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Unformatted text preview: TOTAL INTERNAL REFLECTION For an electromagnetic wave in a high index region incident upon a low index re—
gion at an angle of incidence (01) greater than the critical angle [0c 2 sin’1(n2 / 711)],
total internal reﬂection occurs. The total internal reﬂection process inherently includes
a phase shift upon reﬂection and a lateral shift in the apparent point of reﬂection — the
G005 Hanchen shift. For total internal reﬂection an exponentially decaying ﬁeld exists
in the lower refractive index material. If there is a small drop in refractive index across
the boundary, then the evanescent ﬁeld will extend over larger distances. If there is a
large change in the refractive index across the boundary, the ﬁeld will decay abruptly.
The Goos-H'anchen shift may be viewed as the total internal reﬂection occurring at the
1 / e depth of the decaying ﬁeld. The phase shift may be viewed as being due to the ex—
tra distance light has to travel (into the evanescent ﬁeld) in order to be reﬂected. These effects are apparent in the energy ﬂow upon total internal reﬂection depicted in Fig. 1. X :2: ‘\ ”$“‘\ -0 4), 4.2x, 0 0 2x, 0.4x, 0.6x. 0.8).. 1.0M 1.2M
Danna along muvha (m A. mun) Fig. 1 Poynting vector energy ﬂow lines for total internal reﬂection for TE polarization
with n1 = 1.5, 712 = 1, and 01 = 45°. A. Mahan and C. V. Bitterli, Appl. Opt. 17, 509
(1978). FRESNEL’S EQUATIONS - PHASE SHIFT UPON REFLECTION For an electromagnetic wave in a high index region incident upon a low index re-
gion at an angle of incidence (01) greater than the critical angle [0,: = sin—1(n2 / 711)],
total internal reﬂection occurs. In this case, the amplitude reﬂection coefﬁcients become complex. For TE polarization (Er) _ n1c0391 —— jx/nfsinQQI—ng
TE Ez' n1 cos 01 + j n1 sin2 01 -— n2 and for TM polarization 2 . . : 71200391 — jnl t/nilsmzﬂl—ng
TM Ei n3 cos 01 + j m n? sin2 01 — 72% where j =' \/—1. The magnitude of these amplitude reﬂection coefﬁcients is unity in each case. The phase shift (4)) upon total internal reﬂection is seen to be _ 2tan_1<\/n§1 sin2 01 — n2?) ¢TE -
n1 cos 01 and vim/n:1 sin2 01 — 723) = 2 tan"1
¢TM ( n3 cos 01 ...
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