# Cause of Diffraction of Light

1. Sep 25, 2010

I know the concept of diffraction of light, but not yet understood why light should bend passing the edges of objects ! I mean what makes it bend?

2. Sep 25, 2010

### NobodySpecial

It doesn't bend as such - the light is spreading out in all directions, diffraction just causes some of the light that was around the corner to be added constructively while some of the light that was going straight-on is destructed.

3. Sep 25, 2010

### Idoubt

To understand this you have to think of light as a wave and not a ray. For example suppose you have a tub of water and you place some sort of wooden surface vertically in it with a slit.

If you disturb the water you will see waves propagating in all directions on the surface.

When the waves hit the wood it will pass through the slit, but it wont emerge through the slit on the other side as a straight line, it will still spread in concentric circles ( well semi-circles ) on the other side of the surface with the slit as the centre.

I believe some experiments like this can be seen on youtube have a look it might help

4. Sep 25, 2010

### granpa

each part of the wave front acts as a wave generator.
http://en.wikipedia.org/wiki/Huygens–Fresnel_principle

edit:Immediately after posting this I did a google seach and this post was the first result.
24 seconds after posting this it was already in google search results.

Last edited: Sep 25, 2010
5. Sep 30, 2010

The water waves are made up of material where as light waves are electromagnetic waves. So that cant be example of diffraction.

6. Sep 30, 2010

### Andy Resnick

Why do you think (surface) water waves don't diffract? what about sound waves?

7. Sep 30, 2010

### Idoubt

Other than the fact that its electric and magnetic vectors oscillating for light and particles for waves there is no fundamental difference.

The main point is that all waves show this property. And when light is viewed as an electric and magnetic waves perpendicular to each other explains diffraction.

The interference pattern which is shown in young's double slit experiment comes from the path difference of two rays of light, ie for a dark spot the electric vector of one ray is in the opposite direction as the electric vector of the other.

Even for sound waves this happens. If you have two speakers there are acoustic "sweet spots" where the sound is max and also silent points

8. Oct 1, 2010

I have never said that water waves and sound waves don't diffract.
But their properties are different and the properties of light are very much different.

water waves diffract because of property of fluid that when they touch the edge of an object there is a phenomenon of adhesion with the edge of object and hence they bent towards the direction of object.

Regarding sound waves, the waves travel by striking the medium. And so when there is an object in their path they strike to that object also and the sound is scattered through every direction.

This cant be similar to light waves. Because light waves neither have adhesion properties nor they strike to the medium in their way.
There is only waves and photons traveling, when we say light is propagating.

In the Hyugen's double slit model, its only explained how the diffraction occurs and not WHY the waves bent while passing through an edge of an object.

There may be lack in my knowledge of laws but please try to understand my point.

9. Oct 1, 2010

### Andy Resnick

I think you are confusing mechanisms with properties.

Surface water waves are indeed caused by interfacial energy- that's why spreading oil on water abolishes surface waves. And sound waves are indeed caused by variations in the density of a compressible medium.

But diffraction is a property of waves- not the constituent particles. Diffraction is caused by the spatial truncation of a wavefront.

10. Oct 1, 2010

That's not my query.
Please could anyone tell me that why should light bend towards a specific direction when it passes through an edge of an object?

11. Oct 2, 2010

### Idoubt

Light doesn't bend around an object, it spreads. A slit is simply 2 edges after all.

I don's understand why you are insisting that the wave nature of light and that of waves propagating through mediums have to different. The means of propagation is different but that doesn't change the nature of the wave.

12. Oct 2, 2010

### cragar

You could explain diffraction in terms of the uncertainty principle .
If we try to force light through a small opening, we know to well where its at so we get an uncertainty in its momentum and it starts to spread out .

13. Oct 2, 2010

### Born2bwire

I do not think that works since diffraction is an established macroscopic classical property. The properties that give rise to the Heisenberg Uncertainty Principle are not present in the physics or mathematics of classical wave physics. However, classical wave physics are able to properly describe diffraction. Not to mention the fact that we do not need a hole for diffraction, a wedge is perfectly suitable.

