This question appears somewhat similar to other questions asking about why wavelength affects diffraction (a concept which I'm still not 100% sure on...) however my query is different and not answered that I can find. (The focus of my question is to what degree slit size affects diffraction in terms of the wavelength, not how or why) I was wondering, to what degree do the wavelength and the size of the slit have to be similar for diffraction to be reasonably observable (for example in the classic wave tank example) Does diffraction become negligible at 100x the wavelength? 1000x it? And is this different for longitudinal and transverse waves?
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$\begingroup$ Wavelength doesn't affect diffraction at all, especially if you handle all your spatial values in units of wavelength. $\endgroup$– Carl WitthoftCommented May 3, 2016 at 19:40
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2$\begingroup$ I'm sure it does? Waves diffract most significantly when the gap they diffract through is the same as their wavelength dont they? Or is the whole A level physics course a lie? $\endgroup$– JacobGunnCommented May 3, 2016 at 19:47
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$\begingroup$ @JacobGunn, first of all, remember that the diffraction will never be negligible, because there'll always be single-edge diffraction from each edge of the slit regardless of the slit's size...though the angle of that single-edge diffraction is always quite less compared to that of "double-edge" diffraction when the aperture-width is comparable to the wavelength. Second of all, I think you lose that pronounced diffraction if the slit is even 2x the wavelength, though I'm not sure about that and I don't have any actual figures. $\endgroup$– David ReishiCommented May 3, 2016 at 20:16
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1$\begingroup$ See physics.stackexchange.com/q/95126 and physics.stackexchange.com/q/125903. $\endgroup$– sammy gerbilCommented May 4, 2016 at 2:59
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1$\begingroup$ Possible duplicate of Relationship between slit size and wavelength in diffraction $\endgroup$– John RennieCommented May 4, 2016 at 7:06
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1 Answer
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Whether the amount of diffraction is 'negligible' depends on how you define this criterion.
The first order minimum in the diffraction pattern from a single slit occurs where $\sin\theta = \lambda/d$ where $d$ is slit width, $\theta$ is diffraction angle and $\lambda$ is wavelength. If $d = \lambda$ the central lobe of the diffraction pattern will spread out $90$ degrees above and below the axis, filling the whole screen. If $d = 2\lambda$ the central lobe will spread to $30$ degrees above and below the axis. To achieve $\theta = 1\ \mathrm{degree}$ ($\sin\theta = 0.01745$) we need $d = 60 \lambda$ approx.
It makes no difference if the wave is longitudinal or transverse. The same formulas apply to both, unless polarisation is involved, because longitudinal waves cannot be polarised.
answered May 4, 2016 at 3:28
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$\begingroup$ Hi sammy. Note we have mathjax enabled on this site, so you can use latex-style markup to make math flow smoother. $\endgroup$– user10851Commented May 4, 2016 at 3:52
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$\begingroup$ @ChrisWhite : Thanks, I am aware of that but I've never learnt how to use LaTex. I'm probably too set in my ways to change at my age! $\endgroup$ Commented May 4, 2016 at 4:03
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$\begingroup$ @Sammy gerbil, I'm curious. What do you say when someone who's obviously knowledgeable says, "No, there's absolutely no relation between wavelength and diffraction." Do you say, "Excuse me, dear Sir or Madam, but I believe you may be using the word 'diffraction' to mean interference?" $\endgroup$ Commented May 4, 2016 at 16:32
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$\begingroup$ @DavidReishi : People can take offence when you helpfully point out their mistakes, so I try to restrain myself when tempted to correct them. Carl Witthoft was saying that diffraction scales with wavelength, so I presume that is what he meant by saying that it does not depend of wavelength. $\endgroup$ Commented May 4, 2016 at 16:55
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$\begingroup$ @Sammy gerbil, I was actually referring to the phenomenon generally in which physicists confuse diffraction and interference. It's very prevalent. I understand that Carl Witthoft may have been saying something else. $\endgroup$ Commented May 4, 2016 at 17:03