Why Is the Paraboloidal Wavefront Approximation Valid?

  • Thread starter Thread starter yucheng
  • Start date Start date
  • Tags Tags
    Wave Wave optics
yucheng
Messages
232
Reaction score
57
Homework Statement
Try plotting....
Relevant Equations
$$U(\vec{r}) \approx \frac{A_0}{z} exp(-jkz) \exp(-jk\frac{x^2 + y^2}{2z})$$
We can either plot the real part of the complex amplitude, or the wavefront.

However, how is wavefront meaningful for varying amplitude? In order to plot the paraboloid, we must vary ##z##, which varies the amplitude ##\frac{A_0}{z}##. Unless the amplitude is varies little, i.e. ##1/z## approximately constant within ##\Delta z = \lambda##?

In the book Fundamentals of Photonics, Saleh & Teich, the author mentions that the phase of the second exponential function serves to bend the planar wavefronts into paraboloidal surfaces i.e. ##frac{x^2 + y^2}{2z} = \text{const}##, however, shouldn't it be ##z + frac{x^2 + y^2}{2z} = \text{const}## when plotting surfaces of constant phase i.e. wavefronts?

The result should look like this.

Thanks in advance!
 
Physics news on Phys.org
From the link, they are considering spherical waves where $$U(\vec r) = \frac{A_0}{r}e^{-jkr}.$$ The phase is ##kr##. The true wavefronts are all spherical. For large distances from ##r = 0##, small patches of the wavefronts can be approximated by paraboloid-shaped wavefronts. (For very large distances, the wavefronts can be approximated by plane wavefronts.)

Consider moving out along the ##z## axis to some point ##P## with coordinates ##(x, y, z) = (0, 0, z_0)##. The phase of the wave at that point will be ##kz_0##. We look for points in the vicinity of ##P## for which the wave has the same phase. We assume ##z_0## is large enough so that points in the vicinity of ##P## will have coordinates ##(x, y, z)## satisfying ##x \ll z## and ##y \ll z##. For these points, $$r = \sqrt{x^2 + y^2 + z^2} \approx z + \frac{x^2+y^2}{2z} .$$ The condition that the phase ##kr## at these points be the same as at ##P## is $$k\left[ z + \frac{x^2+y^2}{2z}\right ] = k z_0$$ From this show that points near ##P## on the wavefront passing through ##P## have coordinates ##(x, y, z)## which satisfy $$z-z_0 \approx -\frac{x^2+y^2}{2z_0} .$$ Describe the shape of the locus of points ##(x, y, z)## satisfying $$z-z_0 = -\frac{x^2+y^2}{2z_0} .$$
 
TSny said:
$$z-z_0 \approx -\frac{x^2+y^2}{2z_0}$$
I guess this is the key. But why would it be a valid approximation though? Let me try a quantitative argument...
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top