Single Slit Diffraction

To find the actual/complete diffraction pattern we need to consider each point in the slit as a source of light and find the resultant amplitude (actually resultant electric field displacement) at the concerned point on the screen. This involves some interesting maths and logic. But before finding that infinte sum, lets look at a simplified finite case.

Here, we have shown only five points. Let each ray to inclined at \(\theta\) angle and the slit width be \(d\). Suppose the electric displacement be a, wavelength of light be \(\lambda\) and the angular velocity be \(\omega\). As shown in the figure, after points 0′, 1′, 2′, 3′ & 4′ all the rays travel same distance upto the screen placed at infinity. Thus their resultant at the screen is same as their resultant at those points (i.e. the sum of the displacements at those points). At the starting points 0, 1, 2, 3 & 4 on the slit each ray can be represented by : \(y = a \sin(\omega t)\) but since the path from original poisition to the primed (′) position is different for different rays, and since they start at same phase, they are out of phase when they reach the primed positions. The path difference between cosecutive rays is \(\frac d 4 sin\theta = d′ \sin\theta\)

So corresponding phase diff is: \(\Delta \beta = d′ \sin\theta . \frac {2\pi} {\lambda}\) Thus now the rays can be represented at primed position as:

\begin{align} y_0 &= a \sin(\omega t) \\ y_1 &= a \sin(\omega t + \Delta \beta) \\ y_2 &= a \sin(\omega t + 2 \Delta \beta) \\ y_3 &= a \sin(\omega t + 3 \Delta \beta) \\ y_4 &= a \sin(\omega t + 4 \Delta \beta) \end{align}

And the resultant is \(y = \sum_{i=0}^{i=4} {y_i}\)

Now, we just need to find the sum. Of course if we had actual data like, θ = 30°, a = 10 A°, λ = 642nm, ... we could just put the values and get the result :smiley: . But here we have to do this with all variables. Also, if there were 2 or 3 terms only we could have expanded the \(sin(\omega t + \Delta \beta)\) terms and got the sum (like in the case of Young's double slit experiment's expressions). But, in this case we have to later expand/generalize this to infinte terms as the segment size goes to zero. So, we use a tool called phasor. I hope you are familiar with it. If you aren't don't worry, the main thing about phasors is that they represent expressions of the form \(y = a \sin(\omega t)\) graphically. And their sums, for e.g. (1) can also be represented graphically as :

(Here β = NΔβ where N = number of segments)

We might find a formula for the length of a diagonal joining N th sides (here 5th side) of regular polygon whose each external angle is Δβ and sum of those side is Na. Buts lets go to the real suff.

Dividing the slit into very small equal segments of width \(\Delta d = \frac d N\). We get phase difference of \(\Delta \beta = \Delta d \sin\theta\) between rays of each segment. So, as \(N \rightarrow \infty Δd \rightarrow 0 \text{ and } a → 0\). The \((N + 1)\) sides of the polygon tends to a circular arc. As shown below :

After all these conceptual talk, lets proceed to some small calculations. By geometry we can say,

\begin{align} \frac {A} 2 &= r sin(\frac \beta 2) & [\text{from triangle made by radius, half cord and normal}] \\ Also, \\ r \beta &= A_0 & [\text{by defination of circular measure}]\\ \therefore \frac A 2 &= \frac {A_0} \beta \sin(\frac {\beta } 2)\\ A &= A_0 \frac {\sin(\frac {\beta} 2)} {\frac {\beta} 2} \\ &= A_0 \frac {\sin(\frac {d \sin\theta} 2)} {\frac {d \sinθ} 2} \\ \end{align}

Finally we know the amplitude/electric field displacement at the screen. Plotting this looks like:

Now, the intensity can be calculated as :

\begin{align} &\text{We know, } &I &\propto A^2 \\ &\text{So, intensity at the centre, θ=0 } &I_0 &\propto A_0^2 & \\ &\therefore &I &= I_0 \frac {A^2} {A_0^2} \\ & &I &= \frac {\sin²{\frac {β} {2}}} {(\frac β 2)^2} \\ \end{align}

We can find the maxima and minima by differentiating intensity with respect to β and equating to zero, we get :

\begin{align} &1) & \sin{\frac β 2} &= 0 \\ & & \implies β &= 0 , 2π, 4π, 6π, ... \\ &2) & \frac β 2 &= \tan(\frac β 2) \\ \end{align}

The first set of values represent minima (except central maxima β = 0), β = 2π is first minima, β = 4π is second minima and so on. A quick search at wolframalpha solves the second equation with following values

β =	remark
0	Central Maxima
±8.9868 ~ 3π (-4.6%)	1st Secondary Maxima
±15.4505 ~ 5π (-1.6%)	2nd Secondary Maxima
±21.8082 ~ 7π (-0.8%)	3rd Secondayr Maxima

These values for maxima, get nearer to odd multiples of π for higher order minimas.

I would like to conclude this with the following graph: