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998_Physics ProblemsTechnical Physics

# 998_Physics ProblemsTechnical Physics - Chapter 35 and = 1...

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Chapter 35 339 P35.64 δθ θ =−= ° 12 10 0 . and nn 11 22 sin sin θθ = with n 1 1 = , n 2 4 3 = . Thus, θ 1 1 1 21 10 0 == ° −− sin sin sin sin . bg b g . (You can use a calculator to home in on an approximate solution to this equation, testing different values of 1 until you find that 1 36 5 . . Alternatively, you can solve for 1 exactly, as shown below.) We are given that sin sin . 4 3 10 0 =− ° . This is the sine of a difference, so 3 4 10 0 10 0 1 sin sin cos . cos sin . −° . Rearranging, sin . cos cos . sin 10 0 10 0 3 4 °= ° F H G I K J sin . cos . . tan 10 0 10 0 0 750 1 ° °− = and 1 1 0740 365 ° tan . . af . P35.65 To derive the law of reflection , locate point O so that the time of travel from point A to point
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