problem35_45

problem35_45 - . 2 π Thus the path length difference for...

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35.45: a) If the two sources are out of phase by one half-cycle, we must add an extra half a wavelength to the path difference equations Eq. (35.1) and Eq. (35.2). This exactly changes one for the other, for , and 2 1 2 1 m m m m + + since m in any integer. b) If one source leads the other by a phase angle φ , the fraction of a cycle difference is
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Unformatted text preview: . 2 π Thus the path length difference for the two sources must be adjusted for both destructive and constructive interference, by this amount. So for constructive inference: , λ ) 2 ( 2 1 + =-m r r and for destructive interference, . λ ) 2 2 1 ( 2 1 + + =-m r r...
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This note was uploaded on 01/31/2009 for the course PHYS 221 taught by Professor Hu during the Spring '08 term at Texas A&M.

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