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Physics
Classical Physics
Optics
The Mystery of the Missing Pi Phase Shift in Gaussian Beam Interference
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[QUOTE="davidbenari, post: 5473070, member: 511036"] 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. [/QUOTE]
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Forums
Physics
Classical Physics
Optics
The Mystery of the Missing Pi Phase Shift in Gaussian Beam Interference
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