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enter image description here

The explanation for the single slit diffraction pattern is that when a ray at the top and a ray at the center of the slit, at $\frac{b}{2}$ where $b$ is the slit width, marked by red in the left figure, has a path difference of $\frac{\lambda}{2}$, it destructs totally. This explanation can be shifted to a pair of rays below the first 2 and shows that all rays destruct, leading to a dark fringe.

In the right figure, the next dark fringe is created when the top ray and the ray at $\frac{b}{4}$ destructs since they have a path difference of $\frac{\lambda}{2}$, leading to the first quarter destructing the second quarter. Similarly, the third quarter destructs the fourth quarter, and a dark fringe is shown.

However, I don't understand how this principle is not applied to constructive interference. In the second figure, the top ray and the ray at $\frac{b}{2}$ should interfere constructively since they have a path difference of $\lambda$. When it is applied to all other rays below it shows that it would create a bright fringe??

Thanks for helping out.

Flyingsparky
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  • The related trigonometric equations are good for calculating the positions of the fringes. In reality no interference takes place because the photons of the EM radiation at that energies do not interact. Simply the quantized interaction with the slits edges deflects the photons in fringes. (The interference of water waves indeed takes place because water is a medium which gets locally compressed and displaced and goes back elastically at the next moment.) See https://physics.stackexchange.com/a/209412/46708 – HolgerFiedler Jun 15 '21 at 17:41

4 Answers4

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So, to understand this, you have to keep the big picture in mind. Remember, what's happening is that a plane wave is coming upon a barrier with a non-pointlike slit in it (so, a slit of a certain length).

For a "pointlike" slit (smaller than the wavelength), this is what happens; the slit acts as a source of a spherical wave:
enter image description here

Note that its the wave phenomena that's the "real thing", the rays are just imaginary lines perpendicular to the wavefronts.

For a wider slit, the behavior is more complicated, but, according to the Huygens–Fresnel principle, the situation can be described by treating every point along the slit as an emitter of such spherical waves ("wavelets"), that then interfere (superpose) with each other. The amplitude of the actual (resulting) wave at any point is a superposition of all these wavelets.

enter image description here

So, every point is emitting rays in every direction (within the half plane). Now, the assumption for the explanation you posted is that the screen (where you observe the diffraction pattern) is very far away, meaning the distance to it is much larger than the size of the slit .

That means that the rays originating from every part of the slit that are hitting the same point on the screen are nearly parallel (just how we treat rays of sunlight as nearly parallel). All of the rays in the picture below are meant to be the rays that hit the same point at the other end:

enter image description here

So here, you're just trying to figure out what's happening at a single point on the screen, by summing contributions from all of these rays. Remember, though, it's the value of the field along the ray that is making the contribution (think of each ray as being a slice through its corresponding wavelet; the contribution comes from what the amplitude of the wavelet happens to be at the other end, where it hits the screen).

enter image description here

This image does not depict some alternate setup, this is the same experiment, at the same time. It's just that now you're considering rays that all hit a different point on the screen. Again, you're summing contributions from all of them.

Note that the picture only shows some representative rays, but that there are infinitely many of them; there are rays in-between, not just the ones depicted here. (The contributions from the wavelets are, really, infinitesimal, and the sum is an integral.) All of them, together, add up to zero.

However, I don't understand how this principle is not applied to constructive interference. In the second figure, the top ray and the ray at $\frac{b}{2}$ should interfere constructively since they have a path difference of $\lambda$.

These particular examples were chosen so that all the rays combine to form a minimum. Yes, considering random pairs, some add up constructively, some cancel out, some produce an arbitrary value. However, you're interested in the total sum. Now, rather then working with the actual values, you can make use of the phase-difference in your reasoning. The idea is that you can always pick a pair that interferes destructively (so, zero net contribution), take it out of the total sum, then pick another such pair. So if every pair cancels out, then everything cancels out.

It's the same logic as here:

enter image description here

Note that you don't have to know the actual values if you know that, in this particular arrangement, all the addends 6 positions apart cancel.

Filip Milovanović
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Indeed the waves from sources $a$ and $c$ are in phase add constructively and the waves from sources $b$ and $d$ are in phase and add constructively.

enter image description here

However, the waves from sources $a$ and $c$ are exactly out of phase with the waves from sources $b$ and $d$ and so add destructively, $\Rightarrow$ zero wave amplitude $\Rightarrow$ a minimum.

Farcher
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This principle is applied to constructive interferences:

When the difference in path equals $m\lambda$, with $m$ being an even integer ($m = 0, 2, 4...$) etc., then the interferences are constructive, i.e., they add up:

https://simple.wikipedia.org/wiki/Interference

Serge Hulne
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    Exactly, that's what I thought too, that when $m$ is an even integer in $\frac{b}{2}sin\theta=m\frac{\lambda}{2}$ it would lead to constructive interference but apparently when thinking of it in the way shown in the right figure and in my second paragraph it shows that it's actually a dark fringe, which is what is actually present based on experiments. So different workings of the same principle lead to two different answers, which is why I was asking. – Flyingsparky Jun 15 '21 at 02:03
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It works both. Around single hole you will observe rings

enter image description here

of interchanging positive and negative interference.

enter image description here

Dims
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