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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 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
= 15.000 .
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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- Fall '08