987_Physics ProblemsTechnical Physics

# 987_Physics - 328 P35.31 The Nature of Light and the Laws of Geometric Optics Taking to be the apex angle and min to be the angle of minimum

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328 The Nature of Light and the Laws of Geometric Optics P35.31 Taking Φ to be the apex angle and δ min to be the angle of minimum deviation, from Equation 35.9, the index of refraction of the prism material is n = + sin sin min Φ Φ bg 2 2 Solving for min , min sin sin sin . sin . . . = F H G I K J −= °− ° = ° −− 2 2 2 2 20 25 0 50 0 86 8 11 n Φ Φ af a f . P35.32 Note for use in every part: Φ+ °− + = ° 90 0 90 0 180 23 .. θθ so 32 =− Φ . At the first surface the deviation is αθ θ 12 . At exit, the deviation is βθ θ 43 . The total deviation is therefore δαβθ θ θ θ θ θ =+=+−−=+− 142314 Φ . FIG. P35.32 (a) At entry: nn 22 sin sin = or θ 2 1 48 6 150 30 0 = ° F H G I K J sin sin . . Thus, 3 60 0 30 0 30 0 ° . . At exit: 1 50 30 0 1 00 4 .s i n. i n °= or 4 1 300 486 = ° sin . sin . . so the path through the prism is symmetric when 1 48 6 (b) + ° ° = ° 48 6 48 6 60 0 37 2 ... . (c) At entry: sin sin . . . 45 6 28 4 = ° ⇒= ° 3 60 0 28 4 31 6 ° . . At exit: sin . sin . . 44 316 517 = ° a f + ° ° = ° 45 6 51 7 60 0 37 3 . . (d) At entry: sin sin . . . 51 6 31 5 = ° ° 3 60 0 31 5 28 5 ° . . At exit:
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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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