30_Chapter 24 HomeworkCH24 Wave Optics

30_Chapter 24 - 332 CHAPTER 24 From the figure note that R2 = r 2 R t = r 2 R 2 2Rt t 2 which reduces to 2 r 2 = 2Rt t 2 Since t will be very small

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332 CHAPTER 24 From the figure, note that () 2 22 2 2 2 Rr R t rR R t t 2 = +−=+− + which reduces to . Since t will be very small in comparison to either r or R , we may neglect the term , leaving 2 2 rR t =− 2 t 2 t 2 t . For a dark fringe, 2 film m n t λ = so the radii of the dark rings will be 2 2 film m n    = film mR n for 0, 1, 2, m = 24.65 If the signal from the antenna to the receiver station is to be completely polarized by reflection from the water, the angle of incidence where it strikes the water must equal the polarizing angle from Brewster’s law. This is given by 11 water air tan tan 1.33 53.1 p n n θ −− == () () = ° From the triangle RST is the sketch, the horizontal distance from the point of refection, T, to shore is given by ( ) 90.0 m tan 90.0 m 1.33 1 p x 20 m ( = and from triangle ABT, the horizontal distance from the antenna to this point is
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This document was uploaded on 10/20/2011.

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