# Classical light diffraction round corner

• BruceW
In summary: Yes, there is a plane wave, but it is complicated to draw and it is not the most accurate way to think about waves. In fact, most physical descriptions of waves use more complex methods like diffraction or waves on a string.
BruceW
Homework Helper
I know how photons are diffracted round a corner in quantum mechanics. But is there also an explanation in classical electromagnetic theory (i.e. by Maxwell's laws?)
Imagine there was an EM wave traveling purely normal to a slab of material which is highly attenuating, but which has a small hole. (I've drawn a picture, so you can see what I mean.)
Maxwell's equations in vacuum are:
$$\nabla \cdot \vec{E} = 0$$
$$\nabla \cdot \vec{B} = 0$$
$$\nabla \wedge \vec{E} = - \frac{d \vec{B} }{dt}$$
$$\nabla \wedge \vec{B} = \mu_{0} \epsilon_{0} \frac{d \vec{E} }{dt}$$
And inside the slab, E and B would equal zero, since it is highly attenuating.
So then, am I right in thinking Maxwell's equations predict the wave coming out of the hole to be traveling normal to the slab (i.e. not diffracted)?
But obviously, light is diffracted in this situation. Is there some kind of edge effect that makes the wave change direction? Or are Maxwell's equations simply unable to predict this phenomenon?

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All waves experience diffraction, including water waves, and yes, even classical electromagnetic waves. Thinking about EM waves as linear rays, as you have drawn in the picture (Geometric Optics) is only a simple approximation to Maxwell's equations. If you use the full-wave equations, diffraction is built in. There are actually several ways to derive diffraction from Maxwell's equations with varying degrees of approximation, difficulty, and accuracy. The most common mathematical form is typically Fraunhofer diffraction, which you should look up to learn more about as it is far too involved to recount here.

If you just want a physical picture of why waves diffract, and not the detailed math, there is a neat way of picturing it. You can't make the hole infinitely small. So with a finite size, there will be a wave coming out the top of the hole and one coming out of the bottom of the hole, and they interfere with each other. In this sense, diffraction can be thought of as the interference pattern of all the waves that make it through the hole. http://en.wikipedia.org/wiki/Huygens-Fresnel_principle" was probably the first one to develop the idea of an EM wave interfering with itself.

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Wave mechanics is interesting stuff. Sound waves, water waves, waves on a string, EM waves, etc. have been diffracting, interfering, refracting, and tunneling through potential barriers long before the magic of the quantum world came along.

BruceW said:
So then, am I right in thinking Maxwell's equations predict the wave coming out of the hole to be traveling normal to the slab (i.e. not diffracted)?

I think you're wrong here. Solving the Dirichlet problem of zero field on the slab (say the plane x=0, minus the slit) and on the hemisphere $$x^2+y^2+z^2=\infty \mbox{, x>0}$$, and whatever value of the field on the slit (the plane wave), gives you diffraction solutions.

I've looked up Kirchhoff diffraction, Fraunhofer diffraction, Fresnel diffraction and the Huygens-Fresnel principle on wikipedia. And it says that these equations are caused by diffraction of a spherical wave. My question was for a plane wave.

RedX said:
I think you're wrong here. Solving the Dirichlet problem of zero field on the slab (say the plane x=0, minus the slit) and on the hemisphere
x2+y2+z2=∞, x>0
, and whatever value of the field on the slit (the plane wave), gives you diffraction solutions.
So there must be zero field at x=infinity ? Then I guess a perfectly plane wave is physically impossible? That would explain why diffraction happens (because the wave can't be a perfect plane wave).

BruceW said:
So there must be zero field at x=infinity ? Then I guess a perfectly plane wave is physically impossible? That would explain why diffraction happens (because the wave can't be a perfect plane wave).

It's not just x=infinity, but whenever the distance from the slit to the point is infinity. Also the screen itself must have zero field.

I just worked it out real quick, but you can start with a plane wave at the aperture, and assume a plane wave solution at infinity (rather than zero field), and you still get Fresnel-Kirchhoff diffraction.

Maybe I made a mistake in the calculations, but I think the physical reason for this is that diffraction actually predicts plane waves at infinity with no fringe patterns. Because the angle between a point at infinity and the normal to the slit is always zero, so you get constructive interference everywhere at any point at infinity, or just a plane wave at infinity. It's only at points with finite distance from the slit where you get fringes, because then you can have angles with respect to the normal of the slit.

Anyways, is there really such a thing as plane wave? Don't you have to have an infinite current sheet to produce a plane wave (I can't remember, but I'm basing this off the electric field for an infinite charged plane, which doesn't depend on distance)? So for all practical purposes, fields go to zero at infinity.

edit:

actually, no - a plane wave at an aperture will go to zero at infinity. i think you have to be careful about boundary conditions at infinity, because you can always have charges just beyond infinity that can make fields whatever you want at infinity, and vanish anywhere finite from the slit since they are infinitely far away. anyways, i was probably careless there or something.

