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Unformatted text preview: 600 = 4 d (1 . 50) 4502 (18) d = 600nm (19) Therefore d = 600nm (20) I get m 360 = 9 , m 450 = 7 and m 600 = 5 . To get bright spot, I can ask when is light constructively interfering? Since we want to look for visible spectrum, its safe to pick m = 6 (since its between 450 and 600). Then 6 = 4 600(1 . 50)  (21) 7 = 4(600)(1 . 50) (22) = 514 (23) Thus = 514nm (24) Other m might work, but its getting close to edge of visible spectrum. Last one requires no calculation since n 3 = 1 . 6 would only cause extra phase shift of to ray 2. Which means old maxima is new minima, and old minima is now new maxima. 2...
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This note was uploaded on 10/05/2009 for the course PHYSICS 7C taught by Professor Lin during the Spring '08 term at Berkeley.
 Spring '08
 LIN
 Quantum Physics

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