already passed the slits by the time the detector is turned on

*the detector is positioned so, that the detector is between the slits and the screen, in this case the interference pattern is destroyed, because the photon interacts with the detector
It is very important that there will be no interference pattern only in the case when the photon interacts with the detector, and that interaction is inelastic scattering (or absorption).


When the electron suffers inelastic scattering, it is localized; this means that its wavefunction collapses and after the measurement act, it propagates roughly as a spherical wave from the region of interaction, with no phase relation at all with other elastically or inelastically scattered electrons,” Frabboni said. “The experimental results show electrons through two slits (so two bright lines in the image when elastic and inelastic scattered electrons are collected) with negligible interference effects in the one-slit Fraunhofer diffraction pattern formed with elastic electrons.


https://physics.stackexchange.com/a/452027/132371
Even if the photon has passed through the slits, if it interacts with the detector after it passed through the slits, because the detector is positioned after the slits, and this interaction is inelastic scattering (or absorption), the wavefunction of the photon will collapse, and the phases of the photons (shot after each other) will be lost. Thus there will be no interference pattern only if the photon interacts with the detector.
So the answer to your question is, if the detector is positioned on the slits and is turned on after the photon has passed through the slits (or after the photon has already interacted with the screen and left a dot), there will be an interference pattern.
If the detector is positioned between the slits and the screen, and the detector is turned on after the photon has passed through the slits, and the photon interacts with the detector (inelastically or absorbed) then there will be no interference pattern.
