p36_048 - n air = (1 . 00000) 4001 4000 = 1 . 00025 . We...

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48. We apply Eq. 36-25 to both scenarios: m = 4001 and n 2 = n air ,and m = 4000 and n 2 = n vacuum = 1 . 00000: 2 L = (4001) λ n air and 2 L = (4000) λ 1 . 00000 . Since the 2 L factor is the same in both cases, we set the right hand sides of these expressions equal to each other and cancel the wavelength. Finally, we obtain
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Unformatted text preview: n air = (1 . 00000) 4001 4000 = 1 . 00025 . We remark that this same result can be obtained starting with Eq. 36-41 (which is developed in the textbook for a somewhat di±erent situation) and using Eq. 36-40 to eliminate the 2 L/λ term....
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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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