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 60 0 . , application of Snell’s law at point A gives 1 50 1 00 60 0 . sin . sin . β or 35 3 .. From triangle AOB , we calculate the angle of incidence (and reflection) at point B . FIG. P35.70 θβ γ = ° 90 0 90 0 180 bg so γθβ =−= ° − ° = ° 60 0 35 3 24 7 . . Now, using triangle BCQ : 90 0 90 0 90 0 180 ... °− + °− + °− = ° γδθ a f . Thus the angle of incidence at point C is δθ = ° ° 90 0 30 0 24 7 5 30 . . af . Finally, Snell’s law applied at point C gives 100 150 530 .s i n .s i n . φ or = ° sin . sin . . 1 796 . P35.71 (a) Given that 1 45 0 . and 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 βα 90 0 Thus, Snell’s law at the second surface yields
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