p37_054 - 2 = 2 d sin 2 = 2(0 . 94 nm) sin 1 . 15 = 0 . 038...

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54. We use Eq. 37-31. From the peak on the left at angle 0 . 75 (estimated from Fig. 37-38), we have λ 1 =2 d sin θ 1 =2(0 . 94 nm) sin(0 . 75 )=0 . 025 nm = 25 pm . This estimation should be viewed as reliable to within ± 2 pm. We now consider the next peak:
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Unformatted text preview: 2 = 2 d sin 2 = 2(0 . 94 nm) sin 1 . 15 = 0 . 038 nm = 38 pm . One can check that the third peak from the left is the second-order one for 1 ....
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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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