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1001_Physics ProblemsTechnical Physics

1001_Physics ProblemsTechnical Physics - 342 P35.70 The...

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342 The Nature of Light and the Laws of Geometric Optics P35.70 From the sketch, observe that the angle of incidence at point A is the same as the prism angle θ at point O . Given that θ = ° 60 0 . , application of Snell’s law at point A gives 1 50 1 00 60 0 . sin . sin . β = ° or β = ° 35 3 . . From triangle AOB , we calculate the angle of incidence (and reflection) at point B . FIG. P35.70 θ β γ = °− + °− = ° 90 0 90 0 180 . . b g b g so γ θ β = = °− °= ° 60 0 35 3 24 7 . . . . Now, using triangle BCQ : 90 0 90 0 90 0 180 . . . °− + °− + °− = ° γ δ θ b g b g a f . Thus the angle of incidence at point C is δ θ γ = °− = °− °= ° 90 0 30 0 24 7 5 30 . . . . a f . Finally, Snell’s law applied at point C gives 1 00 1 50 5 30 . sin . sin . φ = ° or φ = ° = ° sin . sin . . 1 1 50 5 30 7 96 a f . P35.71 (a) Given that θ 1 45 0 = ° . and θ 2 76 0 = ° . . Snell’s law at the first surface gives n sin . sin . α = ° 1 00 45 0 (1) Observe that the angle of incidence at the second surface is
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