# Calculating net phase in single slit diffraction

I want to calculate the net phase at a point on the screen in Fresnel diffraction. Attached below is a rough diagram if needed.

For this I considered hypothetical $$n+1$$ small slits acting as sources within the original slit namely $$S_0 , S_1 … S_n$$ equally spaced at a distance $$x$$. The thought was to calculate the net phase at a point at distance $$y$$ from the midpoint of the slit and extract the limit as $$n$$ tends to infinity with the constraint that $$nx=a$$ where $$a$$ is the width of the slit. Assuming the phase of the first slit $$S_0$$ is $$\phi$$ , the phase due to the $$r^{th}$$ slit will be $$\phi +\frac{2\pi}{\lambda} \frac{yrx}{D}$$ under the assumption that the distance D is much much greater than the slit width. Therefore the net phase is $$\sum_{r=0}^{n} \phi +\frac{2\pi}{\lambda} \frac{yrx}{D} = \lim_{n\to \infty}[ n\phi + \frac{\pi n(n+1)yx}{\lambda} ]$$ . Upon putting $$nx= a$$ we get $$\lim_{n\to \infty}[ n\phi + \frac{\pi a(n+1)y}{\lambda}]$$ which is clearly blowing to infinity. I think that this problem is because the interference pattern of the double slit actually involves the superposition of 2 diffraction patterns which is why it cannot be used to explain the single slit diffraction pattern. However I do not have clarity over this. Also, if there is a fairly easy proof of the single slit diffraction pattern which is understandable upto a high school (12th grade) level please do share it, I am highly intrigued! Thanks.

• Note that single slit diffraction interference only works for light .... matter waves do not show single slit diffraction interference. Mar 1 at 15:48

P.S. Your solution is correct(i.e.doesn't have logical errors). Note that phases have a period of $$2\pi$$, so even if the sum mentioned in the solution goes to infinity the phase will have a value between 0 and $$2\pi$$.