Huygens' theory, point on a wavefront, secondary wavelets

[Rainbow]

Huygens' theory says that every point on a wavefront serves as a source of secondary wavelet. Doesn't that imply that if we consider any coherent source of light, and intercept its waves on a screen, we'll get a diffraction pattern, as all those secondary wavelets will interfere among themselves?

 

[ice109]

Huygens' theory says that every point on a wavefront serves as a source of secondary wavelet. Doesn't that imply that if we consider any coherent source of light, and intercept its waves on a screen, we'll get a diffraction pattern, as all those secondary wavelets will interfere among themselves?

every single point on the wavefront, as in there is an infinite amount of them on the wavefront

 

[Claude Bile]

Of course Huygens' theory implies interference and diffraction, it wouldn't be much of a theory otherwise.

Claude.

 

[ice109]

Of course Huygens' theory implies interference and diffraction, it wouldn't be much of a theory otherwise.

Claude.

yes but does coherent light interfere when ther is no slit?

 

[Rainbow]

yes but does coherent light interfere when ther is no slit?

That's what I have asked.

 

[JeffKoch]

Huygens' theory says that every point on a wavefront serves as a source of secondary wavelet. Doesn't that imply that if we consider any coherent source of light, and intercept its waves on a screen, we'll get a diffraction pattern, as all those secondary wavelets will interfere among themselves?

They'll all interfere constructively and destructively, and the net effect exactly cancels as long as you account for ALL the points on the wavefront. If you only take part of the wavefront (like, if the light passes through a slit before reaching your screen), then the wavelets won't cancel and you'll see fringes.

 

[Claude Bile]

yes but does coherent light interfere when ther is no slit?

There is no distinction to be made between diffraction and interference. They are both one and the same. Any wavefront will diffract regardless of the presence of a slit, or any other obstacle you might conceive. A plane wave is simply a special case because after summation of the secondary wavefronts, the wavefront does not change shape.

Claude.

 

[ice109]

There is no distinction to be made between diffraction and interference. They are both one and the same. Any wavefront will diffract regardless of the presence of a slit, or any other obstacle you might conceive. A plane wave is simply a special case because after summation of the secondary wavefronts, the wavefront does not change shape.

Claude.

so then when you shine a laser onto a screen you get an interference pattern?

 

[JeffKoch]

so then when you shine a laser onto a screen you get an interference pattern?

...because of the presence of edges in the resonanator cavity mirrors and output aperture, which limit the amount of wavefront available for "secondary wavelet" summation. The fringes are very faint.

 

[ice109]

...because of the presence of edges in the resonanator cavity mirrors and output aperture, which limit the amount of wavefront available for "secondary wavelet" summation. The fringes are very faint.

its a thought experiment, a beam of coherent light, does it interfere or not, with a source with no imperfections.

 

[cesiumfrog]

"A beam of coherent light" is not specific enough. Do you mean an infinite-width beam? A Gaussian beam? A higher mode beam? An ugly top-hat shaped beam? The ideal case is the Gaussian beam, which will produce a Gaussian spot, and will do so because of interference rather than in spite of it.

 

[ice109]

"A beam of coherent light" is not specific enough. Do you mean an infinite-width beam? A Gaussian beam? A higher mode beam? An ugly top-hat shaped beam? The ideal case is the Gaussian beam, which will produce a Gaussian spot, and will do so because of interference rather than in spite of it.

the kinds of beams of light the hygun's theory is concerned with. what i mean is if you take the same coherent light that produces an interference pattern when traveling through a slit or double list, and removed the slits would it interfere? i think not

 

[cesiumfrog]

Think about what difference the slits make to the beam: the only relavent difference is the shape of the beam profile.

So if the beam profile is gaussian, the interference will add up to cause the beam profile to remain gaussian. If the beam profile has sharp edges, the interference will add up to cause the beam profile to become the shape of a "sinc" function.

 

[Rainbow]

There is no distinction to be made between diffraction and interference. They are both one and the same. Any wavefront will diffract regardless of the presence of a slit, or any other obstacle you might conceive. A plane wave is simply a special case because after summation of the secondary wavefronts, the wavefront does not change shape.

Claude.

You mean all the secondary wavelets will interfere, but since there are an infinite number of these wavelets and they all interfere, the destructive interference at some will be overlapped with constructive interference of some other wavelets at that point ultimately giving a uniform intensity over a screen on which the light is intercepted. Right?

 

[JeffKoch]

its a thought experiment, a beam of coherent light, does it interfere or not, with a source with no imperfections.

By "interfere", do you mean "produce light and dark fringes"? Ringing in the sense of fringes is caused by sharp edges, but light propagation (if it's done correctly) must include the possibility of multiple paths that add as waves - so in that sense it does always interfere, it just may not produce characteristic "diffraction" fringes.

 

[billiards]

Isn't ray theory strictly only valid at infinite frequency? Perhaps this isn't a problem in optics, but in global seismology (where frequencies are typically <10 Hz) a recently new finite-frequency (Dahlen, 2000) approach (used for tomography) has been necessary to account for such phenomenon as wavefront healing.

 

[Claude Bile]

You mean all the secondary wavelets will interfere, but since there are an infinite number of these wavelets and they all interfere, the destructive interference at some will be overlapped with constructive interference of some other wavelets at that point ultimately giving a uniform intensity over a screen on which the light is intercepted. Right?

No, the fact there are an infinite number of wavelets is not special, any wavefront can be regarded as an infinite sum of wavelets (hence the need to use integrals), the wavefront itself needs to be spatially infinite in order for no change in the wavefront shape to occur.

Ice109; There is no distinction, mathematically or conceptually to be made between diffraction and interference. All beams (with a couple of artificial exceptions such as the aforementioned infinite plane wave) will diffract (remembering diffraction = interference), with the diffraction pattern being dependent on the original beam shape.

Claude.