p36_087 - 87. (a) The path length difference is 0.5 µm =...

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Unformatted text preview: 87. (a) The path length difference is 0.5 µm = 500 nm, which is represents 500/400 = 1.25 wavelengths – that is, a meaningful difference of 0.25 wavelengths. In angular measure, this corresponds to a phase difference of (0.25)2π = π/2 radians. (b) When a difference of index of refraction is involved, the approach used in Eq. 36-9 is quite useful. In this approach, we count the wavelengths between S1 and the origin N1 = Ln L n + λ λ where n = 1 (rounding off the index of air), L = 5.0 µm, n = 1.5 and L = 1.5 µm. This yields N1 = 18.125 wavelengths. The number of wavelengths between S2 and the origin is (with L2 = 6.0 µm) given by L2 n N2 = = 15.000 . λ Thus, N1 − N2 = 3.125 wavelengths, which gives us a meaningful difference of 0.125 wavelength and which “converts” to a phase of π/4 radian. ...
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