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 12 .s i n s i n θθ = n . The critical angle at the second surface is given by n sin . θ 3 100 = , or 3 1 = F H G I K J sin . n . But 90 0 90 0 180 23 .. °− + °− + = ° bg Φ which gives =− Φ . Φ 2 1 3 FIG. P35.34 Thus, to have 3 1 < F H G I K J sin . n and avoid total internal reflection at the second surface, it is necessary that 2 1 >− F H G I K J Φ sin . n . Since sin sin = n , this requirement becomes sin sin sin . 1 1 F H G I K J L N M O Q P n n Φ or 1 11 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, 1 1 F H I K sin sin cos n ΦΦ . P35.35 For the incoming ray, sin sin 2 1 = n . Using the figure to the right, 2 1 50 0 166 27 48 violet = ° F H G I K J sin sin . . . 2 1 50 0 162 28 22 red = ° F H G I K J sin sin . . For the outgoing ray, 32 60 0 . FIG. P35.35 and sin sin 43 = n : 4 1 3252 6317 violet = ° sin . sin . . 4 1 3178
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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