ch33-p073 - 73. Let 1 = 45 be the angle of incidence at the...

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73. Let θ 1 = 45° be the angle of incidence at the first surface and 2 be the angle of refraction there. Let 3 be the angle of incidence at the second surface. The condition for total internal reflection at the second surface is n sin 3 1. We want to find the smallest value of the index of refraction n for which this inequality holds. The law of refraction, applied to the first surface, yields n sin 2 = sin 1 . Consideration of the triangle formed by the surface of the slab and the ray in the slab tells us that 3 = 90° –
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