Why does wavelength affect diffraction?

Question

I have seen many questions of this type but I could nowhere find the answer to "why". I know this is a phenomenon which has been seen and discovered and we know it happens and how it happens. But my question is why would wavelength affect the amount of diffraction? I am looking for a very simple logical explanation rather than a complex mathematical answer. Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? I need an answer that will answer "why" does diffraction depend on wavelength of light.

Image sources: http://www.olympusmicro.com/primer/java/diffraction/index.html

Answer 1

Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray?

Don't think of bending. Think of diffraction like this: if you have a plane wave incident on a slit, then you can think about the space in the slit as being a line of infinitely many point sources that radiate in phase.

If you are looking straight down the slit, then all those point sources are in phase. There's not much unusual going on here.

However, if you move a bit to the side, then all those point sources aren't in phase. They are, really, but since they are not at equal distances to you, the radiation from each is delayed by a different amount. Depending on your position, the point sources interfere constructively or destructively, and this is what yields the diffraction pattern.

If you look closely at this image, it appears it was generated by an approximation of four point sources in the slit.

Now, the number of these point sources there are, and the maximum difference in phase between them, is a function of the size of the slit, obviously. If the slit is wider, then when viewed from some direction slightly off center, the phase difference from the left-most source and the right-most source will be greater, because the difference in distance between them is greater.

Compare a small slit:

To a bigger slit:

The significance of the size of the slit is apparent, right?

Well, changing the wavelength is equivalent to changing the size of the slit. If we make the slit bigger, and make the wavelength bigger by the same amount, then the difference in distance between the sources is greater, but the rate of change in the wave function is slower, so the phase difference between the two extremes of the slit is the same.

But, if we just make the wavelength smaller, and leave the slit the same, the rate of change in the wave function is faster, which is equivalent to making the slit bigger without changing the wavelength.

Images from Wikipedia

Answer 2

My answer will be quite close to that of PhotonicBoom although a bit more graphical. When it comes to light phenomena, there are different ways to comprehend them: we can use a wave picture (Hyugens-Fresnel), we can use the most modern picture we have (QED) or we can use something a bit more intermediate which is the picture of light rays travelling from one point to another.

We all kind of know that in air or vacuum, light travels (whatever that means) in straight line from a source point $S$ to an observation point $M$.

This follows from Fermat's principle of shortest path followed by the light from one point to another that can be depicted by the following drawing

Now, this picture of single light ray taking the shortest path as such is not enough to explain, as far as I know, the phenomenon of diffraction.

As someone else said in his answer, we need to account for the fact that a proper theory of light should be looking at the square of an amplitude which arises from the wave nature of light.

This is illustrated by the well known experiment of Young's slits (depicted in the top left panel of the next figure). Incindentally, the light ray picture and its optical length is still very useful when looking at interferences and it is common to represent them explicitly in a graphical way.

Now, starting from the double slit experiment, one can try to add another wall with two holes in it and see what those light paths look like (top right panel of figure 2). One can add another wall and even more holes in these walls and yet again observe what would the paths over which one would have to sum to get the total light intensity at point M. Eventually, continuing this process of adding walls and filling (densely) with holes, one finds that we need to sum over an infinite number of light paths as depicted in bottom left panel of figure 2 even in the limit where is no wall anymore.

Hence, to explain diffraction with a light ray picture, one needs to imagine that the intensity received at point M is actually the sum of the intensities coming from an infinite number of light rays linking the source point to the observer a little bit like in the following picture (bigger version of a picture in figure 2)

Here what happens is that the sum over all paths is weighted such that paths that are close to the shortest one have a huge weight and paths that have a very long optical length have a very tiny weight in the overall sum (I have accounted for these weights by lightening the colours as the lengths of the paths become big). Hence in air or vacuum the total intensity received is overwhelmly due to those paths that are close to a straight line behaviour and Fermat's principle is recovered.

Now, what happens when I put a wall with a hole in it?

More or less what is depicted here

We see that in this case, all the paths that use to contribute a lot to the intensity cannot be used anymore and only those that pass through the hole remain; hence diffraction.

