p36_087.pdf

87. (a) The path length diﬀerence is 0 . 5 µ m = 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)2 π = π/ 2radians.

(b) When a diﬀerence ofindex ofrefraction is involved, the approach used in Eq. 36-9 is quite useful.

In this approach, we count the wavelengths between S 1 and the origin

N 1 = Ln

+ L n

where n = 1 (rounding oﬀ the index ofair), L =5 . 0 µ m, n =1 . 5and L =1 . 5 µ m. This

yields N 1 =18 . 125 wavelengths. The number ofwavelengths between S 2 and the origin is (with

L 2 =6 . 0 µ m) given by

N 2 = L 2 n

=15 . 000 .

N 2 =3 . 125 wavelengths, which gives us a meaningful diﬀerence of 0 . 125 wavelength

and which “converts” to a phase of π/ 4radian.

Thus, N 1 −