Fourier optics

Fourier transforms

The Fourier Transform can be used to represent any well behaved function $f(x)\text{.}$

$$f(x)=\frac{1}{\pi }{\int }_0^{\infty }\left[A(k)\cos (kx)+B(k)sin(kx)\right]dk$$ where $$A(k)={\int }_{-\infty }^{\infty }f(x)\cos (kx)dx$$ $$B(k)={\int }_{-\infty }^{\infty }f(x)\sin (kx)dx$$ I can now substitute for $A$ and $B$ in the original expression and write:

$${\displaystyle \begin{array}{c}f(x)=\frac{1}{\pi }{\int }_0^{\infty }\cos kx{\int }_{-\infty }^{\infty }f(x^{\prime })\cos k{x}^{\prime }d{x}^{\prime }dk+\\ +\frac{1}{\pi }{\int }_0^{\infty }\sin kx{\int }_{-\infty }^{\infty }f(x^{\prime })\sin k{x}^{\prime }d{x}^{\prime }dk\end{array}}$$ and then use $$\cos (x^{\prime }-x)=coskx\cos k{x}^{\prime }+\sin kx\sin k{x}^{\prime }$$

$$f(x)=\frac{1}{\pi }{\int }_0^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })\cos (k[x^{\prime }-x])d{x}^{\prime }dk$$ Since the inner integral is an even function we can write $$f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })\cos (k[x^{\prime }-x])d{x}^{\prime }dk$$ Now consider the fact that $$\frac{i}{2\pi }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })\sin (k[x^{\prime }-x])d{x}^{\prime }dk=0$$ because $\sin$ is an odd function, ie. $${\int }_{-\infty }^{\infty }\sin (k[x^{\prime }-x])dk=0$$ So we could have written $${\displaystyle \begin{array}{c}f(x)=\frac{1}{2\pi }{\int }_{\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })\cos (k[x^{\prime }-x])d{x}^{\prime }dk+\\ +\frac{i}{2\pi }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })\sin (k[x^{\prime }-x])d{x}^{\prime }dk\end{array}}$$ or $$f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })e^{ik(x^{\prime }-x)}d{x}^{\prime }dk$$ or $$f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }g(k)e^{-ikx}dk$$ where $$g(k)={\int }_{-\infty }^{\infty }f(x)e^{ikx}dx$$ is the Fourier transform of $f(x)$ .

Symbolically we write $$g(k)=Ϝ\{f(x)\}$$ $$f(x)={Ϝ}^{-1}\{g(k)\}={Ϝ}^{-1}\{Ϝ\{f(x)\}\}$$

Now these concepts are easily extended to two dimensions:

$$f(x,y)=\frac{1}{{(2\pi )}^2}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }g(k_x,k_y)e^{-i(x{k}_x+y{k}_y)}d{k}_{x}d{k}_y$$ where $$g(k_x,k_y)={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x,y)e^{i(x{k}_x+y{k}_y)}dxdy\text{.}$$

This tells us is that any nonperiodic function of two variables $f(x,y)$ can be synthesized from a distribution of plane waves each with amplitude $g(k_x,k_y)$ .

Lets consider Fraunhofer diffraction through an aperture again. For example consider a rectangular aperture as show in the figure. If $\overrightarrow{{\varepsilon }_A}$ is the source strength per unit area (assumed to be constant over the entire area in this example) and $dS=dxdz$ is an infinitesmal area at a point in the aperture then we have:

We can define a source strength per unit area $$d\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}}{R}{e}^{i(\omega t-kr)}$$

Notice that I flipped the sign in the exponential from what I used in the earlier lectures on diffraction. This does not change the physics content ofwhat we are doing in any way, however it allows our notation to follow standard convention. $${\displaystyle \begin{array}{c}\vec{E}={\int }_{}^{}{\int }_{Aperture}^{}\frac{\overrightarrow{{\varepsilon }_A}}{R}{e}^{i(\omega t-kr)}dzdy\\ =\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(\omega t-kR)}}{R}{\int }_{}^{}{\int }_{Aperture}^{}{e}^{ikyY/R}dy{e}^{ikzZ/R}dz\end{array}}$$

If we define $k_y=kY/R$ and $k_z=Z/R$ and we see that

$${\displaystyle \begin{array}{c}\overrightarrow{E}=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(\omega t-kR)}}{R}{\int }_{}^{}{\int }_{Aperture}^{}{e}^{iy{k}_y}{e}^{iz{k}_z}dydz\\ \overrightarrow{E}=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(\omega t-kR)}}{R}{\int }_{}^{}{\int }_{Aperture}^{}{e}^{i(y{k}_y+z{k}_z)}dydz\end{array}}$$

That is, it is equal to the Fourier transform. In fact one can define an "Aperture Function" $A(y,z)\text{.}$ Such that

$$\overrightarrow{E}={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }A(y,z)e^{i(y{k}_y+z{k}_z)}dydz$$

For a rectangular aperture $A(y,z)=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(\omega t-kR)}}{R}$ inside the aperture and zero outside it. The aperture function can be much more complex (literally) allowing for changes in source strength and phasethrough the aperture. The resulting E field is the Fourier transform of the aperture function.