Single-Slit Diffraction -- Deriving the relation for the minima

[Clemencio]

TL;DR Summary

Single Slit Diffraction

Hi everyone.

I'm studying single-slit diffraction, and a question came up: to derivate the relation for the minima (dark fringes), the slit is devided in two parts, and it's assumed the distance between the light rays is a/2. Why is this distance chosen?

I was wondering about other options. For exemple, to take two rays at the edges of the slit (distance of "a"), or other arbitrary distance, such as 3/4.

Why should I assume that rays separated by a/2 will form the 1st diffraction minimum?

Thanks!

 

[Philip Koeck]

TL;DR Summary: Single Slit Diffraction

Hi everyone.

I'm studying single-slit diffraction, and a question came up: to derivate the relation for the minima (dark fringes), the slit is devided in two parts, and it's assumed the distance between the light rays is a/2. Why is this distance chosen?

I was wondering about other options. For exemple, to take two rays at the edges of the slit (distance of "a"), or other arbitrary distance, such as 3/4.

Why should I assume that rays separated by a/2 will form the 1st diffraction minimum?

Thanks!

Think of Huygen's wavelets. If the wave from the upper edge is 180° out of phase with the central one then the same is true for all other pairs of waves coming from points a distance of a/2 from each other.
In total they will cancel each other in pairs.

 

[sophiecentaur]

TL;DR Summary: Single Slit Diffraction

the slit is devided in two parts

That's one of those bits of book-work that seem to come out of nowhere and it's seldom explained why. The choice of taking an (arbitrary) pair of true point sources, separated by half the slit width, is because the interference patterns for those two points is the same as for the next pair of points across the slit, and the next and the next pair etc.. When you get to the end, you have accounted for all the points. If the first pair of points has a null in a particular direction then so will all the pairs of points.

TL;DR Summary: Single Slit Diffraction

Why should I assume that rays separated by a/2 will form the 1st diffraction minimum?

No. That's the wrong way of looking at it. The direction of the first null will be when the difference in path lengths is λ/2. There's no assumption here - it just happens in a particular direction. And those path length differences will be the same for all pairs.

Think of Huygen's wavelets.

Personally, I had an initial problem with Huygens - to such an extent that I needed to have the whole business right in my head before I could actually make Huygens make sense. It may not be politically correct to think in terms of 'rays' but that diagram doesn't actually have wavelets on it; it just shows the field values at different points along the 'ray'. Trying to make sense of a mass of circular wave fronts can be a bit of an effort.
That's only a personal opinion and it maybe a product of my ancient education content but it did work for me.

 

[sophiecentaur]

If the first pair of points has a null in a particular direction then so will all the pairs of points.

Note: if you work with different pairs of points the geometry says that those pairs can't all have the same spacing so their two point interference patterns will all be different and the nulls will be in all different directions - telling you nothing useful.

 

[Clemencio]

The choice of taking an (arbitrary) pair of true point sources, separated by half the slit width, is because the interference patterns for those two points is the same as for the next pair of points across the slit, and the next and the next pair etc.. When you get to the end, you have accounted for all the points. If the first pair of points has a null in a particular direction then so will all the pairs of points.

Thank you much so much! I looked for this explanation in many books and online classes, and this is the first time someone convincely explained it!

 

[sophiecentaur]

Thank you much so much! I looked for this explanation in many books and online classes, and this is the first time someone convincely explained it!

We aim to please!! I really puzzled me when I first came across it but it's the same with many bits of book work. They just say "do it" but they don't come clean and say it's a 'trick'.

 

[Clemencio]

We aim to please!! I really puzzled me when I first came across it but it's the same with many bits of book work. They just say "do it" but they don't come clean and say it's a 'trick'.

May I bother you with one last time about this topic? The question is: If I try to use the same calculation method to determine the maxima - dividing the slit in 2,4...- it doesn't work. For example, if you calcula for a path difference of 1 wavelength (for a/2), you obtain a value that coincide with the second minimum (using a/4). But we know that the first secondary maximum is between the first two minima. So, why one cannot use the method to calculate the maxima?

 

[sophiecentaur]

So, why one cannot use the method to calculate the maxima?

This method for minima is 'quick and dirty' because it's only when you get minima which are all coincident that you can say you have found a net minimum. The method tells you nothing abut the actual shape of the fringe pattern elsewhere. If you want to do that then you need to do the full diffraction calculation, taking the vector sum of all contributions from each elemental part of the slit. You need to consider elements with a quarter wavelength spacing or less and then the sum (integral) is pretty near the real diffraction pattern.

 

[Clemencio]

I see, the method have serious limitations, though it provides an intuitive understanding of the phenomeon - at least, in the basic level. Thanks once more!