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1076_Physics ProblemsTechnical Physics

1076_Physics ProblemsTechnical Physics - Chapter 38 P38.26...

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Chapter 38 417 P38.26 sin . θ = 0 350 : d = = = × λ θ sin . . 632 8 1 81 10 3 nm 0.350 nm Line spacing = 1 81 . m µ P38.27 (a) d = = × = × = 1 3 660 2 732 10 2 732 10 2 732 4 6 lines cm cm m nm . . λ θ = d m sin : At θ = ° 10 09 . λ = 478 7 . nm At θ = ° 13 71 . , λ = 647 6 . nm At θ = ° 14 77 . , λ = 696 6 . nm (b) d = λ θ sin 1 and 2 2 λ θ = d sin so sin sin sin θ λ λ λ θ θ 2 1 1 2 2 2 = = = d . Therefore, if θ 1 10 09 = ° . then sin sin . θ 2 2 10 09 = ° a f gives θ 2 20 51 = ° . . Similarly, for θ 1 13 71 = ° . , θ 2 28 30 = ° . and for θ 1 14 77 = ° . , θ 2 30 66 = ° . . P38.28 sin θ λ = m d Therefore, taking the ends of the visible spectrum to be λ v = 400 nm and λ r = 750 nm , the ends the different order spectra are: End of second order: sin θ λ 2 2 1 500 r r d d = = nm . Start of third order: sin θ λ 3 3 1 200 v v d d = = nm . Thus, it is seen that θ θ 2 3 r v > and these orders must overlap
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