emcelhannon Posted April 10, 2013 Share Posted April 10, 2013 How thick would the layer of bismuth oxide on pure Bi metal need to be to produce green, and what would the progression of colors be as the layer thickens? I think we would apply Fresnell's Equations, but I'm not sure how. The refractive index of Bi2O3 is 2.5/0.4 according to a Chinese distributor of the yellow powder form. I'm not sure if that applies on the thin oxide layer on pure metal. Many thanks to whoever can help me understand. Ernie Link to comment Share on other sites More sharing options...
Griffon Posted April 10, 2013 Share Posted April 10, 2013 (edited) Constructive interference in a thin film of thickness d and refractive index n for a particular wavelength λ is given by d = Nλ / 2n for N = 1, 2, 3 .... So with increasing thickness the colours progress through blue, green, yellow, red. Repeating with each order of N = 1, 2, 3 .... The equation above applies for light reflected perpendicularly from the surface. At a reflecting angle θ the equation for constructive interference is d = Nλ / 2ncos(θ) Edited April 10, 2013 by Griffon 1 Link to comment Share on other sites More sharing options...
Enthalpy Posted April 11, 2013 Share Posted April 11, 2013 http://stoner.phys.uaic.ro/old/ANALE/Anale_1999_2000/An_Univ_Iasi_1999_2000_17.pdf third hit by Google. 1 Link to comment Share on other sites More sharing options...
emcelhannon Posted April 16, 2013 Author Share Posted April 16, 2013 Many thanks, It will take me a couple of days to digest this. E. Link to comment Share on other sites More sharing options...
John Cuthber Posted April 16, 2013 Share Posted April 16, 2013 Figure two shows how the refractive index (n) of three different sample varies with wavelength. Near 550nm the refractive index varies from about 1.6 to 1.9 for the 3 samples. So you don't know what value to put into the equation Griffon has provided. This sort of problem is usually solved empirically. Link to comment Share on other sites More sharing options...
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