Diffraction wavelength relationship [closed]

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?

closed as off-topic by Carl Witthoft, user36790, John Rennie, CuriousOne, honeste_vivereMay 4 '16 at 16:18

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• Wavelength doesn't affect diffraction at all, especially if you handle all your spatial values in units of wavelength. – Carl Witthoft May 3 '16 at 19:40
• 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? – JacobGunn May 3 '16 at 19:47
• @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. – David Reishi May 3 '16 at 20:16
• – sammy gerbil May 4 '16 at 2:59
• Possible duplicate of Relationship between slit size and wavelength in diffraction – John Rennie May 4 '16 at 7:06

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.