# Single Slit Diffraction Experiment vs Double Slit Interference Experiment- Formula Derivation

I understand that in the interference pattern of a double slit experiment, if the path difference is an integer multiplied by lambda, we have maxima. If it is 1/2 or 3/2 or 5/2 etc.. we get a minima.

However, what I don't get is: In a single slit experiment, we usually started out by assuming that at the point at the top of the slit, there is a light that has a path difference of 1 lambda with the one at the bottom of the slit. From there, we can derive the formulas: If slit width is a, then the distance between a point at the middle of the slit with a point at the top of the slit is only a/2. Thus, $\frac{a}{2} \sin(\theta) = \frac{\lambda}{2}$ since it will be destructive (because the path difference between the top and the bottom point is 1 lambda, then the path difference between a point at the top and at the middle should be lambda/2).

This is all clear to me.

But, all of this ONLY HOLDS IF the assumption is true: that the path difference between the point at the top of the slit and at the bottom of the slit is 1 lambda. Now, WHY exactlt do we assume this?

I mean, I can say that I assume the path difference to be a quarter of lambda, or 3/7 of lambda, or any arbitrary number for that matter. Thus, the path difference between the middle point of the slit and the top point of the slit will also become an arbitrary number of lambda, and thus the equation won't hold, right?

Is my logic mistaken?

The single source produces waves that travel towards the slit. The wavefronts represent peaks of the wave (in phase with each other). According to Huygens' principle, each point on the wave-front can be treated as an individual source. Thus, as this wavefront passes straight through the slit, all the points on it are individual coherent sources. So they cannot have any phase difference. Therefore it is incorrect to say that "points on the slit" have a phase difference. The constructive and destructive interference depends on the path difference only.

• Hmm... I see what you mean... but if we suppose change the light source into an incoherent one, would a pattern also still appear? – user205891 Sep 6 '18 at 15:07
• @user205891: "Coherent" refers to multiple sources being in phase with each other. It can't be applied to a single source. – user7777777 Sep 6 '18 at 15:12
• Hmm... but in doing the experiment, we typically use a laser (which is coherent), right? So, if we do it with, say, a standard flashlight, would it still work? – user205891 Sep 6 '18 at 15:15
• @user205891: Theoretically, it will work as long as there is a single wavelength being emitted (that's why we use lasers). A standard flashlight typically emits a range of wavelengths, so you will see different patterns. – user7777777 Sep 6 '18 at 15:19

at the point at the top of the slit, there is a light that has a phase difference of 1 lambda with the one at the bottom of the slit.

This is wrong. It becomes right if you talk about the phase difference at the screen at the position of the first minimum of light coming from the top vs. coming from the bottom of the slit.

The usual reasoning is that then the phase difference between top and middle is $\pi/2$, leading to destructive interference. The same holds for light from "slightly below top" together with light from "slightly below middle" and so on. For each point source along the slit, there is another one with phase difference $\pi/2$.