p36_068 - × 7 = 28 of these The solutions at m = 0 and m =...

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68. To explore one quadrant of the circle, we look for angles where Eq. 36-14 is satisFed. θ = sin 1 d for m =0 , 1 , 2 ... where mλ/d cannot exceed unity. ±or m =1 ... 7 we have solutions that are “mirrored” in every other quadrant; so there are 4
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Unformatted text preview: × 7 = 28 of these. The solutions at m = 0 and m = 8 are “special” in that they have twins (at 180 ◦ and 270 ◦ , respectively) and their multiplicity is 2, not 4. Thus, we have 28+2(2) = 32 points of maxima....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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