Slit diffraction: time between emission and detection of photon

[m.e.t.a.]

For this question I am considering a slit diffraction experiment set up as follows:

{Monochromatic source} ------> {Single slit} ------> {Diffraction grating with $N$ slits} ------> {Screen with small movable detector}

The monochromatic light source emits photons one at a time. The principal interference maximum occurs at position $x=0$ on the screen. The detector is placed at some point, $x$, on the screen where the probability of detecting the photon is non-zero (also: $x \ne 0$). The detector detects all photons which arrive between positions $x$ and $x + \Delta x$.

Photons are emitted one by one at a slow rate. Every time a photon is emitted, a stopwatch is started. If the photon is detected at the detector then the stopwatch is stopped and that time measurement, $T$ is logged. If the photon is not detected then no measurement is recorded and the experiment is run again with a new photon.

The experiment is repeated many times. Finally, a probability distribution is plotted: {$T$} vs. {probability of $T$}. (I presume that this probability distribution will be approximately Gaussian in shape, although its exact shape is not important here.) This probability distribution will be centred around some mean value of $T$, ${T_{mean}}$.

Suppose that the experiment is run three times with different numbers of slits:

(i) $$N=1$$

(ii) $$N=2$$

(iii) $$N \to \infty$$


My question: will ${T_{mean}}$ vary in each case? And if it will vary, how so?

(This is a stripped-down version of a longer question I posted a few days ago, https://www.physicsforums.com/showthread.php?p=2695689#post2695689.)

 

[azntoon]

Yes, {T_{mean}} will vary in each case. As the number of slits increases, the diffraction pattern becomes more complex and the probability of detecting the photon at a given position on the screen increases. This means that the mean time for detection, {T_{mean}}, will decrease. As N \to \infty, the probability of detecting the photon at any point on the screen is effectively 1, so {T_{mean}} will tend to 0.