# Probability of a photon in Young's Double slit

I recently read an answer and I was not able to make sense of it. The answer said Let us consider the famous double-slit experiment with photons. With the usual set-up, we denote the number of photons passing through by N and we will denote the number of photons which hit the film between y and y+Δy by n(y). The probability that a photon will be detected between y and y+Δy at a time t is given by: Py(y,t)≡limN→∞(n(y)N) If we consider this from an classical electromagnetic point of view, then the above quantity is known as the intensity I of the electromagnetic wave which is well known to be proportional to: I(y,t)∝|ψ(y,t)|2Δy

Now I did not understand why the probability is what is given in the answer so if someone could clarify that and why is that a probability. If the interference pattern is observed what do u mean by probability,the photon is there isn't it. And secondly why is the probability equal to the intensity of the pattern. I am just beginning Quantuam Mechanics and I am not able to understand this .

• If you do the experiment with a very low intensity of light you are dealing with a small number of photons. The intensity pattern on the film will be noisy at first but gradually build up to the expected pattern. So in this case it's useful to consider the expected pattern as a probability. – Not_Einstein Jun 21 '20 at 17:56