# Why aren't maximums at 1 wavelength for single slit interference?

I can understand why there are destructive points in single slit interference. It is because the path length difference between the "pairs" of rays have a path length difference of $$\lambda/2$$. What I don't get is why this argument doesn't work for the maximum.

I watched this video (relevant part starts at 5:00) by Khan academy, which attempted to provide a solution but I don't quite get it. The presenter says that just because they are one wavelength apart doesn't mean that all the waves are in phase (that is, all the crests line up between the "pairs"). However, why would this be? Shouldn't all the infinitesimally small waves sources be coherent and inphase? After all, don't we bother to make the light coherent before sending it through the single slit and isn't this assumption made for double slit? Could someone please explain to me why there isn't a maximum crest when the path length difference is $$\lambda$$?

• Single slit diffraction can be calculated using classical wave theory or even quantum mechanics. I would say quantum mechanic gives the correct answer but it is fairly complex to understand. The classical approaches only give an approximation but when examined under close scrutiny tend to fall apart. Feynman took a path integral approach which is closer to QM. Aug 2, 2019 at 0:09

Good question! This is a bit subtle since the maxima are not in fact at the midpoints between the minima.

The argument for finding where the minina are relies on adding them up in pairs that are $$\pi$$ out of phase (half a wavelength path difference) so that they all cancel nicely. This argument works.

OPs proposed argument (and the one in the video) is that you can pick pairs that interfere constructively to find the maxima. This does not work, because each pair acts as a single source coming from the midpoint of the pair. But then that has a slight phase difference with the next constructively interfering pair. The net result is you don't get perfect constructive interference this way.

For reference, for a slit of width $$w$$, the light intensity at the screen goes as:

$$I \sim |\text{sinc}(kw\sin(\theta)/2)|^2=\left|\text{sinc}\left(\pi\sin(\theta)\frac{w}{\lambda}\right)\right|^2$$

where the sinc function is defined as $$\text{sinc}(x)=\sin(x)/x$$ as usual. The sinc function has extrema at solutions to $$x=\tan(x)$$ which are at $$x=q-q^{-1}+\mathcal{O}(q^{-3})$$ where $$q=(n+1/2)\pi$$ is where you'd naively expect the maxima to be. So in fact the maxima are a little closer in to the centre than you'd expect, although for very high orders the correction is negligible.

• (Thanks for correcting me.) May 11, 2019 at 16:09
• Thank you but I'm not quite following what you meant here: "because each pair acts as a single source coming from the midpoint of the pair. But then that has a slight phase difference with the next constructively interfering pair.". Why does every pair act as a single source? May 13, 2019 at 7:49
• @JohnHon because the pair are in phase with each other, their resultant is a sine wave (constructive interference). So they look like a single source from some effective position (which I think should be their midpoint). As do the next pair with $\lambda$ path difference, but the effective single source that these constitute is out of phase with the first pair. May 13, 2019 at 9:04

On the left I have redrawn the diagram in the video with the emerging rays from eight sources $$A,\,B\,C\,D\,E\,F\,G$$ and $$H$$ parallel to one another.
Those parallel rays can be made to superpose in the focal plane of a lens.

In the video the lecturer has the waves from source $$E$$ travel an extra wavelength $$\lambda$$, equivalent to a phase difference of $$2\pi$$, as compared to the waves from source $$A$$.
Thus when the waves superpose the do so constructively and the phasor diagram to illustrate this is shown on the right and labelled $$A$$ and $$E$$ with phasor $$A$$ being taken as the reference phasor.

Now look at the waves from sources $$B$$ and $$F$$.
They are again $$2 \pi$$ out of phase and add constructively but they differ in phase from the reference phasor $$A$$ by $$\frac \pi 4$$ and $$\frac{5\pi}{4}$$ and are shown in blue in the phasor diagram.
So although the waves from sources $$B$$ and $$F$$ add together constructively they are not do so with waves from sources $$A$$ and $$E$$.

One can now repeat the process for the remaining pairs of sources and looking at the phasor diagram, noting that the resultant phasor is zero, come to the conclusion that $$w\sin \theta = 2 \lambda$$ is indeed the condition for a minimum.

Finding maxima is much more difficult and just to illustrate the point just consider having four sources.

The first minimum is easy to find and is illustrated with the phasor diagram on the left.
The phase difference between adjacent sources being $$90^\circ$$.

One might think that the first maximum occurs when the phase difference between adjacent sources is $$120^\circ$$ as illustrated in the middle diagram.

However it is when the phase difference is slight smaller than $$132^\circ$$ that there is a maximum.

You may find the webpage 4.2: Intensity in Single-Slit Diffraction of interest but note the link given in that page which points to a challenging problem related to finding the accurate positions of the maxima as opposed to the approximate position used on the webpage.