Fraunhofer Diffraction

$$E(x^{\prime },y^{\prime })\propto \int \int t(x,y)\cdot \exp (-ik\frac{x{x}^{\prime }+y{y}^{\prime }}{z})dxdy$$ $$E(k_x,k_y)\propto {\mathcal{F}}^{(2)}(t(x,y))=T(k_x,k_y)=\int \int t(x,y)\exp (-i(k_{x}x+k_{y}y))dxdy$$

See 2D Fourier Transform, Fourier Transform

Derivation Fresnel Diffraction

Recall Fresnel Diffraction, which will be our starting point.

$$E(x^{\prime },y^{\prime })\propto \int \int \underset{\text{Aperature Fn}}{\underbrace{t(x,y)}}\cdot \underset{\text{Electric Field}}{\underbrace{E(x,y)}}\cdot \underset{\text{Spherical Wave}}{\underbrace{\frac{\exp (ikr)}{r}}}dxdy$$ $$\text{where }r=\sqrt{z^2+(x^{\prime }-x{)}^2+(y^{\prime }-y{)}^2}$$

$z$ being signficantly larger than the other distances allows a useful approximation for the denominator of the Spherical Wave

$$z\gg x^{\prime },x,y^{\prime },y⟹r=z$$

However, small changes in $r$ will lead to big changes in the exponential function of the equation so we can’t apply the approximation there. Instead,

Overall we end up with Equation 1

$$E(x^{\prime },y^{\prime })\propto \int \int t(x,y)\cdot \exp (-ik\frac{x{x}^{\prime }+y{y}^{\prime }}{z})dxdy$$

Let $k_x=\frac{k{x}^{\prime }}{z}$, $k_y=\frac{k{y}^{\prime }}{z}$ Allowed due to $z\gg \dots$ assumption See 2D Fourier Transform Equation 2

$$\therefore E(k_x,k_y)\propto {\mathcal{F}}^{(2)}(t(x,y))$$

Examples

Fraunhofer Diffraction from a Slit

Fraunhofer Diffraction from a slit is just the Fourier Transform of the $\text{rect}$ function, which is the sinc function. The Irradiance is ${\text{sinc}}^2$ $t(x)=\text{rect}\left(\frac{x}{w}\right)$ $E(k_x)\propto \mathcal{F}\left(t(x)\right)$

$$E(k_x)\propto \text{sinc}\left(\frac{w{k}_x}{2}\right)E(x^{\prime })\propto \text{sinc}\left(\frac{w{k}_{x^{\prime }}}{2z}\right)$$ $$I(k_x)\propto {\text{sinc}}^2\left(\frac{w{k}_x}{2}\right)I(x^{\prime })\propto {\text{sinc}}^2\left(\frac{w{k}_{x^{\prime }}}{2z}\right)$$

Also, through a square hole, would just be the same idea, but for $x^{\prime }$ and $y^{\prime }$