Why is it that diffraction is greatest when slit is close to size of wavelength?

[gladius999]

i think its to do with huygens sources?

 

[cybersysop]

i think its to do with huygens sources?

The waves bounce off of the sides of the slit and interact with each other. The smaller the slit the more they bounce or diffract

 

[mikeph]

We were always taught using the Heisenberg uncertainty principle applied to the photon in a direction perpendicular to the propagation vector:

If the photon passes through the hole, the wavefunction collapses into a position eigenstate which has a very undetermined momentum along this particular axis. Therefore some come out the other side with considerable transverse momentum. The smaller the gap is, the greater the uncertainty in momentum and the wider the diffraction can get.

I am not fully satisfied with this answer though and would love to hear one using a theory of classical waves rather than QM.

 

[Edgardo]

When a plane wave is incident on the slit Huygen's principle says that new waves are created in the slit. You can now calculate the interference pattern behind the slit by adding up all the little waves that are created in the slit.

This is done by writing down an integral (see [tex]\psi_{\mathrm{rad}}(\theta,\phi,r)[/tex] here).

You can approximate the integral and end up with a Fourier transformation (see here).

For a more thorough discussion read Eugene Hecht's optics book. Also look up Fraunhofer diffraction.

 

[cybersysop]

When a plane wave is incident on the slit Huygen's principle says that new waves are created in the slit. You can now calculate the interference pattern behind the slit by adding up all the little waves that are created in the slit.

This is done by writing down an integral (see [tex]\psi_{\mathrm{rad}}(\theta,\phi,r)[/tex] here).

You can approximate the integral and end up with a Fourier transformation (see here).

For a more thorough discussion read Eugene Hecht's optics book. Also look up Fraunhofer diffraction.

Created or transformed? I thought they were transformed. Is that correct? I like the use of Photons in one of the examples above. Photons are cool. Is this the same question as the one whereas the mere act of watching can change the outcome of the diffraction of the wave patterns and bounce?

 

[dulrich]

Diffraction is a wave phenomena and applies to quantum mechanics, light, sound, water waves -- any oscillations in a medium. The formula for diffraction is

[tex]\sin \theta = \lambda / W[/tex]

But this is really only an approximation valid for relatively small angles. You can see this because there is no physical reason why the wavelength cannot exceed the width of the slit, but the left hand side of the equation cannot be greater than one.

But usually this equation is used when dealing with light. The wavelength of light is so small that its diffractive (wave) nature is only evident when the slits involved are in the micrometer range or less.

 

[cybersysop]

I thought diffraction was when a wave encounters and obstacle such as a slit and thus diffraction of the wave destructing because of that obstacle. Diffraction grating regarding light (photons) is difficult for me to get my brain around, because to me light is not really a wave in the true sense. I am not stating this is correct but when light can be one particle or can travel were there are none, it is hard for me to view that as a wave.

 

[dulrich]

I think we agree on diffraction for waves in general.

But photons are quantum things -- they have both wave and particle properties. Diffraction is one of those classic wave properities.

 

[cybersysop]

Thanks for the equation dulrich, your are correct. I have a thought regarding this part of your previous post; "You can see this because there is no physical reason why the wavelength cannot exceed the width of the slit" I think or theorize, the reason that the wavelength does not ever exceed the slit is because the slit absorbs the spreading energy of the wave itself leaving only the momentum of the wave moving forward through the slit so it no longer has the ability to grow beyond the slit dimensions. I do not know who to formulate this but some sort of contained momentum caused by the slit is clearly evident, don't you think? Do you or anyone know how to put this into a formula?

 

[dulrich]

Actually, my point was kinda the opposite. The wavelength can exceed the width. There's really no reason it couldn't -- they are completely independent, right? But the question of what actually happens when the wavelength does exceed the width, I am slightly unclear on. I imagine a pool of water with a water wave and a very small slit. I think the wave on the other end would look just like it was coming from a point source (which is kind of like maximum diffraction: angle equal to 90 degrees). The majority of the original wave is simply reflected back off of the barrier.

 

[cybersysop]

I understand the water wave expanding after a slit and it was your point. You did point that out very well. I was thinking more about light in the form of a beam which is technically a wave in this particular case (pardon the pun) of which made it through the slit nano seconds before diffraction and after with a single slit considering 90 degrees. Light must have momentum with particles and or waves. I don't think light comes through the single slit in the same manner as water. I should have elaborated more, my bad! look at; http://www.walter-fendt.de/ph14e/singleslit.htm

 

[dulrich]

Oh. I see your question more clearly now. Initially all the momentum is to the right (say). After diffraction some of the photons acquire momentum upward. How? What force is pushing the photon up?

Well, remember the other side: some of the photons acquire momentum down, too. In fact the amount of momentum going up is the same as going down. If we look at the total momentum, these different directions cancel leaving only the original right-ward momentum.

Remember that momentum is conserved (in the absence of external forces), but it also has direction. Think of a pool table. The cue ball only has momentum in one direction but after the break there is momentum in all directions. Nonetheless, the total momentum is conserved.

 

[cybersysop]

Thank you dulrich. That answers my question well. Is there some way to quantify the conserved momentum with a formula? I don't have the skill to do that, but would really like to see one.
Any help would be greatly appreciated.

 

[dulrich]

I'm not sure this really answers your real question, but the conservation of momentum looks like:

[tex] \sum \vec{p}(0) = \sum \vec{p}(t) [/tex]

In this case, all the momentum is in the [itex]x[/itex]-direction, so [itex]\sum p_y = 0[/itex] both before and after. Since the [itex]y[/itex] component is zero, any contribution in the positive direction (i.e., up) must be compensated by a contribution in the negative direction (i.e., down).

 

[dulrich]

As I think about it some more, maybe your question is for a formula for how much momentum goes up (and therefore how much goes down as well). This is a much more difficult question. For example, on the pool table, the amount of new momentum (I'm not sure if there is a technical name for this) generated in the across the table depends on the internal forces and configuration of the balls. This is just 15 balls and very complicated. If we talk about light, I'm not sure what to think. I haven't really thought about it in these terms before. The analysis may be complicated but should result in the classical diffraction formula I mentioned above.

Sorry I can't help on that front.

 

[cybersysop]

No, you did very well and actually answered it. It is a time consuming effort to be sure but the formula is very good. Thank you very much! OK, so that's the summation. It almost looks like a network in model. The slit is a very cool deal in physics with some real world application. With photons, on the other (far) side is the opposite as the (near) side. It almost seems like an equal and opposite universe. one side is 45* up and on the other side it is 45* down with momentum conserved in the absence of an external force. If there was a fracture or slit in the conserved momentum eight ball, we now can figure out the result given the data? Way cool, thanks! As a 50+ year old high school drop out, even I can understand that. dulrich R U a Professor?

 

[cybersysop]

I'm not sure this really answers your real question, but the conservation of momentum looks like:

[tex] \sum \vec{p}(0) = \sum \vec{p}(t) [/tex]

In this case, all the momentum is in the [itex]x[/itex]-direction, so [itex]\sum p_y = 0[/itex] both before and after. Since the [itex]y[/itex] component is zero, any contribution in the positive direction (i.e., up) must be compensated by a contribution in the negative direction (i.e., down).

It is nice to see a little calculi here in the formula as well. Is the slit an axiom, logically yes.
That would certainly explain allot! I see now!