1060_Physics ProblemsTechnical Physics

1060_Physics ProblemsTechnical Physics - Chapter 37 2nt = m...

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Chapter 37 401 P37.58 (a) Minimum: 2 2 nt m = λ for m = 012 ,,, Maximum: 2 1 2 1 nt m =′ + F H G I K J for ′ = m for λλ 12 > , ′+ F H G I K J < mm 1 2 so ′= 1. Then 2 1 2 21 nt m m == F H G I K J 22 1 =− so m = 1 2 bg . (b) m = =→ 500 2500 370 192 2 af . (wavelengths measured to ± 5 nm ) Minimum: 2 2 nt m = 2140 2370 . af a f t = nm t = 264 nm Maximum: 1 2 15 nt m + F H G I K J = . 2 1 40 1 5 500 .. a f t = nm t = 268 nm Film thickness = 266 nm . P37.59 From the sketch, observe that xh dh d =+ F H G I K J = + 2 2 2 4 2 . Including the phase reversal due to reflection from the ground, the total shift between the two waves is δ 2 2 xd . d d/ 2 x x h FIG. P37.59 (a) For constructive interference, the total shift must be an integral number of wavelengths, or
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