Application One: Diffraction

Problem Introduction

• what is diffraction: inteference patterns that light makes passing through apertures

• how to make:

  • light from a distant source, which means the aperture plain is a wavefront i.e. wave have the same phase at all points of the aperture

  • plain with apertures

  • image plain at some distance

• assumptions:

  • light is an oscollating E/M field

  • light is monochromatic (only one frequency)

  • light of the aperture plain as:

  $$E_0\cdot e^{2\pi i\nu t}$$

where $E_0$ is strength of the field on the aperture plain and $\nu$ is frequency (monochromatic)

Near-Field vs. Far-Field

Distance between the aperture plain and the image plain (measured relative to the wavelength) determines diffraction

• near-field Fresnel diffraction

• far-field Fraunhofer diffraction

Huygens Principle

Each point on a wavefront can be regarded as a source i.e. all the sources on the wavefront will be integrated

Question: what is the strength of the wave on point $P$?

Solutioin

After introducing coordinates on the aperture plain, it is clear that the main effect in light going from $X$ to $P$ over a certain distance is: change in phase. The distance $r$ can be represented as $\frac{r}{\lambda }$ cycles, which means a phase change of $\frac{2\pi r}{\lambda }$. Therefore the light magnitude of $P$ resulted from a tiny source of point $x$ of the aperture plain is:

$$dE=E_0\cdot e^{2\pi i\nu t}\cdot e^{-2\pi i\frac{r}{\lambda }}dx$$

therefore the total field is the integral over the aperture:

$$\begin{array}{rl}{E}_P& =\int E_0\cdot e^{2\pi i\nu t}\cdot e^{-2\pi i\frac{r}{\lambda }}dx\\ & =E_0\cdot e^{2\pi i\nu t}\int e^{-2\pi i\frac{r}{\lambda }}dx\\ & =E_0\cdot e^{2\pi i\nu t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-2\pi i\frac{r}{\lambda }}A(x)dx\end{array}$$

where $r$ (not $t$) depends on $x$ and $A(x)$ is the aperture function:

$$\begin{array}{r}A(x)=\{\begin{array}{ll}1& x\in \text{aperture}\\ 0& \text{otherwise}\end{array}\end{array}$$

with Fraunhofer approximation:

$$r=r_0-xsin\theta$$

The integral can be re-written as:

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-2\pi i\frac{r}{\lambda }}A(x)dx& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2\pi i\frac{xsin\theta -r_0}{\lambda }}A(x)dx\\ & =e^{-2\pi i\frac{r_0}{\lambda }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{2\pi ixp}A(x)dx\end{array}$$

where axillary variable $p$:

$$p=\frac{sin\theta }{\lambda }$$

$$\begin{array}{rl}\therefore E_P& =C{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{2\pi ixp}A(x)dx\\ & =C\cdot {\mathcal{F}}^{-1}A(p)\end{array}$$

$$\begin{array}{}◼& \end{array}$$

Conclusion: for far-field diffraction, the intensity of the light is the magnitude of the inverse Fourier transform of the aperture function.

Back to Fourier Transform.