Slit Fraunhofer Diffraction

One Slit

The Fraunhofer diffraction pattern of one infinitely long slit using the exact analytic solution is $$I(x) \propto sinc^{2} \left ( \frac{\pi Wx}{\lambda z} \right )$$ where $W$ is the slit width, $\lambda$ is the wavelength, and $z$ is the aperture-to-image distance.

Two Slits

Tthe Fraunhofer diffraction pattern of two infinitely long slits using the exact analytic solution is $$I(x) \propto cos^{2}\left ( \frac{\pi Sx}{\lambda z} \right ) sinc^{2}\left ( \frac{\pi Wx}{\lambda z} \right )$$ where $W$ are the slit widths, $S$ is the slit separation, $\lambda$ is the wavelength, and $z$ is the aperture-to-image distance.

N Slits

The Fraunhofer diffraction pattern of a slit grating using the exact analytic solution is $$I(x) \propto sinc^2\left ( \frac{\pi W x}{\lambda z} \right ) \left ( \frac{\sin\left ( \frac{N \pi S x}{\lambda z} \right )}{N\sin \left (\frac{\pi S x}{\lambda z} \right )} \right )^2$$ where $W$ are the slit widths, $S$ is the slit separation, $\lambda$ is the wavelength, $N$ is the number of slits, and $z$ is the aperture-to-image distance. Be careful when implimenting this formula in something like MATLAB (or C#). If $N\sin \left (\frac{\pi S x}{\lambda z} \right ) \to 0$ the value of the fraction blows up to infinity. However $$\lim_{N\sin \left (\frac{\pi S x}{\lambda z} \right ) \to 0} \left ( \frac{\sin\left ( \frac{N \pi S x}{\lambda z} \right )}{N\sin \left (\frac{\pi S x}{\lambda z} \right )} \right )^2 = 1$$ I've implimented this function in C# as $$I(x) \propto \left\{\begin{matrix} sinc^2\left ( \frac{\pi W x}{\lambda z} \right ) \left ( \frac{\sin\left ( \frac{N \pi S x}{\lambda z} \right )}{N\sin \left (\frac{\pi S x}{\lambda z} \right )} \right )^2 & \left |N\sin \left (\frac{\pi S x}{\lambda z} \right ) \right | > 0.01\\ sinc^2\left ( \frac{\pi W x}{\lambda z} \right ) & \left |N\sin \left (\frac{\pi S x}{\lambda z} \right ) \right | < 0.01 \end{matrix}\right.$$