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I just realized that if instead, there were a linear superposition of plane waves (all going in the same direction, but with different wavelengths), then the electric and magnetic fields could be made to go to zero at infinity, but be non-zero near the aperture.
Or you could say that the wave is very slowly attenuated (i.e. traveling through air).
But anyway, in a real experiment, as long as the EM field is a plane wave (going in one direction) in the region local to the aperture, then surely it doesn't matter how I made that wave initially, or what happens to the wave after it has traveled far from the aperture?

The wave is only a plane wave before it goes through the aperature. Once it goes through, it can't be a plane wave anymore because a plane wave by definition is a wave with a constant-valued wavefront across an infinite flat plane. So the question is, what kind of wave does is become when it goes through the aperature? Mathematically, this is answered by solving Maxwell's equations. Conceptually, this is answered by saying any wave (even a plane wave for that matter) is a sum of small spherical waves interfereing with each other. That is why the Huygens-Fresnel principle still applies even if you start with a plane wave.

RedX said:
Anyways, is there really such a thing as plane wave? Don't you have to have an infinite current sheet to produce a plane wave (I can't remember, but I'm basing this off the electric field for an infinite charged plane, which doesn't depend on distance)?

No, all you need is an infinitely strong dipole infinitely far away.

Well, a perfect, infinite plane wave is non-physical. But that does not mean that the concept is useless. If the relative sizes are right, a plane wave can be a very good approximation. For instance, a radio tower broadcasts its signal. Miles away, your car antenna picks up the waves. To your car antenna, the waves look more or less like plane waves, because the size of the car antenna is so small compared to the size of the signal's wavefront after traveling many miles. Or more to this problem, if you shine a laser beam through a hole, and the hole is much smaller than the beam spot size, then the waves at the hole are essentially plane waves.

I think I get the answer to this problem now. The electric and magnetic fields must be continuous, and since they are zero in the slab, they must go to zero smoothly at the edges of the aperture. This requires making the wave equation solution to be at least partly non-planar. The ratio of 'planar to not-planar' is related to the aperture size divided by the wavelength (which is why diffraction only happens when the wavelength is of at least similar size as the aperture).
So I guess my initial misunderstanding was because I thought the wave equation could be solved independently for each point in space, but actually this isn't true.

BruceW said:
I think I get the answer to this problem now. The electric and magnetic fields must be continuous, and since they are zero in the slab, they must go to zero smoothly at the edges of the aperture. This requires making the wave equation solution to be at least partly non-planar. The ratio of 'planar to not-planar' is related to the aperture size divided by the wavelength (which is why diffraction only happens when the wavelength is of at least similar size as the aperture).
So I guess my initial misunderstanding was because I thought the wave equation could be solved independently for each point in space, but actually this isn't true.

Correct, you need to remember that what you are solving, as you posed it in your OP, is a set of differential equations. The material properties of your problem impose boundary conditions on your differential equations which will effect the waves in all space. But it may be more intuitive to look at the problem in terms of waves alone.

One other thing to point out is that a spherical wave can be decomposed into a superposition of plane waves and vice-versa. So if you can demonstrate a wave phenomenon using spherical waves, then you know that it should also hold for plane waves.

raktimroy said:
gg.mvfv

means what?

## 1. What is classical light diffraction round corner?

Classical light diffraction round corner is a phenomenon in which light waves bend or spread out as they pass through a narrow opening or around an obstacle. This occurs due to the interaction of light with the edges of the opening or object, causing interference patterns to form.

## 2. How does classical light diffraction round corner work?

When light waves encounter a corner or narrow opening, they interact with the edges, causing them to diffract or bend. This bending of light waves creates interference patterns, which can be observed as bright and dark fringes. The exact pattern formed depends on the size and shape of the opening or object, as well as the wavelength of the light.

## 3. What is the significance of classical light diffraction round corner?

Classical light diffraction round corner is a fundamental property of light and plays a crucial role in many scientific and technological applications. It allows us to understand the behavior of light, and its interaction with matter, which is essential in fields such as optics, astronomy, and medical imaging.

## 4. How is classical light diffraction round corner different from other types of diffraction?

Classical light diffraction round corner differs from other types of diffraction, such as single-slit or double-slit diffraction, in that it occurs when light waves encounter a sharp edge or corner. In contrast, single-slit and double-slit diffraction occur when light waves pass through a narrow opening or slit.

## 5. How is classical light diffraction round corner related to the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle states that it is impossible to know the exact position and momentum of a particle simultaneously. This principle also applies to light waves, where the position and momentum of a photon cannot be known with certainty. Classical light diffraction round corner is one of the phenomena that illustrates this principle, as the exact path of a light wave cannot be predicted due to its wave-like behavior.

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