Is diffraction at a maximum when the width of the gap is similar in length to the wavelength of the light? (ie. W ≈ λ, where w=width)
Or is it a maximum when the width is smaller than the wavelength? (i.e W < λ)
Thanks for the help!
Depends on what you mean by "diffraction at a maximum".
Diffraction is best thought of as "spread" of a wave travelling forwards. i.e. A wave travelling forwards will "spread" after passing through a slit. The amount of this spreading is characterised by λ/W: as λ/W becomes large (W < λ), the wave passing the slit approach a uniform radial wave.
A commonly confused interpretation of diffraction is that passing a wave through a slit causes an interference pattern. You should avoid this interpretation, since an interference pattern (of peaks and troughs) only exist if W > λ :
If you follow the derivation for the simplest single slit diffraction (
wikipedia), we arrive at the formula sin(θ)=λ/W, where θ is the angular location of the first minimum of the interference pattern. You can then see that, if λ/W>1 => W<λ, the above relation does not have a solution, then the first antinode is not visible.
The behaviour of a single-slit experiment is then:
- λ/W < 1 ==> λ < W, interference pattern is produced. The interference pattern becomes narrower as W increases (less smearing). Conversely, as W approaches λ, the interference pattern broadens with the central peak becoming wider and wider (more smearing), until...
- λ/W ~ 1 ==> λ ~ W, interference pattern widens so much that the entire 180degree field of view is filled by the central peak. Maximum amplitude at the center, zero amplitude along the walls.
- λ/W < 1 ==> λ > W, approach uniform radial wave, maximum amplitude at the center, amplitude along the walls are non-zero.