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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.

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  • $\begingroup$ 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. $\endgroup$
    – Colin K
    Commented Jan 8, 2012 at 22:21
  • $\begingroup$ 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!;-) $\endgroup$
    – Simon S
    Commented Jan 8, 2012 at 22:24
  • $\begingroup$ @SimonS Sir how you arrived at that result , was it same as the method given in the answer ? $\endgroup$
    – Orion_Pax
    Commented Apr 14, 2022 at 10:58

1 Answer 1

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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)

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  • $\begingroup$ Right. I centered my slit/transmission function in the middle of the diagram, but yes, I should have squared the answer. Thanks. $\endgroup$
    – Simon S
    Commented Jan 8, 2012 at 22:49
  • $\begingroup$ The squaring is optional really. It gives you the intensity you would measure, but the complex valued field is meaningful on its own as well, just not as an observable. $\endgroup$
    – Colin K
    Commented Jan 8, 2012 at 23:25
  • $\begingroup$ yohBS, I'm not sure if the homework tag was on the problem when you posted this, but just in case, I wanted to make sure you're aware of our policy that we shouldn't be posting complete answers to homework questions. Any time that happens the answer may be temporarily deleted. $\endgroup$
    – David Z
    Commented Jan 9, 2012 at 2:07
  • $\begingroup$ The homework tag was not there, if I recall correctly, but even if it was, I was not aware of this policy. I will abide in the future... $\endgroup$
    – yohBS
    Commented Jan 9, 2012 at 22:42
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    $\begingroup$ For the record, this was not a homework problem for me and I did not put that tag there. I simply wanted to understand it for myself. My days of undergrad physics are well and truly over. $\endgroup$
    – Simon S
    Commented Jan 29, 2012 at 19:05

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