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I have been going through the paper

Diffraction from an irregular screen with applications to ionospheric problems. H.G. Booker, J.A. Ratcliffe and D.H. Shinn. Philos. Trans. Royal Soc. A 242, 579 (1950).

The authors state that

it is possible to represent a wave emerging from a diffracting screen in terms of an angular spectrum of plane waves, and it can be related to the Fourier analysis of the wave disturbance over the aperture.

Is this 'angular spectrum' a mathematical quantity by itself, or does it have something to with the electromagnetic spectrum we know of?

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  • $\begingroup$ Have you googled it? Angular spectrum is just a fancy way of saying 3d Fourier transform. The 'angular' part comes from the fact that finite and fixed speed of light limits the magnitude of the wavevector, so all components of the Fourier spectrum correspond to wavevectors of the same magnitude, but different orientation (different angles) e.g. see Goodman "Fourier Optics" $\endgroup$ – Cryo Sep 16 '19 at 21:47
  • $\begingroup$ Yes, I had but I didn't get a proper satisfactory answer. The authors in the above mentioned paper also adds that, "The angular spectrum produced by a spatial distribution of electric field is given by the Fourier transform of the field, f(x)". I was hence wondering whether you could explain more about your wordings "Angular spectrum is just a fancy way of saying 3d Fourier transform". $\endgroup$ – budding physicist Sep 17 '19 at 12:51
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It's quite frequent to use the word 'spectrum' whenever a given quantity $f(q,t)$ (where $q$ is some generalized coordinate and $t$ is time) can be decomposed as a linear combination of 'simpler' functions $g_\kappa(q,t)$, $$ f(q,t) = \int A(\kappa) g_\kappa(q,t) \mathrm d\kappa, $$ in which case $A(\kappa)$ is known as the 'spectrum' of $f(q,t)$. (Though not generically ─ it's always used with a qualifier unless the context makes that qualifier redundant.)

  • The standard electromagnetic spectrum is specifically the frequency spectrum, which corresponds to the temporal Fourier transform, $$f(q,t) = \int A(q,\omega) e^{-i\omega t}\mathrm d\omega.$$
  • The angular spectrum is generally defined only for monochromatic waves, and it corresponds to decomposing a complex-valued amplitude function of position $f(\mathbf r)$ as a linear combination of plane waves propagating along different directions $\hat{\mathbf n}$, i.e. $$ f(\mathbf r) = \iint_{\mathbb S^2} A(\hat{\mathbf n}) e^{i\hat{\mathbf n}\cdot \mathbf r} \mathrm d\Omega,$$ with $\mathrm d\Omega$ the solid-angle differential at unit vector $\hat{\mathbf n}$ taken over the entire unit sphere $\mathbb S^2$.
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Let's say we can work with scalar optical fields for now. Vectorial treatments are possible, but are usually more involved, and are often un-necessary

So you have the scalar monochromatic optical field:

$A\left(\mathbf{r},t\right)=A_\omega\left(\mathbf{r}\right)\exp\left(i\omega t\right)$

Where $\omega$ is the angular frequency. The 3d Fourier transform for this field is:

$\tilde{A}_\omega=\int d^3 r \exp\left(i\mathbf{k}.\mathbf{r}\right)A_\omega\left(\mathbf{r}\right)$

$A_\omega=\frac{1}{\left(2\pi\right)^3}\int d^3 k \exp\left(-i\mathbf{k}.\mathbf{r}\right)\tilde{A}_\omega\left(\mathbf{k}\right)$

Now the optical field will satisfy the Helmholtz equation:

$\left(\nabla^2+\left(\frac{\omega}{c}\right)^2\right)A_\omega=0$

Where $c$ is the speed of light, so:

$\left(-k^2+\left(\frac{\omega}{c}\right)^2\right)=0$ for each wavevector $\mathbf{k}$

It follows, that the only way the Helmholtz equation can be satisfied is if:

$\tilde{A}_\omega\left(\mathbf{k}\right)=\frac{\left(2\pi\right)^3\mathcal{A}_\omega \left(\mathbf{\hat{k}}\right)\delta\left(k-\omega/c\right)}{\left(\omega^2/c^2\right)}$

Where $\mathbf{\hat{k}}=\mathbf{k}/\sqrt{\mathbf{k}.\mathbf{k}}$. It is convenient to introduce a different quantity $\mathcal{A}$ to draw attention to the restriction imposed by the Helmholtz equation (for example, it will have different units).

Then

$A_\omega\left(\mathbf{r}\right)=\int d^2 \Omega_k \exp\left(-i\frac{\omega}{c}\mathbf{\hat{k}}.\mathbf{r}\right)\mathcal{A}_\omega\left(\mathbf{\hat{k}}\right)$

Where the integral is over the solid angle (i.e. all possible orientations of $\mathbf{\hat{k}}$)

So now you can interpret this as $A_\omega$ being made of a superposition of different plane-waves propagating in different directions given by the orientation of $\mathbf{\hat{k}}$ - hence the angular spectrum ($\mathcal{A}_\omega$)

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  • $\begingroup$ May I know then whether you could clarify what angular spectrum produced by a spatial distribution of electric field mean? Is it obtained by superimposing different plane-waves propagating in different directions as given by the orientation of k vector in this case also? $\endgroup$ – budding physicist Sep 17 '19 at 14:50
  • $\begingroup$ Angular spectrum .... is a Fourier transform of a field in 3d space. It is obtained by Fourier transform, but instead of 3d Fourier transform it is a 2d Fourier transform over a plane or a surface (Helmholtz equation removes one degree of freedom). I have no desire to copy text books here, sorry, so the best I can do is to suggest J. Goodman, "Introduction to Fourier Optics" Ch3 (quite accessible); Born & Wolf "Principles of Optics", Ch 8.3 (slightly old style of exposition); A. J. Devaney, "Mathematical Foundations of Imaging ...", Ch 4 (most complete and modern approach, quite in-depth). $\endgroup$ – Cryo Sep 18 '19 at 10:11

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