# Derivation of the fraunhofer diffraction formula

Re the derivation in this PDF (Section 14.5: Single-slit diffraction, page 13).

My question is: how do we know that the wave orginating from the point at the top of the slit and the wave orginating from the middle one are exactly $\pi$ out of phase ?

Besides, is there a better mathematical formulation of this? Since I think the division of the points in halves is quite arbitrary, the same argument may apply with abitrary number of parts.

• The PDF says: At the first minimum, each ray from the upper half will be exactly 180 out of phase with a corresponding ray from the lower half. The key point is the phrase at the first minimum since at the first minimum the light rays have to sum to zero. – John Rennie Sep 12 '16 at 10:32
• I still don't quite understand, sorry. :( – Tung Nguyen Sep 12 '16 at 11:04
• The light rays have to sum to zero, but it doesn't guarantee that these two specific light rays sum to zero. – Tung Nguyen Sep 12 '16 at 11:25

## 1 Answer

Let's draw a diagram of the light hitting the slit and being diffracted by some angle: The light ray at the bottom of the slit has a phase lag, $\phi$, compared with the ray at the top of the slit because it has to travel farther. Let's assume that the angle happens to be the one where the phase lag is $2\pi$.

Now let's ask what the phase lag is for a light ray coming from somewhere in the slit between the top and bottom ray: The light ray comes from a distance $x$ measured from the top of the slit so $0\le x \le a$. It is (hopefully) obvious from the diagram that the phase lag of this light ray is:

$$\phi(x) = 2\pi\frac{x}{a} \tag{1}$$

Now consider two light rays, one coming from the position $x$ and one coming from $x+a/2$. For example this could be the two rays you describe in the question, one from the top of the slit ($x=0$) and one from the middle ($x=a/2$), but we'll stick with the general case of any value of $x$.

The phase lag of the ray from $x$ is given by equation (1) above, and the phase lag of the ray from $x = a/2$ is given by:

\begin{align} \phi(x+a/2) &= 2\pi\frac{x+a/2}{a} \\ &= 2\pi\left(\frac{x}{a} + \frac{a/2}{a}\right) \\ &= 2\pi\left(\frac{x}{a} + \frac{1}{2}\right) \\ &= 2\pi\frac{x}{a} + \pi \\ &= \phi(x) + \pi \end{align}

So what we've found is that any two rays separated by a distance $a/2$ have a phase difference of $\pi$. This answers your question as to why a ray from the top of the slit and the centre of the slit have a phase difference of $\pi$, but it's a more general result.

The importance of this result is that two waves with a phase difference of $\pi$ interfere destructively and cancel each other out. So for every ray in the range $0 \le x \le a/2$, i.e. from the top half of the slit, there is a corresponding ray at $x+a/2$, i.e. from the bottom half of the slit, that has a phase difference of $\pi$ and therefore interferes destructively.

This means that all the light being emitted at the angle I've drawn interferes destructively and the total intensity is zero. This angle, i.e. the angle where the phase lag across the whole slit is $2\pi$, corresponds to a dark region in the diffraction pattern.

• Thanks for your help! In my understanding you proved a bit more rigorously that: – Tung Nguyen Sep 12 '16 at 16:26
• If phase lag accross the whole slit is 2pi, then we have a destructive interference. Shouldn't we prove it the other direction: if we have a destructive interference, then the angle should be 2pi? Otherwise we only have a necessary but not sufficient condition for the minima pattern? (or in other words, that the condition sin (theta)=2pi/a accounts for all the minima ? ) – Tung Nguyen Sep 12 '16 at 16:32
• Nice diagrams. What did you use to create them? – garyp Sep 12 '16 at 16:33
• @garyp: Google Draw. It's a bit basic and requires a bit of messing around for complicated diagrams but it's free and works well enough. You can export to a range of formats, but I usually just screengrab and paste into Paint. – John Rennie Sep 12 '16 at 17:09
• @xpmrz: destructive interference requires that for every ray there is a corresponding ray with a phase difference of $\pi$. If you extend the method I've used above this requires that the phase shift across the whole slit be $2\pi n$, where $n$ is an integer greater than or equal to one. – John Rennie Sep 12 '16 at 17:12