Calculating Fraunhofer diffraction patterns

How does one calculate the Fraunhofer diffraction pattern for the following arrangement of slits:

|...|...........|...|

..a.....3a......a

(Four slits arranged linearly, spaced a distance a, 3a and a apart.)

The width of the single slits can be neglected, so that the transmission function can be expressed as sum of delta-functions.

• I'm adding the homework tag because this is a classic homework question. If it's not, please add more detail and remove the tag. Jan 8, 2012 at 22:21
• Yes, fair enough. I think I have figured out the solution. $$U(\theta) = 2 \left( cos(5\pi \frac{\sin\theta}{\lambda}) + cos(3\pi \frac{\sin\theta}{\lambda}) \right)$$ If anyone violently disagrees with that, I'd be very happy to know about it!;-) Jan 8, 2012 at 22:24
• @SimonS Sir how you arrived at that result , was it same as the method given in the answer ? Apr 14 at 10:58

The Fraunhofer diffraction pattern is simply the (square absolute value of) the Fourier transform of the transmission function. If you put your first line at $x=0$ then the transmission function is $$f(x)=\delta(x)+\delta(x-a)+\delta(x-4a)+\delta(x-5a)$$ and its Fourier transform is $$1 + e^{iak}+e^{4ika}+e^{5ika}$$ Which can be simplified to $$\left(\cos\left[\frac{3 a k}{2}\right]+\cos\left[\frac{5 a k}{2}\right]\right) \left(\cos\left[\frac{5 a k}{2}\right]+i \sin\left[\frac{5 a k}{2}\right]\right)$$ Taking the square of the absolute value, you get $$\left(\cos\left[\frac{3 a k}{2}\right]+\cos\left[\frac{5 a k}{2}\right]\right)^2$$ (I omitted all along multiplicative factors)