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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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 Spring '08
 Any
 Physics

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