Triple slit experiment If the experiment is done with single photons and  a "detector" at one of the slits the interference pattern breaks down.
What happens if three slits are used with single photons and a "detector" at, say, the right hand slit. Does the interference pattern occur in the two remaining slits or does the interference pattern breaks down for all the slits ?
Thanks for any response 
 A: In much the same way that putting a photon-absorbing detector at one slit in a double-slit experiment gives you the single-slit interference pattern back, putting a detector at one slit out of three will give you the two-slit pattern back.
It might help to think of the detector as blocking photons, in the same way that the material in which the slits are cut blocks photons. A detector doesn't have magical powers different from other physical objects, and blocking one slit doesn't do anything to the remaining slits. You could get the same visual result by putting chewing gum in one slit, you just wouldn't get any data from your gum. :)
A: You can think of the wave function of the three slit problem as consisting of three mutually nonorthogonal parts: $\left|L+C+R\right>$. Because the three have overlap, the square of their sum contains interference terms, so 
$\begin{equation}\left<L+C+R|L+C+R\right>=\left<L|L\right>+\left<C|C\right>+\left<R|R\right>+2\left<L|C\right>+2\left<L|R\right>+2\left<C|R\right>\end{equation}~$.
Now we add a detector to the system that can decide with certainty whether the particle was in L or not. This detector has two states, $D_L$ and $D_{CR}$, which are by design mutually orthogonal. The new wave function is $\left|D_LL+D_{CR}(C+R)\right>$. Now the square is 
$\begin{align}
& \left<D_L|D_L\right>\left<L|L\right>
 + \left<D_{CR}|D_{CR}\right>\left<C|C\right>
 + \left<D_{CR}|D_{CR}\right>\left<D_R|D_R\right>\left<R|R\right> \\ &+2\left<D_L|D_{CR}\right>\left<L|C\right>+2\left<D_L|D_{CR}\right>\left<L|R\right>+2\left<D_{CR}|D_{CR}\right>\left<C|R\right> \\ &= \left<L|L\right>+\left<C|C\right>+\left<R|R\right>+2\left<C|R\right>\end{align}$
Thus the sum of the single slit pattern of the left slit plus the double slit pattern of the centre and right slits is observed. 
Interference can only be observed if there is wave function overlap. A set-up that can with certainty distinguish the wave functions requires entanglement of the particle wave function with the detector wave function in such a manner that this overlap is eliminated.

A: As detection necessarily decoheres the input signal, one gets a combination of the pattern with a single slit (for the observed slit) and of the pattern with two slits (for the unobserved slits).
A: In response to the request for a reference regarding an actual triple slit experiment with photons, please check Science Vol. 329, Issue 5990, pp. 418-42. The title of the paper is Ruling Out Multi-Order Interference in Quantum Mechanics, and the result is that there is no new, exotic interference term ("$I_{ABC}$") due to a third slit. The only interference terms are $I_{AB}$,$I_{BC}$,$I_{AC}$ as you would expect from squaring of probability amplitudes of going through the slits $A$,$B$, and $C$ individually, and the authors measure these separately. 
I feel others have already answered the question, but to add my own, you can think of a blocked slit $A$ as a detector on $A$ (mother nature has detected the photon for you ;)) and the resulting pattern on the screen (if you see anything on the screen, you know for sure that the light has not passed through $A$) is the double slit pattern from $B$ and $C$ = $P_{B} + P_{C} + I_{BC}$, using the symbols from the reference. 
EDIT: regarding the nature of detection, it doesn't really matter whether it was a hard block on slit A or a photodetector. If an actual photodetector were placed on A, the signal from it will just show a single slit pattern.
A: I agree with arnold, the detector need not interfere with the photon as it passes through the detector; e.g. delayed response quantum erasure experiments have gotten around this. And really, you can't talk about blocking a photon, because if it exhibits an interference pattern, it probably didn't pass through both slits, just a man made probability wave to estimate its final position. A detector doesn't "block" a particle, it just means that the particle had to exist to serve the purpose of the detector at that point in space and time, to service consciousness and measurement.  For if you were to erase the which path information detected by the detector, after it passes through the slits, the interference pattern comes back. I am of the opinion, that a photon doesn't exist in any form until it absolutely had to, it's just in superposition of infinite possibility, until required by some conscious purpose.  Quantum mechanics and the "laws of nature" just sound a bit like a marvelous type of technology if you ask me.  But I ramble. 
