Gaussian beam interference

  • #1
461
16

Main Question or Discussion Point

The superposition of a gaussian beam and a plane wave generates a pattern of rings whose phase shift before and after the focal plane (of the gaussian beam) is ##\pi##. This means that if you measure interference before and after you'll see the minimums and maximums of intensity invert.

We can create the superposition via

##(e^{-ikz}+\frac{w_0}{w(z)}e^{-r^2/w(z)^2}e^{-i(kz+\frac{kr^2}{2R(z)}-\psi(z))} ) * (e^{ikz}+\frac{w_0}{w(z)}e^{-r^2/w(z)^2}e^{i(kz+\frac{kr^2}{2R(z)}-\psi(z))} )##

The obtained expression is

##1+\frac{w_0^2}{w(z)^2}\exp(-2r^2/w(z)^2) + 2 \frac{w_0}{w(z)} e^{-r^2/w(z)^2} \cos (\frac{kr^2}{2R(z)}-\psi(z))##

Remember ##R(z)=z[1+(z_R/z)^2]## and ##\psi(z)=\textrm{arctan}(z/z_R)##

MY PROBLEM IS:

This equation doesn't predict the pi phase shift!

Consider the argument of the cosine function.

##R(z^+)= R##

At ##z^+##

The argument is

##\frac{kr^2}{2R}-\pi/2##

at ##z=-z^+##

The argument is

##\frac{-kr^2}{2R}+\pi/2##

Now we know ##cos(\theta)=cos(-\theta)## therefore there is no pi phase shift.

But I know as a matter of fact that there should be a pi phase shift( I've observed it!).

But I don't understand what's happening here mathematically.

Any help will be very much appreciated.
 

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
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784
By ##\pi## phase change, they must mean the Gouy phase ##\psi(z)##. As you go from ##-\infty## to ##\infty##, this term undergoes a change of ##\pi##.
 
  • #3
461
16
Yeah, but the interference term isn't changing by ##\pi## which would explain the inversion of fringes. How else can one explain this inversion of fringes ? The sources I've seen just say "it's due to the gouy phase as can be seen from the interference term".
 
  • #4
461
16
Is there any way to write a coordinate free gaussian beam so as to make ##k## negative some times? This would fix a lot of things!
 

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