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Unformatted text preview: θ = sin − 1 p 2(2 a a cm ) 5 a a cm P = sin − 1 p 4 5 P Such that: θ = 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 λ n 2 is required, where: λ n = λ n This can be accomplished by adding the diamond block at the upper slit. The number, N , of wavelengths, λ n , crossing a length L is: N = L λ n = ⇒ L = Nλ n = Nλ 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
- Light, Wavelength, θ, diamond block