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Thickness of the oxide layer on bismuth that produces 550nm wave length


emcelhannon

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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

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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 by Griffon
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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.

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