Now, this happens for all frequencies of light for which this hole is not opaque for some reason (its size too small compared to the wavelength or something like that). However, when it comes to the notion of length of the travel path (related to the weight each path is given inside the sum), it turns out that it depends on the wavelength of the light in such a way that for the same path followed by the light, the weight associated to it will be smaller for blue than red color, hence red color can reach further than blue for the same hole size.

Two/three comments are here in order:

• This description is not wrong I believe but is not hyper rigorous

• It is mostly inspired by a quote in Zee (QFT in a Nutshell) and by Richard Feynman description of this phenomenon in QED, it is just that I believe that this actual phenomenon has nothing to do with QED and classical description (based on light rays) should be possible.

Comments and questions are very much welcome to improve this answer.

Answer 3

Huygen's principle alone will not answer your question, however the Huygen-Fresnel principle modifies this to include wavelength. It states that every point in an unobstructed beam acts as a secondary source of wavelets with the same wavelength as the primary wave. The amplitude of the optical field at any point is then the superposition of all the wavelets. The superposition takes into account both the amplitude and the phase of light.

On your diagram the intensity values oscillates and you get a peak when the wavelets constructively interfere and a trough when the waves destructively interfere. Consider the two wavelets coming from each edge of the slit. When they hit the screen they will have travelled a distance $d_1$ and $d_2$ respectively. Both distances depend on the position you are measuring the intensity on the screen. The phase difference between the the two waves is then $\phi=(d_1-d_2)*2*\pi/\lambda$. You will get a peak in intensity when $\phi=0, 2\pi,4 \pi, \ldots$ and destructive interference when $\phi= \pi, 3 \pi, 5\pi \ldots$

It is the interference of the wavelets, which depends on wavelength, that is causing the wavelength dependancy in the diffraction pattern.

For more information look up Fraunhofer diffraction from a single slit.

Answer 4

The answer lies in QED and is quite complicated mathematically. As a simple explanation this is what happens:

Photons follow all possible paths from the source to the screen and each path has its own probability amplitude associated with it. Summing up all these paths cancels out most terms in the summation and you end up with your desired final "classical" path.

The path is wavelength depended and thus different wavelength photons have different ending points on the screen. I'm no expert in QED so someone with more knowledge can expand on this.

Check out the wikipedia link here

Answer 5

There are a lot of good answers already here, but I think I'll add a few pictures that show the ways I have to force myself to think to keep track of the differences between light waves and light rays when thinking about diffraction patterns. Above we have some light coming through the center of the slit. If we draw it like this it seems to make sense with some of the things we're told about light, but it kind of only half tells two different stories. It is drawn to look like a wave (with amplitude, wavelength, etc.) but it comes straight through (like a ray), so it muddles up what is happening. If we keep in mind these shortcomings we can keep working though.

If we add more rays, we get a picture that seems to make some intuitive sense; it looks like we have more light coming through and where more light goes the brighter the screen will be, but that intuition is very wrong when we have very small slits! If we do that we're completely ignoring the wave properties of light, and completely ignoring poor Huygens!

If we want to do right by Huygens, we have to think about many points acting as "secondary emitters" and imagine how multiple waves coming from multiple points will interact further away from the slit.

If we imagine two points emitting from either side of the slit and investigate what happens if we look directly in front of the slit we can see a couple things. First of all, because we're looking at a point dead center in front of the slit that means the point is the same distance from the two emitters. In the drawing that distance is 1.5 wavelengths. Because the waves have the same wavelength and travel the same distance, we know that the crests and troughs of the wave will line up when reaching the green line. Because the waves line up, they are "in phase", they add together and make a wave with twice the amplitude. The drawing shows a black and white wave to match the colors of the peaks and troughs in the images already posted by Phil Frost. Remember, light waves are moving, like in the animated drawings, so we don't want to imagine white as bright spots and black as dark spots in the diffraction pattern. These areas with high-contrast black and white nearby mean we have large amplitudes, so these large striped areas will be the bright spots! What if we look at a different point? Well, if that point isn't the same distance from both of the emitters we're imagining, then they might not be "in phase" anymore when they reach the green line. In fact, in the drawing one wave is low and the other is high, they're very much in opposite phase. If we add two opposite phase waves together, they will cancel each other out. This creates a combined wave with a very small amplitude. When drawn in black and white, this would look all grey and washed out- so that means the grey areas in that other picture must be the dark spots.

