10.3 An array of apertures

Consider an array of rectangular slits of widths a and b with spacing d in the y direction, illuminated by a plane wave. We want to derive the Fraunhoferdiffracted field $E(k_x,k_y)$ . The simplest way to do this problem is to use Fourier optics: The array is aconvolution of a single rectangular slit with an array of Dirac delta functions.

That is the aperture function can be written $$A=A_0(f\otimes h)$$ where the symbol $\otimes$ represents convolution and $${\displaystyle \begin{array}{c}f=1when|x|\le a/2and|y|\le b/2\\ =0otherwise\end{array}}$$ $$h=\sum _{n_y=0}^{N-1}\delta (y-n_{y}d)\text{.}$$ We now have that $$E=\frac{{\varepsilon }_A}{R}{e}^{-i(kR-\omega t)}Ϝ\{f\}Ϝ\{h\}$$ where $R$ is the distance from the origin to the point of measurement. and $F\{\}$ represents a Fourier transform. Now we have $$Ϝ\{f\}={\int }_{-b/2}^{b/2}{\int }_{-a/2}^{a/2}{e}^{i(k_x{x}^{\prime }+k_y{y}^{\prime })}d{x}^{\prime }d{y}^{\prime }$$ or rearranging, $$Ϝ\{f\}={\int }_{-b/2}^{b/2}d{y}^{\prime }{e}^{i{k}_y{y}^{\prime }}{\int }_{-a/2}^{a/2}d{x}^{\prime }{e}^{i{k}_x{x}^{\prime }}\text{.}$$ These are integrals we have done a number of times already: $$Ϝ\{f\}=bsinc(k_{y}b/2)asinc(k_{x}a/2)$$ Next we transform the sums of the delta functions $$Ϝ\{h\}=\sum _{n_y=0}^{N-1}{\int }_{-\infty }^{\infty }{e}^{i{k}_{y}y}\delta (y-n_{y}d)dy\text{,}$$ which upon integration becomes $$Ϝ\{h\}=\sum _{n_y=0}^{N-1}{e}^{i{k}_y{n}_{y}d}\text{.}$$ We have also performed this sum before and it gives $$F\{h\}=e^{i{k}_y(N-1)d/2}\frac{\sin (N{k}_{y}d/2)}{\sin (k_{y}d/2)}\text{.}$$ So we have $$E=\frac{{\varepsilon }_A}{R}{e}^{-i(kR-\omega t)}{e}^{i{k}_y(N-1)d/2}\frac{\sin (N{k}_{y}d/2)}{\sin (k_{y}d/2)}bsinc(k_{y}b/2)asinc(k_{x}a/2)$$ or if we define $$k{R}_c=kR-(N-1)d{k}_y/2$$ $$E=\frac{{\varepsilon }_{A}ab}{R}{e}^{-i(k{R}_c-\omega t)}\frac{\sin (N{k}_{y}d/2)}{\sin (k_{y}d/2)}sinc(k_{y}b/2)sinc(k_{x}a/2)\text{.}$$