One way to look at it classically is from the induced source perspective. The electromagnetic wave, when it impinges upon the scatterer, induces currents on the surface and inside the scatterer. If we are using plane waves as our incident field and if the scatterer was an infinite perfectly conducting plane, then the induced currents would be line currents running parallel to each other. Now if we were to suddenly truncate this plane, that would mean that at the corner edge, we have a single line current running along the edge. This current element will send out a cylindrical wave in all directions since it is not reflected (mixing a bit here, technically if we think about it as a source picture then we ignore the presence of the scatterer and just place the currents in the homogeneous background) back by the presence of the scattering plate. Now, once we take into account all of the currents together, we find that far from the edge in the middle of the plate the currents make a reflecting plane wave and on the other side a plane wave that cancels out the incident wave. This is done by the phase differences in the excited currents that work together to cancel out the radiation on the other side. But at the edge, we no longer have currents on both sides of our edge current. Thus, the edge current's contribution on the other side of the plate is not cancelled out and we see this as the diffracted field.

So a very rough picture of the problem is that the induced currents on the plate cause specular reflection of the plane wave. But at the edge, since we do not have a continuum of currents to cancel out the radiation from the edge current on the shadow side of the plate, the edge current puts out a cylindrical wave that is only partially cancelled. The cylindrical part radiates out as the diffracted contribution.

14. Oct 2, 2010

### Staff: Mentor

In fact, they are. From Fourier analysis, a wave packet (whether classical or quantum-mechanical) must obey the relationship

$$\Delta x \Delta k \ge \frac{1}{2}$$

In QM this becomes the HUP via $p = \hbar k$.

15. Oct 2, 2010

### Born2bwire

But this creates a problem because we experience diffraction with monochromatic plane waves. When we talk about the diffraction patterns that arise in quantum mechanics via the Heisenberg Uncertainty Principle, the examples generally do so under the auspices of a particle like the electron or photon. In this case we relate the position and momentum of the particle and we describe the resulting diffraction in terms of its wavefunction. But how do we relate this to a problem that isn't localized in space and remains monochromatic?

If we are talking about the Fourier analysis, then we are relating the spatial domain with that of the wavenumber. In this case the localization of the wave results in a wave packet that must be produced by a bandwidth of frequencies. That isn't what happens with the a slit source because the transmitted wave is still monochromatic. This is due to the fact that we can use a plane wave source (or other similar infinite source) which allows us to have a Fourier transform that is a delta function regardless of how we constrain it in space via scatterers.

16. Oct 2, 2010

### Andy Resnick

Be careful- we do not observe diffraction from a monochromatic *plane* wave- a plane wave extends to +/- infinity in the transverse direction. We *do* observe diffraction from a monochromatic wave that consists of a plane wave truncated by an aperture.

17. Oct 2, 2010

### Born2bwire

You do not need an aperture to diffract the wave. You can have a finite plate or wedge and still achieve diffraction. Making it an infinite object of course makes it much easier to calculate a solution or approximate one (ie. UTD, Sommerfeld half-plane solution, etc.) but the diffraction is a localized effect. We can easily create a finite scatterer and observe the typical diffraction behavior and doing so is a common problem in modeling high frequency scattering.

EDIT: Or perhaps you are misunderstanding my meaning. For example, Idoubt posted a picture of what I would consider is the diffraction of a monochromatic plane wave due to an infinite half-plane perfect electrical conductor (a problem that Sommerfeld solved almost a century ago and is of particular interest to us since a half-plane represents an interesting problem for a finite plate diffraction approximation).

Last edited: Oct 2, 2010
18. Oct 2, 2010

### Idoubt

These two lectures should give you a better idea about diffraction

The first one shows the classical approach and the second one the quantum mechanical

MIT OCW lectures are the best I've ever seen btw, it will be definitely worth your while to see as many of them as you can.

Last edited by a moderator: Sep 25, 2014
19. Oct 2, 2010

### Andy Resnick

we could be misunderstanding each other. A plane wave (that is, a solution e^ikz or e^ikr) formally has infinite extent and so does not diffract. We agree that when this is not the case: aperture, scatterer, 1/2 plane screen, etc, diffraction occurs. Bessel beams are another example of a non-diffracting field, but one that has finite spatial extent.

20. Oct 2, 2010

### Born2bwire

Ok then, yeah we are just misunderstanding.