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If the width of one slit is doubled, but that it is still much narrower than the distance to the screen, what happens to the intensity distribution that we see on the screen?

I have been reading some sources that state where we had minima, are no longer of $0$ intensity (i.e. destructive interference) but are just $1/9^{th}$ of the maximum intensity.

I had one idea why this might be the case, but I don't know if it's "mathematically rigorous" or even right. Since we know that the intensity is proportional to area of slit, and the areas are in ratio of 1:2, then we can say that the amplitudes is in a ratio of $1:\sqrt{2}$ (since amplitudes are the square root of intensity more or less). At the maxima, the 2 waves get added, so the maxima will be $1+\sqrt{2}$ whereas at the minima the amplitudes will be $1-\sqrt{2}$, so the intensities would be of a ratio $\frac{(1-\sqrt{2})^{2}}{(1+\sqrt{2})^2}$ which isn't 1/9.

I am trying to formulate a mathematical description/proof of this but can't. Any guidance is much appreciated.

EDIT: could it be perhaps that slit width is not proportional to intensity, and is in fact proportional to amplitude?

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  • $\begingroup$ Which 'sources'? If you provide that context then it will be easier to answer. $\endgroup$ Commented Mar 31, 2021 at 17:19
  • $\begingroup$ It was a question from a physics textbook. I can post the picture but I don't think it's much help. $\endgroup$
    – jambajuice
    Commented Mar 31, 2021 at 17:21
  • $\begingroup$ Maybe don't include the image into the post, but do include a link to the image as well as a complete reference to the textbook. Context always helps. $\endgroup$ Commented Mar 31, 2021 at 17:22
  • $\begingroup$ Well, what is the Fourier transform of the new slit configuration? $\endgroup$
    – Jon Custer
    Commented Mar 31, 2021 at 18:02
  • $\begingroup$ I don't really know much about Fourier transforms, not in the case of slits anyway. $\endgroup$
    – jambajuice
    Commented Mar 31, 2021 at 18:06

1 Answer 1

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First consider the two slits alone, than open the second half of one slit. The interference of the two equal slits remains. However, the third which contributes 1/3 of the amplitude gives, at the place of the minimum, almost a maximum, though it will be a little less than 1/9 of the intensity since there is some negative interference of the half slit.

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  • $\begingroup$ Hmm I suppose. But to give it some mathematical treatment, would one just perform a Kirchhoff Diffraction Integral? $\endgroup$
    – jambajuice
    Commented Mar 31, 2021 at 19:48
  • $\begingroup$ I would use Huygens–Fresnel since I am not familiar with Kirchhoff Diffraction Integral $\endgroup$
    – trula
    Commented Mar 31, 2021 at 19:56

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