# Hyugens principle and laser wave dispersion

1. Mar 22, 2008

### Ulysees

According to wikipedia below:

> Huygen's principle states that each point of an advancing wave front is in fact the center of a fresh disturbance and the source of a new train of waves; and that the advancing wave as a whole may be regarded as the sum of all the secondary waves arising from points in the medium already traversed.

http://en.wikipedia.org/wiki/Photon_dynamics_in_the_double-slit_experiment [Broken]

Then how is it possible that laser light moves in a narrow beam with very small dispersion sideways?

Can radiowaves be produced to be like that? Cause that would make an extremely high directivity antenna for satellite communications etc.

Last edited by a moderator: May 3, 2017
2. Mar 24, 2008

### Andy Resnick

Laser beams disperse just like any other beam. Perhaps you are wondering how any well-collimated beam can exist, given Huygen's principle?

3. Mar 24, 2008

### Ulysees

Sure. It was mentioned:

Exactly. In fact I'd like to simulate it with sound waves if possible. Maybe with a large number of speakers driven differently or something.

4. Mar 24, 2008

### Staff: Mentor

A laser beam that is (say) 1 mm wide is still about ten thousand times wider than a single wavelegth which is on the order of $10^{-7}$ m.

5. Mar 24, 2008

### Ulysees

So what happens at the edge then? If we illuminate a camera chip directly with a laser beam, what does the edge look like?

6. Mar 24, 2008

### Staff: Mentor

7. Mar 24, 2008

### Ulysees

Alright. It's hard to imagine why with all the smoothness at the edge it still maintains the direction.

Does the Hyugens principle apply to sound waves too? Can the same directionality be achieved with sound waves?

8. Mar 25, 2008

### Andy Resnick

FWIW, you are in good company, some of the finest scientific minds in history have written explanations of this. I have been trying to think of a simple way to explain this- here goes:

First, consider an abstract representation of the source- a hole in an opaque screen, located at z = 0. You have no idea what happens at z < 0, all you know is that at z = 0, the intensity is 1 in the hole and 0 everywhere else. If that's all there was, then indeed, you would expect the light to wildly diverge at z>0. But it doesn't- why?

Because light is a *vector* field rather than a scalar field. The field in the hole is not simply 1- it's got a direction associated with it. Intuitively, you can see that if the vectors all point along z, then the light should continue to go along z. If the vectors point 'out', then the divergence will be large (or conversely, can converge). If you like, it's possible to work in funny coordinates that take this into account- look up the "Rayeligh length" for a Gaussian beam.

The edge is a problem- the field is discontinuous, which implies that something diverged. Since infinities do not actually exist, what this means is that the field 'spreads out' in order to maintain a smooth function.

In practical terms, to make a perfectly collimated beam, the beam diameter must be infinitely large. Gaussian beams (beams from laser cavities) can be made well-collimated by passing them through a spatial filter (to make a pinhole source), and then to a lens. *But*, the best performance is when the pinhole clips the Gaussian near the zeros, not arbitrarily in the middle. This is to minimize the 'spreading' effect at the edge (which I now call diffraction).

Does this help?

9. Mar 25, 2008

### Ulysees

Is a Gaussian beam one with a Gaussian density function along a line perpendicular to its direction?

An example of a spatial filter that makes a pinhole source? Is a small hole through an opaque sheet an example? Is yes, wouldn't passing the beam through the hole make it diverge a lot?

That makes sense. If a beam diverges 1 mm per metre, then a lens with a focal distance of 1 metre should make a 1 mm-wide beam perfectly parallel?

I don't understand this bit. Do you mean that illuminating a pinhole with the very edge of the beam, would be ideal?

Last edited: Mar 25, 2008
10. Mar 26, 2008

### Andy Resnick

Ulysees,

Let's go through point-by-point.

1)Is a Gaussian beam one with a Gaussian density function along a line perpendicular to its direction?

Yes, the beam within an optical cavity behaves as Hermite-Gaussian (or Laguerre-Gaussian) functions for the transverse field density. A TEM00 beam has a Gaussian (~exp^-r^2) dependence.

