Diffraction proof

Question

I am talking about the way we generally explain the occurence of bright and dark fringes during diffraction. When we want to show a dark fringe we consider a pair of secondary light sources who have a path difference of λ/2.

Like If the slit width is "a", we see that dark fringe is obtained where asinθ=λ. Suppose we assumed there are 100 light sources and thus where the dark fringe is obtained, 1 and 51, 2 and 52, 3 and 53... these light source pairs destructively interfere because the path difference will be λ/2.

My question is that why are we considering only these pairs which interfere destructively. All the 100 light sources will interfere, so isn't there a possibility that some combinations of light sources interfere constructively?

BTW, I am a 12th class student, so please keep the reply less mathematical and simple

Answer 1

Two nothings is nothing. Reliably. So if you can pair each little virtual source with another that causes destructive interference you can be confident that when you add up the contribution from all the sources it will sum to zero.

Two somethings is not always something, at least when you're dealing with wave interference. So if you come up with some scheme where you pair each little virtual source with another that that give constructive interference you can't be guaranteed the sums of all those pairs will also add to something nonzero.

Answer 2

Welcome to SE. Lets say you have field $\mathbf{E}^{(i)}\left(\mathbf{r}\right)$ from i-th slit on the screen at position $r$. The total field at that position, due to all slits, is then:

$$ \mathbf{E}^{(net)}\left(\mathbf{r}\right)=\sum_i \mathbf{E}^{(i)}\left(\mathbf{r}\right) $$

Since fields (electric or magnetic) add up (follows from linearity of Maxwell's equations in free space). If the amount of power going through all the slits is finite, the sum above is absolutely convergent, so you can choose which order you sum your things in. For example:

$$ \mathbf{E}^{(net)}\left(\mathbf{r}\right)=\sum_{i=0}^N\mathbf{E}^{(i)}\left(\mathbf{r}\right)=\sum_{j=0}^{\lfloor N/2\rfloor} \left[\mathbf{E}^{(2j)}\left(\mathbf{r}\right)+\mathbf{E}^{(2j+1)}\left(\mathbf{r}\right)\right]+remainder $$

If every bracket vanishes, then the sum overall vanishes. This way you get the destructive interference. Remainder term (unpaired slit) could create problems, but the power going through it will scale as $1/N$ (assuming uniform illumination), so it is unlikely to be significant.

For constructive interference you can get a slightly more nuanced picture, since it is not clear that peak in the individual pairs will lead to peak in the overall sum. I would probably aim for a more complete treatment there, including the finite size of the grating, and the size of the slits as well as spacing between them. The treatment becomes much cleaner in the angular spectrum.

At the same time, if the way you construct solution for the constructive interference is to require that fields from successive slits ($\mathbf{E}^{(i)}$ and $\mathbf{E}^{(i+1)}$) differ by the slit-to-screen path of one wavelength, then the field from the slit after the next one ($\mathbf{E}^{(i+2)}$) will differ by two wavelengths, within paraxial approximation (slit spacing is much smaller than screen to slit distance), and same will work for successive slits, so you will get constructive interference (approximatelly)

Answer 3

This is a good and practical question. When you measure the intensity of a sinusoid or something modulated by a sinusoid you are measuring $\sin^2x=\tfrac{1}{2}-\tfrac{1}{2}\cos2x$ which has a broad peak at $x=\pm\tfrac{\pi}{2}$ and a sharp null at $x=0$. The latter is very sharp and its sharpness helps measuring the location more precisely in noise.

The localization of a null of a sinusoid rather than its peak is usually preferred even if the measurement is to involve phase comparison and not intensity; and if it has to be peak measurement in noise it is usually done by measuring the null crossing of its derivative, otherwise how would we know if it is really the peak?

Answer 4

The condition for interferences: two waves constructively interfere if their optical path differ by a multiple of the wavelength, is necessary but not sufficient.

A light source emits wave trains that have length and duration. The interferences only occur if the interferometer splits each wave train into two beams (creating two artificial coherent light sources), make the two beams travel a different path and reunite them on the screen. Two different light sources or two different wave trains emitted by he same source cannot interfere.

You can learn more about coherence length and time there.