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Unformatted text preview: = sin 1 p 2(2 a a cm ) 5 a a cm P = sin 1 p 4 5 P Such that: = 0 . 927 rad = 53 . 1 2 (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 n 2 is required, where: n = n This can be accomplished by adding the diamond block at the upper slit. The number, N , of wavelengths, n , crossing a length L is: N = L n = L = N n = N n = 1 2 (2 cm ) 2 . 4 Finally, in order to cause a microwave maximum at the screen to become a minimum, the length of the diamond block must be: L = 0 . 42 cm . Pretty BIG diamond!!! 3...
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 Spring '10
 Hanlet

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