9.3 Diffraction from a rectangular aperture

We consider diffraction from apertures other than a slit. For example consider a rectangular aperture as shown below. If $\overrightarrow{{\varepsilon }_A}$ is the source strength per unit area (assumed to be constant over the entire area in this example) and $dS=dydz$ is an infinitesmal area at a point in the aperture then we have:

We see from the figure that $$r=\left[x^2+{(Y-y)}^2+{(Z-z)}^2\right]$$ and that $$R^2=x^2+Y^2+Z^2\text{.}$$ Thus we use $$x^2=R^2-Y^2-Z^2$$ to write $$r={\left[R^2-Y^2-Z^2+Y^2+y^2-2yY+Z^2+z^2-2zZ\right]}^{1/2}$$ or $$r={\left[R^2-2Yy-2Zz+y^2+z^2\right]}^{1/2}$$ $$r=R{\left[1-2Yy/R^2-2Zz/R^2+(y^2+z^2)/R^2\right]}^{1/2}\text{.}$$ We are only considering Fraunhofer diffraction so $R,Z,Y$ are much larger than $y$ and $z$ and we can rewrite $$r\approx R{\left[1-2(Yy+Zz)/R^2\right]}^{1/2}$$ and then finally expanding using the binomial theorem and taking only the most significant terms $$r\approx R\left[1-(Yy+Zz)/R^2\right]\text{.}$$

So we have $$d\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}}{r}{e}^{i(kr-\omega t)}dydz$$ or using the fact that $R$ is large $$d\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}}{R}{e}^{i(kr-\omega t)}dydz$$ which we integrate to get the field

$${\displaystyle \begin{array}{c}\vec{E}={\int }_{-b/2}^{+b/2}{\int }_{-a/2}^{+a/2}\frac{\overrightarrow{{\varepsilon }_A}}{R}{e}^{i(kr-\omega t)}dzdy\\ =\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}}{R}{\int }_{-a/2}^{+a/2}{e}^{-ikyY/R}dy{\int }_{-b/2}^{+b/2}{e}^{-ikzZ/R}dz\\ =\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}}{R}{\left[\frac{e^{-ikyY/R}}{-ikY/R}\right|}_{-a/2}^{+a/2}{\left[\frac{e^{-ikzZ/R}}{-ikZ/R}\right|}_{-b/2}^{+b/2}\\ =\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}}{R}\left[\frac{e^{-ikaY/2R}-e^{ikaY/2R}}{-ikY/R}\right]\left[\frac{e^{-ikbZ/2R}-e^{ikbZ/2R}}{-ikZ/R}\right]\end{array}}$$

Now lets define $\alpha =kaY/2R$ $\beta =kbZ/2R$ and $\beta =kbZ/2R$ and we see that

$$\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}}{R}\left[\frac{e^{i\alpha }-e^{-ik\alpha }}{ikY/R}\right]\left[\frac{e^{i\beta }-e^{-i\beta }}{ikZ/R}\right]$$

or rearranging

$$\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}}{R}\left[\frac{e^{i\alpha }-e^{-i\alpha }}{2i\alpha /a}\right]\left[\frac{e^{i\beta }-e^{-i\beta }}{2i\beta /b}\right]$$

$$\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}ab}{R}\left[\frac{e^{i\alpha }-e^{-i\alpha }}{2i\alpha }\right]\left[\frac{e^{i\beta }-e^{-i\beta }}{2i\beta }\right]$$

$$\vec{E}=\frac{\overrightarrow{{\varepsilon }_A}{e}^{i(kR-\omega t)}ab}{R}\left[\frac{\sin \alpha }{\alpha }\right]\left[\frac{\sin \beta }{\beta }\right]$$

So finally we can write the intensity as

$$I(Y,Z)=I(0,0){\left[\frac{\sin \alpha }{\alpha }\right]}^2{\left[\frac{\sin \beta }{\beta }\right]}^2$$

Below is a plot of ${\left[\frac{\sin \beta }{\beta }\right]}^2{\left[\frac{\sin \alpha }{\alpha }\right]}^2$

Below is a plot of $\left[\frac{\sin \beta }{\beta }\right]\left[\frac{\sin \alpha }{\alpha }\right]$