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A point light source S sits at the bottom of a stack of plane-parallel layers. The bottommost layer has refractive index mu0. The n-th layer above the bottom has refractive index mu(n) = mu0 + mu0/(4n - 18). A ray leaves S at an angle of incidence i = 30° measured from the vertical. At the upper surface of which layer does total internal reflection (TIR) first occur?

1. n = 3
2. n = 2
3. n = 1
4. n = 4

Correct answer: n = 3

Solution

The product mu*sin(theta) = mu0*sin30° = mu0/2 is conserved across all interfaces. TIR at the top of layer n occurs when mu(n+1) <= mu0/2, i.e., mu0(1 + 1/(4(n+1)-18)) <= mu0/2. For n = 3, mu(4) = mu0(1 + 1/(16-18)) = mu0(1 - 1/2) = mu0/2, satisfying the critical condition exactly.

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