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988_Physics ProblemsTechnical Physics

# 988_Physics ProblemsTechnical Physics - Chapter 35 P35.34...

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Chapter 35 329 P35.34 At the first refraction, 1 00 1 2 . sin sin θ θ = n . The critical angle at the second surface is given by n sin . θ 3 1 00 = , or θ 3 1 1 00 = F H G I K J sin . n . But 90 0 90 0 180 2 3 . . °− + °− + = ° θ θ b g b g Φ which gives θ θ 2 3 = Φ . Φ θ 2 θ 1 θ 3 FIG. P35.34 Thus, to have θ 3 1 1 00 < F H G I K J sin . n and avoid total internal reflection at the second surface, it is necessary that θ 2 1 1 00 > F H G I K J Φ sin . n . Since sin sin θ θ 1 2 = n , this requirement becomes sin sin sin . θ 1 1 1 00 > F H G I K J L N M O Q P n n Φ or θ 1 1 1 1 00 > F H G I K J L N M O Q P F H G I K J sin sin sin . n n Φ . Through the application of trigonometric identities, θ 1 1 2 1 > F H I K sin sin cos n Φ Φ . P35.35 For the incoming ray, sin sin θ θ 2 1 = n . Using the figure to the right, θ 2 1 50 0 1 66 27 48 b g violet = ° F H G I K J = ° sin sin . . . θ 2 1 50 0 1 62 28 22 b g red = ° F H G I K J = ° sin sin . . . . For the outgoing ray, θ θ
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