p37_084 - 84(a We require that sin = m1,2/d sin 30 where m...

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84. (a) We require that sin θ = 1 , 2 /d sin 30 ,where m =1 , 2and λ 1 = 500 nm. This gives d 2 λ s sin 30 = 2(600 nm) sin 30 = 2400 nm . For a grating of given total width L we have N = L/d d 1 , so we need to minimize d to maximize R = mN d 1 .Thu sw echoo se d = 2400 nm. (b) Let the third-order maximum for
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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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