P36_075 - 75(a A path length difference of /2 produces the first dark band of 3/2 produces the second dark band and so on Therefore the fourth

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Unformatted text preview: 75. (a) A path length difference of λ/2 produces the first dark band, of 3λ/2 produces the second dark band, and so on. Therefore, the fourth dark band corresponds to a path length difference of 7λ/2 = 1750 nm. (b) In the small angle approximation (which we assume holds here), the fringes are equally spaced, so that if ∆y denotes the distance from one maximum to the next, then the distance from the middle of the pattern to the fourth dark band must be 16.8 mm = 3.5∆y . Therefore, we obtain ∆y = 16.8/3.5 = 4.8 mm. ...
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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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