Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 1 / ii / b

Table of contents
- Solution b

Solution

The Rytov approximation at first order writes the total profile as

f_{R} = f_{i} e^{χ_{1}}

and identifies the first logarithmic perturbation with the Born relative field:

χ_{1} = \frac{f _{s, B}}{f _{i}} = - \frac{α}{4} e^{2 i q_{0} z} .

Hence, in the upper half-space,

ψ_{R} (x, z) = e^{i (p x - q_{0} z)} exp (- \frac{α}{4} e^{2 i q_{0} z}) .

Expanding the exponential function to first order shows that its reflected component is

ψ_{r, R}^{(1)} (x, z) = - \frac{α}{4} e^{i (p x + q_{0} z)} .

The exponentiated expression is the Rytov approximation to the total field; only its term linear in

α

is a single specular reflected plane wave.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version