2)An example of a spatial filter that makes a pinhole source? Is a small hole through an opaque sheet an example? Is yes, wouldn't passing the beam through the hole make it diverge a lot?

A proper spatial filter is a microscope objective focused to a pinhole. And yes, the beam can diverge a lot: the numerical aperture of the microscope objective sets the angle.

3) That makes sense. If a beam diverges 1 mm per metre, then a lens with a focal distance of 1 metre should make a 1 mm-wide beam perfectly parallel?

Well, usually the NA is closer to 0.5, so the divergence is much higher. But the raw beam coming out of a laser has a rough divergence of 1 mm/meter (whatever that is in radians). The 'collimated' beam will not be perfectly parallel in either case, because the source (pinhole or exit pupil) has a size. Using smaller pinholes will in the end give a better collimation.

4) I don't understand this bit. Do you mean that illuminating a pinhole with the very edge of the beam, would be ideal?

No, the pinhole needs to be correctly sized. It's always centered on the beam axis. But the size of the pinhole should be matched to the size of the focused beam- specifically, the size should be equal to the size of the central peak of the Airy disk.

11. Mar 26, 2008

### Ulysees

If we imagine a long, thin and straight tube made of light-absorbent material with a light source at one end, will the light coming out the other end be as collimated as the ratio diametre/length suggests?

I think not, but do you know the maths for it, any analytical solution how divergent the output beam would be?

If this works, it could be done with sound too, and make a sound laser.

Last edited: Mar 26, 2008
12. Mar 26, 2008

### Ulysees

And by the way, no need to say laser is monochromatic or coherent, sound can have these properties easily.

13. Mar 27, 2008

### Andy Resnick

Definitely not. The tube is not relevant- the only thing that matters for the propagation of light once it leaves the tube are the field values at the end face of the tube.

"light pencils" cannot exist. That is a fundamental conceptual error unfortunately aided by geometrical optics. A related conceptual error is the concept of "combustion from afar" (Archimedes legend). The two errors are:

1) parallel rays to not exist
2) even if they did, they could not carry energy.

If you can find it, an excellent read is "On the possible and impossible in optics", by G. G. Slyusarev (1962). It's a DTIC document, FTD-TT-62-175 (accession number 886286, I think).

I appreciate the idea of manipulating coherent sound- there's been some interesting work (odd overlap between military and advertising interests) with directional broadcast techniques:

http://www.temple.edu/ispr/examples/ex02_06_23.html [Broken]

Last edited by a moderator: May 3, 2017
14. Mar 27, 2008

### Ulysees

Which is as was expected, but what are the equations here? I'd like to know the maths by which we derive the intensity profile below.

http://img262.imageshack.us/img262/4459/image1ls6.gif [Broken]

The flat part I can do from geometric optics: it has a width given by:

$$w = D(L+d)/L$$

How do we derive the part due to diffusion, above and below the flat part? I think it can only be done numerically with a method like TLM. But do you know an analytical solution? Maybe something based on Hyugens principle, so we get a chance to understand how the principle accounts for collimated waves?

Last edited by a moderator: May 3, 2017
15. Mar 27, 2008

### Claude Bile

Firstly, I think you mean diffraction (not diffusion).

It is possible to get a solution analytically by solving the scalar wave equation and using the paraxial approximation $d^2E/dz^2 = 0$. You do however need to make some assumptions about the solution (that it is beam-like for example). Note too that since the solution is a scalar function, it does not take into account vector effects such as polarisation.

Claude.

16. Mar 28, 2008

### Andy Resnick

I don't understand the picture- there's a point source within a long tube, the interior walls of which are covered by absorbent material?

Usually, when an incoherent source is used, people work in terms of the intensity instead of the field. In this case, sometimes people speak of the 'diffusion' of energy (intensity) rather than diffraction (field). They are similar concepts, but lead to very different and surprising results: for example, using incoherent light when imaging can provide twice the resolution over using coherent illumination.

