ch35-p032 - 32. (a) We can use phasor techniques or use...

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we find 12 0 2 cos( / 2)sin( / 2) EE E t φ ωφ += + where E o = 2.00 μV/m, ω = 1.26 × 10 15 rad/s, and = 39.6 rad. This shows that the electric field amplitude of the resultant wave is 0 2 cos( / 2) 2(2.00 V/m)cos(19.2 rad) 2.33 V/m EE µµ == = . (b) Eq. 35-22 leads to 2 00 4 cos ( / 2) 1.35 I II at point P , and 2 center 0 0 4c o s ( 0 )4 I at the center . Thus, center / 1.35/ 4 0.338 . (c) The phase difference φ (in wavelengths) is gotten from φ in radians by dividing by 2 π. Thus, φ = 39.6/2π = 6.3 wavelengths. Thus, point P is between the sixth side maximum
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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