Hyugens principle and laser wave dispersion

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

Sure. It was mentioned:
very small dispersion sideways

Perhaps you are wondering how any well-collimated beam can exist, given Huygen's principle?
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.
 
jtbell
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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 [itex]10^{-7}[/itex] m.
 
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So what happens at the edge then? If we illuminate a camera chip directly with a laser beam, what does the edge look like?
 
jtbell
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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?
 
Andy Resnick
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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?
 
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Gaussian beams (beams from laser cavities)
Is a Gaussian beam one with a Gaussian density function along a line perpendicular to its direction?

can be made well-collimated by passing them through a spatial filter (to make a pinhole source)
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?

and then to a lens
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?

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)
I don't understand this bit. Do you mean that illuminating a pinhole with the very edge of the beam, would be ideal?
 
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Andy Resnick
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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.
 
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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.
 
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If this works, it could be done with sound too, and make a sound laser.
And by the way, no need to say laser is monochromatic or coherent, sound can have these properties easily.
 
Andy Resnick
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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.
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]
 
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Definitely not. (...) "light pencils" cannot exist. That is a fundamental conceptual error unfortunately aided by geometrical optics.
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.

Image1.gif

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

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

[tex]w = D(L+d)/L[/tex]

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?
 
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Claude Bile
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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?
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 [itex]d^2E/dz^2 = 0[/itex]. 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.

A search on "Gaussian Beams" will yield plenty more information.

Claude.
 
Andy Resnick
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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.

<snip>

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?
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.
 
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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?
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?

It is possible to get a solution analytically by solving the scalar wave equation and using the paraxial approximation [itex]d^2E/dz^2 = 0[/itex]. You do however need to make some assumptions about the solution (that it is beam-like for example).
Thanks. But what is that paraxial approximation? Looks like assuming the E field varies linearly along the axis of the beam:

[itex]E = az + b[/itex]

at x=0, y=0

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

[tex]\nabla^2E+k^2E=0[/tex]

Now define an envelope function, U such that;

[tex]E(r) = U(r)e^{ikz}[/tex]

Substituting this into the Helmholtz equation yields;

[tex]\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[/tex]

The paraxial approximation is that the [itex]d^2U/dz^2[/itex] term reduces to 0 (not the [itex]d^2E/dz^2[/itex] 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;

[tex]\frac{d^2U}{dx^2}e^{ikz} + \frac{d^2U}{dy^2}e^{ikz} + 2ik\frac{dU}{dz}e^{ikz}=0[/tex]

The Gaussian Beam is a solution to this equation.

EDIT: The derivatives should be partial derivatives.

Claude.
 
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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?
 
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Claude Bile
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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 - [itex]x^2 + y^2[/itex] not the position vector);

[tex]U(x,y,z) = E_0e^{iQ(z)r^2}e^{iP(z)}[/tex]

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

[tex]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)}[/tex]

Where;

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

[tex] R(z) = \frac{1}{z}(z^2+z_0^2)[/tex]

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 [itex]d^2U/dz^2[/itex] term is assumed to be negligible.

Claude.
 
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Thank you.

Is there a theoretical maximum on the directivity of an e/m wave? Perhaps measured as the angle for a 3 dB fall in power relative to the centre of the beam, at 1 metre?
 
Claude Bile
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Not sure what you mean by directivity. Do you mean the angle of divergence?

For Gaussian Beams, the angle of divergence increases as the beam waist decreases. The beam waist radius cannot be smaller than [itex]\lambda/4[/itex] due to the diffraction limit, which will, in turn, limit the angle of divergence.

For EM waves in general though, there is no theoretical limit.

Claude.
 
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Not sure what you mean by directivity. Do you mean the angle of divergence?
Yes, but only at a given distance from the source (because, as you said too, the angle increases with distance - the edge is curvy).

And because the profile U(r) is typically similar to Gaussian (ie asymptotically extends to infinity sideways), we need a threshold for the intensity to define an edge. Eg 3 dB below the maximum value of U(r) as I said.

The beam waist radius cannot be smaller than [itex]\lambda/4[/itex] due to the diffraction limit
How do you define the beam waist? There must be a threshold implied since U(r) extends to infinity sideways, but what is the chosen threshold?

For EM waves in general though, there is no theoretical limit.
Don't the same rules apply as with light? Don't EM waves come from sources of finite size, just like light?
 
Claude Bile
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Yes, but only at a given distance from the source (because, as you said too, the angle increases with distance - the edge is curvy).
I see, however the divergence angle does approach a fixed value for large z - which is why the far-field divergence is typically used to characterise the spread of a beam.

And because the profile U(r) is typically similar to Gaussian (ie asymptotically extends to infinity sideways), we need a threshold for the intensity to define an edge. Eg 3 dB below the maximum value of U(r) as I said.
Yes, the 3 dB point is typically defined as the "edge" of a Gaussian beam. If you know the beam width at a certain distance from the beam waist, calculating the angle of divergence is trivial.

How do you define the beam waist? There must be a threshold implied since U(r) extends to infinity sideways, but what is the chosen threshold?
I'm confused by this, the "r" in U(r) is the radial distance from the beam axis. Hence the beam is confined in both x and y. The beam waist is defined as the point where the beam width is a minimum.

Don't the same rules apply as with light? Don't EM waves come from sources of finite size, just like light?
Not all light travels as a Gaussian Beam, isotropic sources (such as fluorescent sources for example) emit spherical waves that spread equally in all directions.

Claude.
 
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Thank you.

I think you forgot something I was asking.

I'd like to know if radiowave EM beams can be made as directional as a light beam.

Does light make more directional beams than microwaves?

Also, is laser the most directional beam we can send?
 

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