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Unformatted text preview: 117. (a) With λ = 0.5 µm, Eq. 3514 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. 3517. One should exercise caution in appealing
to Eq. 3517 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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 Spring '08
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