How Does Light Interact with a Slit in Single Slit Diffraction?

[cragar]

When light goes through a narrow single slit it diffracts. And you can explain this with the uncertainty principle or the wave-like nature of light. But if a photon can only be in one medium at a time then how does it know its going through a slit , How is it interacting with the edges of the slit . It may sound weird how i put it .

 

[VanOosten]

you need to understand quantum effects to answer this. the short answer is that the light both hits the walls and the slit at the same time. there is a great youtube video explaining it here:

 

[Born2bwire]

I never really liked it when they use the uncertainty principle in such a manner. I do not feel that it is a proper treatment though one can easily go through and see why it works due to the Fourier relationships between momentum and position (in wave physics the far-field pattern can generally be described by the Fourier transform of the source or aperture). But then you never see it applied to any other kind of analysis (how do you use this to predict the diffraction due to a half-plane for example).

Sticking with the wave description is much better. The slit represents a potential barrier that would be represented in the potential term in the Schroedinger Equation (metal is an infinite potential or equivalently the wavefunction is zero along and inside the boundary.). Assuming a time-independent problem, we can quickly reformulate the Schroedinger Equation so that it is mathematically the same as a wave equation.

\left[ \nabla^2 - \frac{2m}{\hbar^2} \left( V(\mathbf{r}) - E \right) \right] \psi(\mathbf{r}) = 0

This is like a scalar wave equation in a source free inhomogeneous media. So we already know that a wave will diffract due to a slit (which in wave physics the boundary can be modeled as a Dirichlet boundary condition which is similar to how the metal around the slit is treated in wave mechanics) and thus the solutions to the Schroedinger Equation follow suit.

In other words, the problem becomes

\psi(\mathbf{r}) = 0 where there is metal and elsewhere
\left[ \nabla^2 + \frac{2mE}{\hbar^2} \right] \psi(\mathbf{r}) = 0

 

[cragar]

Thanks for your answers

 

[cragar]

Born 2 b wire , what mass are you referring to in your wave equation?

 

[Born2bwire]

Ah heck. Well the Schroedinger Equation is a non-relativistic equation so it isn't actually valid for photons. So in this context the mass would be the mass of the particle, like an electron. Electrons will display the exact same single and double slit diffraction that photons will and they have even experimentally shown this to be true of much more massive particles like buckyballs.

Unfortunately to show that this is also the case for photons we would have to look at quantum field theory which is a proper relativistic treatment of quantum mechanics. Fortunately it isn't too bad of a problem in this case since we could use a path integral to find the wavefunction.

Drat, I seemed to have returned my copy of Path Integrals by Feynman and Hibbs. If you can lay your hands on a copy of that text (which isn't too common as they only ran one printing I think, pity) then I seem to recall that Feynman does a treatment of either the single or double slit problem using a path integral. The QFT texts I have at hand are sufficiently high level enough that I can't remember how to show the path integral formulation of this problem in a simple way like Feynman can.

EDIT: OH HAPPY DAYS! I just checked and it looks like they put out a softcover reprint of the text with many corrections. Now I can order a $12 book from Amazon and pay twice that to have it shipped to me in Hong Kong. Maybe I can throw in a Kindle... Hmmm... wait time: 3-5 weeks... You win this round Amazon...

 

[cragar]

Ironic i tried to get that book at my schools library about 2 weeks ago and it was checked out.

 

[Born2bwire]

Well, here you go. It is a good textbook but it certainly does not go into great depth that a more modern text like Peskin or Weinberg would.

https://www.amazon.com/gp/product/0486477223/?tag=pfamazon01-20

In 18-32 days I should get the book and hopefully be able to answer more of your question.

 

[sophiecentaur]

you need to understand quantum effects to answer this. the short answer is that the light both hits the walls and the slit at the same time. there is a great youtube video explaining it here:

A very pretty movie.
I think it makes things too simple, however, and takes far too many short cuts on the way to a very sophisticated ending message. Also, can we really forgive him when he he says that the colours reflecting off an oil film are "rainbow" colours - which they are definitely not? (Any amateur artist would tell you that.) They are the result of subtracting certain wavelengths rather than selecting certain wavelengths. This is what so often happens when people want the simple explanation without the rigour of classical approaches beforehand.
Let's face it, Huygens (from a million years before) introduced the idea of phase and wavelength to explain all those introductory examples. His approach gives all the right answers to simple or even very complicated situations.
Obviously, QED rules but why muck about with coloured rotating arrows and talk of adding them together without coming clean and talking of vectors? It's almost like saying we'll do this bit of EM theory without addressing Calculus at all.

 

[Born2bwire]

A very pretty movie.
I think it makes things too simple, however, and takes far too many short cuts on the way to a very sophisticated ending message. Also, can we really forgive him when he he says that the colours reflecting off an oil film are "rainbow" colours - which they are definitely not? (Any amateur artist would tell you that.) They are the result of subtracting certain wavelengths rather than selecting certain wavelengths. This is what so often happens when people want the simple explanation without the rigour of classical approaches beforehand.
Let's face it, Huygens (from a million years before) introduced the idea of phase and wavelength to explain all those introductory examples. His approach gives all the right answers to simple or even very complicated situations.
Obviously, QED rules but why muck about with coloured rotating arrows and talk of adding them together without coming clean and talking of vectors? It's almost like saying we'll do this bit of EM theory without addressing Calculus at all.

Using a stopwatch as a representation of phase is something that Feynman started in his QED book. Personally I agree with you. It is a nice visual that a laymen can understand but I would think that it causes more confusion since it really doesn't convey any reason for the behavior.

EDIT: The more I watch this the more I find that it really is following Feynman's QED book.

 

[sophiecentaur]

A layman may be able to see a stopwatch / arrow but I think the word "understand" may be a bit hopeful, actually. The rules of how these stopwatches add up are very vague. As someone who started life on 'the parallelogram of forces', vectors were always 'my friends' rather than something to struggle with.

Why does a model involving Particles behaving according to wavelike rules have to be better than a model involving Waves, which interact with each other in Quanta of Energy?