Single Slit Diffraction

  1. I know that a smaller width of a gap would lead to "more diffraction", and diffraction effects are at their peak when the wavelength of the diffracted waves is equal to the gap width. But what if the gap were smaller than the wavelength? Can someone describe the diffraction pattern (if any) that would be formed in this scenario?
     
  2. jcsd
  3. Are you sure about that?
     
  4. Well, this is what my A level Physics book says.
     
  5. Would you please quote a bit of the passage or context from the Physics book?
     
  6. jtbell

    Staff: Mentor

    Does the book tell you the equation that gives the angles at which minimum intensity occurs?
     
  7. Not at all.
     
  8. PHP:
     
    "As the gap becomes narrower, the diffraction effect becomes more pronounced. It is greatest when the width of the gap is equal to the wavelength of the ripples."
    Cambridge International AS and A Level Physics Coursebook by David Sang, Graham Jones, Richard Woodside and Gurinder Chadha.
     
  9. Here it is explained well

    [​IMG]

    http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1
     
  10. What I'm questioning is the bolded part above. Think about the shape of the outgoing wavefront in the limit of (gap width) « wavelength.

    Alternatively, examine the equation for the intensity of the single-slit diffraction pattern, given at Hyperphysics and also at Wikipedia:
    [itex]I(\theta) = I_0 \, \mathrm{sinc}^2 \left(\dfrac{d\pi}{\lambda} \sin\theta \right),[/itex]
    where I0 is the original intensity, d is the width of the slit, λ the wavelength, and θ the exit angle. There will be zeros (dark bands) at at angles such that [itex]\sin\theta_m = \lambda / d[/itex], where [itex]\theta_m[/itex] gets bigger as [itex]d/\lambda[/itex] gets smaller (as stated in the first line from the course book).

    But why doesn't the spreading continue to increase for [itex]d < \lambda[/itex]?
     
  11. What I'm questioning is the bolded part above. Think about the shape of the outgoing wavefront in the limit of (gap width) « wavelength.

    Alternatively, examine the equation for the intensity of the single-slit diffraction pattern, given at Hyperphysics and also at Wikipedia:
    [itex]I(\theta) = I_0 \, \mathrm{sinc}^2 \left(\dfrac{d\pi}{\lambda} \sin\theta \right),[/itex] where I0 is the original intensity, d is the width of the slit, λ the wavelength, and θ the exit angle. There will be zeros (dark bands) at at angles such that [itex]\sin\theta_m = \lambda / d[/itex], where [itex]\theta_m[/itex] gets bigger as [itex]d/\lambda[/itex] gets smaller (as stated in the first line from the course book).

    But why doesn't the spreading continue to increase for [itex]d/\lambda < 1[/itex]?
     
  12. jtbell

    Staff: Mentor

    It does continue to increase.

    At ##d/\lambda = 1## i.e. ##d = \lambda##, ##\theta_1 = 90^{\circ}##. The central peak of the intensity distribution exactly fills the "field of view" looking outwards from the slit, with the zero-intensity points at ±90°.

    For ##d/\lambda < 1##, the intensity decreases as you "look" outwards at larger angles from the slit, but it never reaches zero. In effect, you can "see" only a central "slice" of the central peak.
     
  13. It's a little vague, but he's referring to the width of the central peak and the ratio of intensities of the maxima and minima.

    You might want to play with that function a little to convince yourself of it.
    You can do it here: rechneronline.de/function-graphs/

    Try:

    sinc(0.5*pi*sin(x))*sinc(0.5*pi*sin(x))
    sinc(1.0*pi*sin(x))*sinc(1.0*pi*sin(x))
    sinc(2.0*pi*sin(x))*sinc(2.0*pi*sin(x))
    sinc(9.9*pi*sin(x))*sinc(9.9*pi*sin(x))

    As you increase the factor, a value of 1.0 forms a sort of tipping point for that function, where zero intensity is present, but the secondary maxima aren't present yet.
     
    Last edited: Mar 18, 2014
  14. That's what I think too, which is why I found the passage from the course book surprising.
     
  15. Why would those be the metrics? In particular, why the ratio of intensities between maxima and minima? A ratio of 1:1 would just mean the beam has become fully cylindrical, right?

    I'd expect something like the deviation from ray behavior to be a more obvious indication of how "pronounced" the diffraction effect has become.
     
  16. It's not a technical term. If you understand the importance of the slit width to wavelength ratio, then you understand at least as much as he's trying to illustrate. My advice is to ignore the word "pronounced".

    The beam isn't cylindrical, even with a circular hole instead of a slit. There is never a sharp cut off. This is true regardless of the slit width to wavelength ratio. If you follow the link that I gave you should be able to see that by playing with the function.

    This video might help if you haven't seen this phenomena yet:
     
    Last edited by a moderator: Sep 25, 2014
  17. sophiecentaur

    sophiecentaur 14,175
    Science Advisor
    Gold Member

    The sinc function is clearly the best way to express the diffraction pattern of a single slit. What point is there in putting it in a conversational way and then discussing how well those words fit the sinc description? For people who can't cope with the details of the formal Maths for this sort of problem then why not just accept the diagrams corresponding to a few different slit widths? What relevance has the word "pronounced" and why is it worth losing any sleep over? Is one circle any more 'round' than another circle or a rectangle any more rectangular than another rectangle?
    The authors of that text book have managed to demonstrate just how inadequate ordinary language can be as a substitute for some very simple Maths. Let's face it, there is absolutely no non-mathematical explanation for the way a diffraction pattern is formed in detail. (There's a challenge for someone.)
     
  18. The argument of the sinc function scales as 1/d. Nothing else changes in the function. It doesn't get any simpler than that.

    The beam exhibits cylindrical spreading in the limit of a narrow vertical slit. The spreading would be spherical with the limit of a small circular aperture.

    Thanks, very nice video. I'm quite familiar with the phenomenon however. :tongue2:
     
    Last edited by a moderator: Sep 25, 2014
  19. Even in a professional paper on the topic, it would be perfectly normal for there to be a (relatively) conversational description in the text to accompanying the "formal Maths." I wouldn't expect the discussion to describe everything contained within the maths, but at the least it should be useful and technically correct.

    Here what I'm saying is that the conversational description doesn't seem to make sense and in fact may be misleading. Worse yet, the formal Maths don't seem to have been included in the course book.

    I think it's useful to say, in ordinary language, that the diffraction effect becomes more pronounced with decreasing slit width d. If however I say the effect is most pronounced when d = λ, then I think that deserves some further explanation (or at least definition).
     
  20. sophiecentaur

    sophiecentaur 14,175
    Science Advisor
    Gold Member

    I would agree. In this case, More was Less, as it added confusion. Text book authors should always realise that many of their readers are coming to what is in the book for the first time. 'You and I' can look at statements and see them in the light of experience. A student, on his or her own, when presented with some 'helpful' embellishment can easily be thrown into panic.

    Words need to be very carefully selected in a one way statement, whereas, in a conversation, the two parties can arrive at an understanding much more reliably. That's where PF comes in handy.
     
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