# Intensity for Single-Slit Interference Pattern

In the derivation for the equation for the relative intensity of a single-slit interference pattern in my textbook, there is an assumption that I find a bit fishy. I know this equation works, so it must be a correct assumption, but I see no obvious motivation for it.

The electric field vectors are represented as phasors for geometric simplicity. Part C is what I don't get. How do they assume that, upon summing more (and smaller) of the phasors, the curve they create is a part of a circle? In my head, the path difference a pair of neighboring 'rays' may not (actually will not) be the same as the pair of neighboring 'rays' that are right next to it. In fact, if I were to guess the curve formed by joining all those phasors, I would guess it looks something more like a section of a cardioid-like spiral (non-constant curvature).

For the single slit interference problem, the slit is assumed to be made of $N$ equally spaced oscillators, these small oscillators (Huygens Wavelets) produce disturbances/waves (electric field vector) which propagate in all radial directions. The interference of these disturbances from each oscillator produces the Diffraction pattern ~ (Huygens-Fresnel principle).

These Huygens wavelets are of equal amplitudes and driven similarly(driven with same phase) as the plane wave is uniform over the slit.

At any point on the screen there is a phase difference between each Electric field vector because we are looking at them at an angle such that there is a difference in time delay. The phase difference between any two consecutive vectors is $\phi$ [Now copying from Feynman]

Source: http://www.feynmanlectures.caltech.edu/I_30.html

Resultant Electric Field $$R=A[\cos(\omega t)+\cos(\omega t+\phi)+\cos(\omega t+2\phi)+\cdots+\cos(\omega t+(n-1)\phi)],$$ where $\phi$ is the phase difference between one oscillator and the next one, as seen in a particular direction. Now we must add all the terms together. We shall do this geometrically. The first one is of length $A$, and it has zero phase. The next is also of length $A$ and it has a phase equal to $\phi$. The next one is again of length A and it has a phase equal to $2\phi$, and so on. So we are evidently going around an equiangular polygon with $N$ sides Now the vertices, of course, all lie on a circle, and we can find the net amplitude most easily if we find the radius of that circle. Suppose that $Q$ is the center of the circle. Then we know that the $\angle OQS$ is just a phase angle $\phi$. (This is because the radius $QS$ bears the same geometrical relation to $A_2$ as $QO$ bears to $A_1$, so they form an angle $\phi$ between them.) Therefore the radius $r$ must be such that $A=2r\sin(\frac{\phi}{2})$, which fixes $r$. But the large $\angle OQT$ is equal to $n\phi$, and we thus find that $AR=2r\sin(\frac{nϕ}{2})$. Combining these two results to eliminate $r$, we get

$$A_R=A\frac{\sin(\frac{n\phi}{2})}{\sin(\frac{\phi}{2})}$$

The resultant intensity is thus

$$I=I_0\frac{\sin^2(\frac{n\phi}{2})}{\sin^2(\frac{\phi}{2})}$$