Solved: Paraxial Wave Equation - Constant Phase Surfaces

[Confundo]

paraxial wave equation - Solved

Homework Statement

When a laser beam traveling is traveling in one direction, we can make the paraxial approximation.

Question: Find an expression for the surfaces with constant phase in the beam.

Homework Equations

From a previous part of the question, I had to work out the paraxial wave equation
which
\nabla^2_{x,y}G - 2ik\frac{\delta G}{\delta z}
and confirm that the below was a solution
G(r, \omega) =\frac{1}{s^2(z)}exp[-\frac{x^2+y^2}{s^2(z)}]
where s^2(z) = w_0^2 - \frac{2iz}{k}

The Attempt at a Solution

I think I have to find an equation of the form.

G = Re^{i\phi}
phase = \phi
where R = radius of curvature

The complex parts are confusing, the solution/equation isn't spherical so I'm a bit stuck on where to start.

 

[turin]

I considered the more general algebraic problem of finding the expression for the constant phase of

<br /> G=\frac{1}{a-ib}\exp\left(-\frac{c}{a-ib}\right)<br />

I first do the conjugate multiplication thing, so that I'm dealing with sums of real and imaginary parts.

<br /> G=\left(\frac{a}{a^2+b^2}+i\frac{b}{a^2+b^2}\right)\exp\left(-\frac{ca}{a^2+b^2}-i\frac{cb}{a^2+b^2}\right)<br />

Then I factorize, with my eye on the prize.

<br /> G=\frac{1}{\sqrt{a^2+b^2}}\exp\left(i\arctan\frac{b}{a}\right)\exp\left(-\frac{ca}{a^2+b^2}\right)\exp\left(-i\frac{cb}{a^2+b^2}\right)<br />

Finally, I collect factors so that I have the standard polar form.

<br /> G=\frac{1}{\sqrt{a^2+b^2}}\exp\left(-\frac{ca}{a^2+b^2}\right)\exp\left(i\left(\arctan\frac{b}{a}-\frac{cb}{a^2+b^2}\right)\right)<br />

So, a constant phase would require the transcendental equation to be satisfied.

<br /> \arctan\frac{b}{a}-\frac{cb}{a^2+b^2}=\textrm{constant}<br />

Perhaps the paraxial approximation allows the \arctan to be approximated so that the equation becomes solvable ...

 

[Confundo]

Thanks. Very clever piece of factorising there :). The lecturer said not to be concerned with a long equation.

 

[turin]

I wouldn't call that result "a long equation" (even though it is unsolvable). Are you looking for something simpler?

 

[Confundo]

Shouldn't of said long without a reference if I'm studying physics really :). The paraxial approximation allows the small angle approximation to be used.

 

[turin]

The paraxial approximation allows the small angle approximation to be used.

I believe you mean

The paraxial approximation requires the small angle approximation to be used.