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Single-slit diffraction

  1. Dec 14, 2014 #1
    In my physics book, one of the basic quick quizzes checkpoints asks what happens to the central peak in a diffraction envelope when you decrease the wavelength of light (from 650 nm to 450 nm, for reference).

    My understanding is that the width of the peak would decrease, while the number of interference fringes would remain the same.
    This is because the number of interference fringes depends only on the width of the slit and the distance of the screen based on m=d/a.
    The width of the peak decreases, because using the equation asin(theta)=m(lambda), m and a are constant, while changing lambda would change theta, thereby changing the height of the central maximum. The first order minimum would be closer to the center with a smaller wavelength.

    However, my book says the answer is that the width of the peak remains the same (along with the number of interference fringes).

    This seems inherently wrong. Based on my research on the internet, as well as the interactive picture my book included with the question, I feel like the publishers made a mistake in the answer. Clearly, wavelength affects the diffraction pattern. Right? Please help. My brain hurts.
  2. jcsd
  3. Dec 14, 2014 #2
    oops, this question is regarding the combined effects of two-slit and single-slit interference, not just single slit. That makes a pretty big difference, but I am still pretty confused. Any help would be appreciate.
  4. Dec 14, 2014 #3


    Staff: Mentor

  5. Dec 14, 2014 #4
    Thanks, Bill. I don't really see two much in there that answers my question. Here is where I am having trouble grasping the supposed easy question in my book. Interference pattern depends on wavelength. Diffraction pattern depends on wavelength. How does the combined effects not depend on wavelength. This seems counter-intuitive.
  6. Dec 14, 2014 #5
  7. Dec 14, 2014 #6


    Staff: Mentor

    You are stuck in the wave-particle paradigm by thinking in terms of wavelength. Forget it. My link gives the correct analysis.

    That said, in that paradigm obviously the interference pattern will depend on wavelength so your objection leaves me scratching my head.

  8. Dec 14, 2014 #7
    I'm sorry, but that article does not make sense to me. This is supposedly a somewhat simple understanding in my introductory quantum physics class.

    Attached Files:

  9. Dec 14, 2014 #8


    Staff: Mentor

    As far as I can see that's purely a classical wave analysis.

    What I am saying is this wave-particle duality stuff is simply a way-station to the correct theory which doesn't really have it. The link I gave is the correct explanation.

    But as a wave analysis I cant see anything wrong or what your issue is.

  10. Dec 14, 2014 #9
    The chapter I am studying primarily has to do with wave analysis. My problem is that based on the information I have learned, with a combination of single-slit and multi-slit pattern, if the wavelength is decreased, the central maximum should also decrease in width. The interference fringes should also decrease in width, although there are the same number of them within the central diffraction envelop. My book asks this question. It says the answer is that everything remains the same. I just don't see how this could be the correct, based on the equations provided.
  11. Dec 14, 2014 #10
    So if the width of the peak remains constant and the number of fringes remains constant, what WOULD change if the wavelength changes. Surely something has to. I am wondering if there is a mistake in the book.
  12. Dec 14, 2014 #11


    Staff: Mentor

    I would need to go through the book.

    But if the wavelength changes what doesn't change is getting only single particle hits - of course the pattern changes with particle wavelength.

    But I want to emphasise this wave-particle duality is a crock. Its not the real explanation.

  13. Dec 14, 2014 #12
    okay, thank you, I appreciate your help. based on what I have researched online, their answer is incorrect, which is even more frustrating. I will go back and try to comprehend the article you sent me after my final exams.
  14. Dec 15, 2014 #13


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    The single-slit interference pattern in Fraunhofer observation is described as the Fourier transform of the opening, i.e., the amplitude is given by
    $$A(x) \propto \int_{-d/2}^{d/2} \mathrm{d} y \exp(\mathrm{i} k x y/L)=\frac{L}{\mathrm{i} k x}[\exp[\mathrm{i} k x d/(2L)]-\exp[\mathrm{i} k x d/(2L)]]=\frac{2L}{kx} \sin \left (\frac{k x d}{2L} \right ).$$
    The width is usually taken as the distance of the first minimum relative to the main maximum at ##x=0##, which is thus found by
    ##\frac{\Delta x k d}{2L}=\pi \; \Rightarrow \; \Delta x=\frac{2 \pi L}{k d}=\frac{\lambda L}{d}.##
    This shows that the width is proportional to the wave length ##\lambda## and the distance from the slit to the screen and inversely proportional to the width of the slit (the approximations involved require ##\lambda \ll d \ll L##.

    For more details, see
  15. Apr 22, 2015 #14
    here is the answer:
    what you are saying is correct for the width (horizontally). BUT the question is asking about the central peak which is intensity at theta=0 (width vertically!!!, there is no angle so it has nothing to do with wavelength)
    Of course the central peak remains the same!
  16. Apr 24, 2015 #15
    heehe, thank you. this question will forever haunt me as i missed a similar question on the final exam last semester, worth about 1/3 of the points, dropping me a whole letter grade and even more in invaluable self confidence, haaha. at least i know now. pain is the greatest teacher.
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