p36_028 - . This leads to two corresponding angle values: =...

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28. Setting I =2 I 0 in Eq. 36-21 and solving for the smallest (in absolute value) two roots for φ/ 2, we Fnd φ =2cos 1 µ 1 2 = ± π 2 rad . Now, for small θ in radians, Eq. 36-22 becomes φ =2 πdθ/λ
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Unformatted text preview: . This leads to two corresponding angle values: = 4 d . The dierence between these two values is = 4 d ( 4 d ) = 2 d ....
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