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Why is geometric optics the low wavelength limit of the wave theory of light? I can't seem to grasp why either a low or high wavelength limit would be necessary.

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marked as duplicate by Emilio Pisanty, Dan, Qmechanic Aug 11 '13 at 3:38

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Visualise it and it would be clear . –  Dimensio1n0 Aug 11 '13 at 2:29

1 Answer 1

Actually I would answer this almost exactly as @EmilioPisanty answered the a very like question -

http://physics.stackexchange.com/a/65241/26076

this is the way that Born and Wolf "Principles of Optics" do it.

However, I would add as a footnote the following intuitive points:

There is one, unique class of Maxwell Equation solution that fulfills the Eikonal equation exactly and these are the plane waves: their wavevectors are the "rays" in ray optics. These are exactly light travelling in straight lines. However, they are also altogether "delocalized" - spread out evenly over all space, so strictly you can't talk about a ray at a particular place. What one finds in practice though is that if waves are approximately plane so that their phase fronts are flat over several wavelengths or more, they do not tend to spread out much as they propagate. Lasers are the same: the wider your beam, the less it diverges, indeed the divergence angle $\times$ beamwidth product is constant. So you can represent the field coming from a region wherein the phase fronts are flat as a plane wave along the ray with it tail in that region. Another good example is the phased array and dish antennas. They set up a field that is locally plane over many wavelengths, thus achieving a very tight beam. The ray is the spatial analogy of the ideal bandlimited AND time-limited signal often talked about in systems analysis. Theoretically, if something is time-limited, it has infinite bandwidth, but one can do practical analysis by assuming the "ideal" (strictly impossible) case. This is what we do when we draw rays.

Indeed intuitive comments like this lead to the "Ansatz" that Emilio talks of in his other answer: you assume the field is a plane wave multiplied by an "envelope" function that varies slowly in space. The Eikonal equation then results from putting the assumed solution into MAxwell's equations and making approximations.

Does all this sound familiar - it should do: it's exactly how the Heisenberg uncertainty principle works: the momentum operator is just a scalar multiplication in a space that is the Fourier transformed space wherein the position operator amounts to a scalar multiplication. So, the momentum corresponds to wavevectors (the directions of rays) and positions to a place where we can nail the rays' tails down. As long as we don't demand too much precision in either direction or position, rays are a useful model of all waves in this way.

Footnote: I should also say thank you to Emilio Pisanty for bringing to my attention the analogy between the Hamilton Jacobi and Eikonal equation in the answer cited above. I have never really appreciated that until today. Now the standard quantisation procedure Poisson bracket $\mapsto$ Lie bracket suddenly really makes sense to me.

Edit: The following is a code that you can import into Wolfram Mathematica to do diffraction calculations. You can play with the beam width and so forth and simulate how well the beam will propagate without spreading for different beamwidths and wavelengths. Lines beginning in "#" are comments: Mathematica will read the following if you ask it to open a "text file".

#Index Rotation for Rotated Fourier Transform

[Kappa][k_,n_]:=If[k >= n/2,k-n,k]

#Whether the phase of the diffraction operator is imaginary (i.e. whether wave is evanescent) when the domain width is n samples and the wavelength is Lambda (in samples)

PhaseImaginary[k_,n_,[Lambda]_]:=1-([Kappa][k,n]/n [Lambda])^2<0

#Diffraction Phase Calculation: assume wavelength Lambda (in samples) and propagate through axial distance z

DiffractionPhase[k_,n_,[Lambda],z]:=If[z>= 0,2 [Pi] z/[Lambda] (-1+Sqrt[1-([Kappa][k,n]/n [Lambda])^2]),If[PhaseImaginary[k,n,[Lambda]], -2 [Pi] z/[Lambda],2 [Pi] z/[Lambda] (-1+Sqrt[1-([Kappa][k,n]/n [Lambda])^2])]]

#Diffraction Calculation: Take List x (assumed field samples), assume wavelength Lambda (in samples) and propagate through axial distance z (also in samples)

Diffract[x_,[Lambda],z]:= InverseFourier[Table[Exp[I DiffractionPhase[k,n,[Lambda],z]],{k,0,n-1}] Fourier[x]]

#Domain Width in samples

n:=2048;

#Beam Width in samples

w:= 200;

#Beam Offset in samples

[Delta]:=n/4;

#Beam Tilt Relative z axis (Radians)

[Alpha]:=0.7;

#Beam Wavelength (in samples)

[Lambda]:=20;

#Gaussian Beam

x:= Table[Exp[-((2 (k - [Delta])^2)/w^2)] Exp[2 [Pi] I/[Lambda] (k-[Delta]) Sin[[Alpha]]],{k,0,n-1}]

#Square Input Beam

x:= Table[If[Abs[k-[Delta]]

ListPlot[Abs[Diffract[x,[Lambda],1000]]^2,PlotRange->All]

ListPlot3D[Table[Abs[Diffract[x,[Lambda],10z]]^2,{z,0,100}],PlotRange->All]

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