Why Is the Paraboloidal Wavefront Approximation Valid?

  • Thread starter Thread starter yucheng
  • Start date Start date
  • Tags Tags
    Wave Wave optics
Click For Summary
SUMMARY

The discussion centers on the validity of the paraboloidal wavefront approximation in wave optics, specifically referencing the work of Saleh & Teich in "Fundamentals of Photonics." The approximation holds under the condition that the amplitude varies minimally, allowing for the simplification of spherical wavefronts to paraboloidal shapes at large distances. The mathematical derivation shows that for points near a given point P along the z-axis, the coordinates satisfy the equation $$z - z_0 \approx -\frac{x^2 + y^2}{2z_0}$$, confirming the paraboloidal approximation's relevance in practical applications.

PREREQUISITES
  • Understanding of wave optics principles
  • Familiarity with complex amplitude and phase concepts
  • Knowledge of spherical and paraboloidal geometries
  • Basic proficiency in mathematical derivations involving wave equations
NEXT STEPS
  • Study the derivation of wavefronts in "Fundamentals of Photonics" by Saleh & Teich
  • Learn about the implications of the paraxial approximation in optics
  • Explore the mathematical foundations of wavefront propagation
  • Investigate applications of paraboloidal wavefronts in optical systems design
USEFUL FOR

Optical engineers, physicists, and students studying wave optics who seek to understand the practical applications and theoretical foundations of wavefront approximations.

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 $$z-z_0 = -\frac{x^2+y^2}{2z_0} .$$
 
  • Like
Likes   Reactions: yucheng
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...
 

Similar threads

Replies
12
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
1
Views
3K