332CHAPTER 24 From the figure, note that ()2222 22Rr Rt rR Rtt2=+−=+− +which reduces to . Since twill be very small in comparison to either ror R, we may neglect the term , leaving 22rRt=−2t2t2≈t. For a dark fringe, 2filmmntλ=so the radii of the dark rings will be 22filmmn≈=filmmRnfor 0, 1, 2,m=…24.65If 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 11waterairtantan1.3353.1pnnθ−−==() ()=°From the triangle RST is the sketch, the horizontal distance from the point of refection, T, to shore is given by ( )90.0 m tan90.0 m1.331px20 m(=and from triangle ABT, the horizontal distance from the antenna to this point is
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