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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.