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problem33_09

# problem33_09 - sin sin sin sin sin sin index refractive 2 2...

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33.9: a) Let the light initially be in the material with refractive index n a and let the third and final slab have refractive index n b Let the middle slab have refractive index n 1 1 1 sin sin : interface 1st θ n n a a = b b n n sin sin : interface 2nd 1 1 = . sin sin gives equations two the Combining b b a a n n = b) For N slabs, where the first slab has refractive index n a and the final slab has .
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Unformatted text preview: sin sin , , sin sin , sin sin , index refractive 2 2 2 2 1 1 1 1 b b N N a a b n n n n n n n = = =-- of angle on the depends travel of direction final The . sin sin gives This b b a a n n = incidence in the first slab and the indicies of the first and last slabs....
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