73. Let θ1= 45° be the angle of incidence at the first surface and 2be the angle of refraction there. Let 3be the angle of incidence at the second surface. The condition for total internal reflection at the second surface is nsin 3≥1. We want to find the smallest value of the index of refraction nfor which this inequality holds. The law of refraction, applied to the first surface, yields nsin 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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