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Unformatted text preview: âˆ’ 1 p mÎ» d P Substituting values: Î¸ = sin âˆ’ 1 p 2(2 a a cm ) 5 a a cm P = sin âˆ’ 1 p 4 5 P Such that: Î¸ = 0 . 927 rad = 53 . 1 â—¦ (b) If a diamond block of index of refraction n =2 . 4 is placed behind one of the slits, what thickness of diamond would cause the second maximum, to become a minimum? So that the maximum becomes a minimum, a wavelength shift of Î» 2 is required. This can be accomplished by adding the diamond block at the upper slit. The number, N , of wavelengths, Î» , crossing a length L is: N = L Î» = â‡’ L = NÎ» = 1 2 (2 cm ) Finally, in order to cause a microwave maximum at the screen to become a minimum, the length of the diamond block must be: L = 1 cm . Pretty BIG diamond!!! 2...
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 Spring '10
 Hanlet
 Work, Light, Wavelength, narrow parallel slits, 0.927 rad

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