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Unformatted text preview: when cos 2 Ï€ L Î» ( n 2 n 1 ) = 0, or Ï€ L Î» ( n 2 n 1 ) = (2 m +1) Ï€ 2 , i.e., L ( n 2 n 1 ) = (2 m +1) Î» 2 . There is no interference when I = 2 A 2 (i.e., the sum of the intensities of the two individual waves), i.e., when cos 2 Ï€ L Î» ( n 2 n 1 ) = 1 2 , or Ï€ L Î» ( n 2 n 2 ) = Â± Ï€ 4 + mÏ€ , i.e., L ( n 2 n 1 ) = ( m Â± 1 4 ) Î» . The twoâ€“slit experiment (Youngâ€™s experiment) : The twoâ€“slit experiment describes the interference of light emanating from two narrow slits, when at the two slits the waves are in phase. We found above that when two waves, initially in phase, acquire some phase difference Î” Ï† and interfere, the intensity is I =  Ïˆ  2 = 4 A 2 cos 2 Î” Ï† 2 We therefore need to find the phase difference at the screen. Assuming the screen is very far from the slits (i.e., the distance to the screen D d , where d is the separation between the slits, the only length scale in the problem except for the wavelength Î» .) Under this assump tion, the two rays r 1 , 2 are nearly parallel, which simplifies the geometry. The difference in distance is Î” L = d sin Î¸ , and the difference in number of wavelength is Î” N = Î” L Î» = d Î» sin Î¸ . The phase difference then is Î” Ï† = 2 Ï€ Î” N = 2 Ï€ Î» d sin Î¸ = k d sin Î¸ . We find bright interference fringes (constructive interference) when cos 2 ( Î” Ï†/ 2) = 1, that is when Ï€ ( d/Î» ) sin Î¸ = nÏ€ , or d sin Î¸ = nÎ» . We may interpret this last result as the difference in the optical path (which here equals the physical distance) equaling an integral multiple of the wavelength: Î” L = nÎ» , or d sin Î¸ = nÎ» , where n = 0...
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 Spring '09
 LIORBURKO
 Physics, Diffraction, Wavelength, Michelson interferometer

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