# I Gaussian beam interference

1. May 13, 2016

### davidbenari

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.

2. May 13, 2016

### blue_leaf77

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. May 13, 2016

### davidbenari

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. May 14, 2016

### davidbenari

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!