Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I have a monochromatic ribbon beam with $E(x)e^{i(kz-\omega t)}$ being the electric field's amplitude. I want to show that the lowest order approximation in terms of plane waves is

$\mathbf{E}(x,z,t)=\mathbf{\epsilon} \int{d\kappa A(\kappa) e^{i(\kappa x+kz-\omega t)}}$

where $\mathbf{\epsilon}$ is the polarisation direction and $A(\kappa)$ is the Fourier transform of $E(x)$ around $\kappa=0$.

From the result I can understand/identify that the Fourier kernel is $e^{i(\kappa x)}$ but usually when you use a Fourier transformation you go from $f(x)\rightarrow F(\omega)$, from one variable to another that is, but here all of a sudden you just add one new variable in the transformed field. How is that possible?

EDIT There is a new approach to this, but there is a tiny little point that I don't get. Consider the following geometry.

$d=x_1\sin{\theta},\; \sin{\theta}=\dfrac{x_0}{r_{01}}\approx \dfrac{x_0}{z},\; d\approx \dfrac{x_0x_1}{z}$

Consider the travelling wave


Why $r$ becomes $(r_{01}-d)$?

I beieve that $r$ describes the optical path. It feels like the two rays have an optical difference of $d$, but I am not sure if this is the answer.

Any ideas would be more that welcomed!

Solution found here

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.