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As part of analysing the double slit experiment, I've recorded position values for a number of minima and the central peak position.

Also, I've conducted a single slit experiment (with unknown width) and gathered around 50 data points. This is the graph I've plotted to obtain the measured voltage distribution.

enter image description here

Now, I would like to find a function that fits this graph. According to theory, it should be in the form \begin{equation} I(x)=I_{0}\left[\frac{\sin\left(\frac{\pi a x}{\lambda f}\right)}{\left(\frac{\pi a x}{\lambda f}\right)}\right]^{2}, \end{equation} where $I_{0}$ is the intensity at the centre and $\lambda$ is the wavelength. The wavelength $\lambda$, the focal length of the lens f and the separation of the slits d are known (incl. an error).

Then I'd like to take this computed function as an envelope for the intensity distribution of the double slit experiment I conducted. The formula should be:\begin{equation} I(x)=4I_{0}\left[\frac{\sin\left(\frac{\pi a x}{\lambda f}\right)}{\left(\frac{\pi a x}{\lambda f}\right)}\right]^{2} \cos^{2}\left(\frac{\pi d x}{\lambda f}\right) \end{equation}.

Eventually, I'd like to compare the predicted theory with the function obtained by fitting my single slit and double slit data to this general function. If possible, I'd also like to see if the with of the double slits equals the width of the single slit (assuming that both double slit widths are identical). However, I am not entirely sure if there is not a better way to analyse the data obtained by these two experiments.

I would very much appreciate any help. Thank you.

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  • $\begingroup$ Does your data cover more minima/maxima than you are showing? $\endgroup$ – Aaron Stevens Dec 24 '18 at 12:38
  • $\begingroup$ For the double slit experiment, I've 7 minima for that range, the single slit experiment covers only the amount of minima shown. $\endgroup$ – MrDerDart Dec 24 '18 at 13:02
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    $\begingroup$ Is your question about how to fit an equation to the data? If so, it appears that you need to find values for $a$ and $d$ that optimize the fit. Are you asking for help with that? $\endgroup$ – S. McGrew Dec 24 '18 at 16:41
  • $\begingroup$ I agree with @S.McGrew. What are you really asking for here? $\endgroup$ – Aaron Stevens Dec 24 '18 at 18:26
  • $\begingroup$ @S.McGrew Exactly! $\endgroup$ – MrDerDart Dec 25 '18 at 8:39
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First you need to find an optimal value for $a$ using the data shown in your first graph. A brute force way to do that is to:

  1. assume a value for $a$,
  2. calculate V for each value of x,
  3. calculate the difference between the calculated value of V and the measured value of V for each value of x,
  4. Square those differences individually and sum the squares to get ERROR. (Steps 2 through 4 give you a measure of the ERROR in you assumed value for $a$.

Set this up in a spreadsheet to test all reasonable values of $a$, and pick the one that gives the smallest value of ERROR. This is the optimal value of $a$.

In a slightly more sophisticated approach, Newton's method can be used to speed the search by using changes in ERROR vs changes in $a$ to avoid having to test all values of $a$. But a spreadsheet calculation is fast enough that you'll hardly notice the time needed to test all values of $a$.

Once you've found a value for $a$, (presuming the widths of the slits in the two-slit experiment are identical to the width of the slit in the one-slit experiment), you follow the same procedure with your second equation and data from the two-slit experiment, setting the value of $a$ fixed at the optimum value you found above, to search for an optimal value for $d$.

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