# Choice of wavelength... n For diffraction of light, with condition a sin = n n For n=1, this gave a maximum value of for diffraction: sin = 1 a = n Realistically,

## Presentation on theme: "Choice of wavelength... n For diffraction of light, with condition a sin = n n For n=1, this gave a maximum value of for diffraction: sin = 1 a = n Realistically,"— Presentation transcript:

Choice of wavelength... n For diffraction of light, with condition a sin = n n For n=1, this gave a maximum value of for diffraction: sin = 1 a = n Realistically, sin n So separation must be same order as, but greater than, the wavelength of light.

If a < n Say we are dealing with light of wavelength 500nm n If the separation of the slits were 300nm, then from ( /a) > 1, so the only solution occurs if n=0 (equivalent to the direct, undiffracted beam).

If a >> n Say we are dealing with light of wavelength 500nm n Again using: If the separation of the slits were: 5000nm 10000nm20000nm n=1 =5.74º =2.86º =1.43º n=2 =11.54º =5.74º =2.86º n=3 =17.46º =8.63º =4.30ºetc.

If a >> n Thus, as a increases, the different diffraction orders tend towards a continuum - it would become very difficult to distinguish between them. n Try and work through similar examples for Braggs Law and X-ray diffraction. For example, look at a wavelength of 3.0Å and a d-spacing of 1.2Å. n Then consider a wavelength of 0.71Å and d-spacings of 10Å and above.

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