ch35-p117 - 117. (a) With λ = 0.5 µm, Eq. 35-14 leads to...

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Unformatted text preview: 117. (a) With λ = 0.5 µm, Eq. 35-14 leads to θ = sin −1 b3gb0.5 µmg = 48.6° . 2.00 µm (b) Decreasing the frequency means increasing the wavelength — which implies y increases, and the third side bright fringe moves away from the center of the pattern. Qualitatively, this is easily seen with Eq. 35-17. One should exercise caution in appealing to Eq. 35-17 here, due to the fact the small angle approximation is not justified in this problem. (c) The new wavelength is 0.5/0.9 = 0.556 µm, which produces a new angle of θ = sin −1 b3gb0.556 µmg = 56.4° . 2.00 µm Using y = D tan θ for the old and new angles, and subtracting, we find b g ∆y = D tan 56.4°− tan 48.6° = 149 m. . ...
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