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Author Topic: Super quick question on diffraction (unit 4 light)  (Read 1060 times)  Share 

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The_Nster

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Super quick question on diffraction (unit 4 light)
« on: October 22, 2015, 04:26:43 pm »
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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!

jyce

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Re: Super quick question on diffraction (unit 4 light)
« Reply #1 on: October 24, 2015, 10:35:18 pm »
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My understanding is that the first statement you gave applies to the diffraction of LIGHT, while the second statement applies to SOUND.

Mao

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Re: Super quick question on diffraction (unit 4 light)
« Reply #2 on: October 27, 2015, 08:53:45 pm »
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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.
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