17. Mar 28, 2008

### Ulysees

Exactly. The curved edges of the beam that are labeled "diffusion" are not rays of light, they are just the edges beyond which the intensity is below a threshold.

Claude, I definitely didn't mean diffraction, diffraction requires different materials, eg glass and air, or a varying diffraction index causing the rays to bend. "Diffusion" is not accurate either, because it is normally used for the scattering due to air molecules. I want to understand the spreading due to the wave nature of light, in a perfect vacuum. What is that spreading called and how is it calculated?

Thanks. But what is that paraxial approximation? Looks like assuming the E field varies linearly along the axis of the beam:

$E = az + b$

at x=0, y=0

What equation would you be solving analytically under this assumption? Maxwell's?

Last edited: Mar 28, 2008
18. Mar 30, 2008

### Claude Bile

I think you do mean diffraction (it sounds like you are confusing diffraction with refraction?).

With regard to the paraxial wave approximation (I made a slight error in my previous post by the way...);

First, if you assume a monochromatic wave, the wave equation (which can be derived from Maxwell's equations) reduces to the Helmholtz equation;

$$\nabla^2E+k^2E=0$$

Now define an envelope function, U such that;

$$E(r) = U(r)e^{ikz}$$

Substituting this into the Helmholtz equation yields;

$$\frac{d^2U}{dx^2}e^{ikz} + \frac{d^2U}{dy^2}e^{ikz} + \frac{d^2U}{dz^2}e^{ikz} + 2ik\frac{dU}{dz}e^{ikz}=0$$

The paraxial approximation is that the $d^2U/dz^2$ term reduces to 0 (not the $d^2E/dz^2$ term as I mistakenly said in my previous post). This approximation is, in essence, saying that since the solution is beam-like, the envelope is slowly varying with z.

This yields the Paraxial Wave Equation;

$$\frac{d^2U}{dx^2}e^{ikz} + \frac{d^2U}{dy^2}e^{ikz} + 2ik\frac{dU}{dz}e^{ikz}=0$$

The Gaussian Beam is a solution to this equation.

EDIT: The derivatives should be partial derivatives.

Claude.

19. Mar 30, 2008

### Ulysees

That's what I was looking for, thank you.

So if instead you do not assume the envelope U(r) varies linearly along z, you get something close to a Gaussian U(r) which can only be derived numerically.

It seems strange to me that the paraxial approximation is close to the real thing. Isn't the approximation saying that the envelope U(r) does NOT approach zero at z=infinity asymptotically, but linearly, like below?

U(r)=(az+b)G(r),

G(r)=Gaussian

Do actual light beams look more like an inverse square function of z as we approach infinity?

Last edited: Mar 30, 2008
20. Mar 31, 2008

### Claude Bile

Well, the paraxial approximation is essentially saying that a beam diverges at a fixed angle for large z, which is a fairly accurate approximation.

If you assume a solution of the form (noting that r = distance from the beam axis - $x^2 + y^2$ not the position vector);

$$U(x,y,z) = E_0e^{iQ(z)r^2}e^{iP(z)}$$

The Gaussian solution can be derived analytically. The actual solution is of the form;

$$U(x,y,z) = E_0\frac{w_0}{w(z)}e^{-itan(z/z_0)}e^{-r^2/w^2(z)}e^{ikr^2/2R(z)}$$

Where;

$E_0$ is the E-field amplitude.
$w_0$ is the spot radius at the beam waist.
$w(z)$ is the spot radius as a function of distance.
$z_0$ is the Rayleigh Range (or the Confocal parameter) of the beam.
$R(z)$ is the Radius of Curvature of the beam as a function of distance, where;

$$R(z) = \frac{1}{z}(z^2+z_0^2)$$

The 1st exponent is the Guoy (longitudinal) Phase term, the 2nd exponent is the transverse beam amplitude and the 3rd exponent is the transverse phase term. Also, note that there is some curvature in the beam profile, however it is very small, which is why the $d^2U/dz^2$ term is assumed to be negligible.

Claude.