We really shouldn't stop at just one or two tiny emitters, but it gets harder to think about. The picture realy does get closer and closer to reality if we add more and more points. We can either get a computer to try to do all the calculations, or we can do some math and take limits for $N=\infty$ emitters and doing the calculus to integrate the result of all of those waves adding together.

The things we need to remember though, are that the "bright spots" we'd see on the wall when doing single slit diffraction experiment correspond to the high-contrast large-amplitude areas of the greyscale drawing. Also, the light of one wavelength doesn't "bend" any more all secondary emitters are always emitting spherically in all directions all the time- the reason the "bright spots" change places is because changing the wavelength changes the number of peaks and troughs for the trips from both emitters - and if their phases aren't exactly in phase again, then the brightness has to change at that location.

I think those are some common tripping points that occur for people when combining the ideas of waves, rays, diffraction patterns and interference patterns. So hopefully it helps someone out.

If your question was more hoping to get an answer to "why can even vacuum act like one of Huygen's imaginary secondary emitters?" then I guess my answer doesn't help at all.

Answer 6

A wave is a perturbation in a system that propagates. The wavelength is the typical length along which a wave is coherent, which means that what happens at some position affects the wave behaviour in the vicinity if this point at distances of a few wavelengths. The reason for that is that the medium in which the wave propagates has some rigidity and the wavelength is actually determined by the rigidity of the medium.

As a consequence in diffraction, the wave field must cancel at the edge of the hole, which results in the wave feeling it. For small holes, all the wave energy feels the presence of the edges. For large holes, most of the wave field is at a distance much larger than $\lambda$, so only a small fraction of the wave is affected by the edges and the diffraction effect is not noticeable.

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EDIT In answer to your comment, I will try to give a simplified explanation concerning coherence.

A very common picture in optics to figure out light trajectories is the following

In many situations, this is enough to understand the phenomenon (lens, mirrors, etc.). To understand what happens in diffraction, prefer the following picture

in which you notice that the rays are now beams of a certain width, depending on the color. The width you should consider is roughly the wavelength $\lambda$.

Using this picture, what happens to a light when crossing a hole. If the hole is much larger than $\lambda$, (almost) nothing happens : the beams have enough space to cross the hole

If the hole is smaller than $\lambda$, the beam will "bend" in order to keep its length and fit into the hole, as is sketched on the following picture, zoomed on the hole (I have drawn the beam as a train of stems that bend when entering the hole)

The reason for the bending is what I called coherence. It is due to the conservation of energy: if you could shrink the stems, you would lower the wavelength and thus increase the energy, which is impossible.

Answer 7

The interference between all the rays emitted from the aperture to a fixed point on the screen can be constructive or destructive, depending on the various path lengths involved (measured in wavelengths). If you change wavelengths, the path lengths (measured in wavelengths) change. What is constructive interference between paths at one wavelength can be destructive at other wavelengths. This is why the effect is wavelength dependent.

The path lengths I speak of are illustrated below, where the red lines are the paths from the aperture to one point on the screen. The red line lengths are all fixed, but the electromagnetic wave phase depends on the physical distance divided by the wavelength, so the path lengths can be thought of as changing with wavelength, as far as the electromagnetic phase is concerned.

Answer 8

The question "why does the wavelength affect diffraction", I think, could be best answered by looking at the two extreme cases. Assuming a narrow opening is illuminated:

1. If the wavelength is much smaller than the width of a slit, wave effects can be completely ignored, because interference effects won't play a role. Consequently, the light waves will pass through the opening like a ray.

2. If the wavelength is much larger than the width of a slit, again, no diffraction pattern will be observed. However, the slit now acts as a point source, i.e. the narrow opening becomes the source of a new wave (Huygen's principle).

Consequently, longer wavelengths are redirected more strongly than shorter wavelengths, and hence diffraction is wavelength dependent. Diffraction is understood as the interference pattern of all waves behind the narrow opening. Considering the two extreme cases above, I would then argue that blue light is less diffracted as red light.

Answer 9

The greater the wavelength the heavier the wave. If you think of it visually, the heavier the wave the more energy needed to move the wave in a different direction. As a result, the greater the wavelength, less diffraction.