This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Snell’s Law Implications An electromagnetic plane wave with wavevector E1 is propagating in region 1, a
homogeneous isotropic dielectric of refractive index 77.1 as shown in the diagram. This
ray is incident upon a plane interface between region 1 and region 2. The 2 axis is nor—
mal to this interface. The second region is a homogeneous isotropic dielectric of refrac
tive index 712. The incident ray is refracted into region 2 where it now has a wavevector
E2. The refraction of this wave is described by Snell’s law which states that in the plane
of incidence sindl n2 , = — = constant.
31.7102 711 That is, the ratio of the sines of the incidence and refraction angles is a constant. This ratio is equal to the same constant regardless of the angle of incidence. That is,
siné’l/ 317162 is not a function of 61 (or 62). Note that this is not true, in general, for
birefringent materials. Snell’s law does not hold, in general, for these materials. A ray
incident upon a birefringent material is refracted into two rays at two different angles
of refraction! In general, neither refracted ray is described by Snell’s law. However, for
both isotropic and birefringent materials, the incident and refracted waves are phase
matched. That is, the tangential component of the wavevector along the interface is the same for the incident wave and for the refracted wave. Thus, 7;; sinﬁl = sindg,
where = 27m1/A, = 27rn2/A, and /\ is the freespace wavelength. Snell’s law occurs in isotropic materials when phase matching is applied to the plane of incidence
(the plane containing F1 and the normal to the interface, However, phase matching applies to any plane normal to the interface. Using the above information, determine whether each of the following mathemat—
ical statements is true or false. Circle the correct response. For each answer, provide a complete written justiﬁcation and / or a mathematical proof of your choice. (151 = $2 (tl‘ue)(false)
ktl : ktg (true)(false)
kxl = x2 (true)(false)
kzl = 22 (true) (false)
km 2 kHz (true)(false) n1 sinﬁnl :2 71231710232 (true) (false) Snell’s Law Implications The incident wavevector is 79: = 27:11 (cos¢ 32'an 5: + singb sin01 1) + 00801 2)
and the refracted wavevector is [6—2 = 27:“ (cosqb 3757162 2”: + singb sin02 1} + c0392 2)
(1) (J51 ; ¢2 The refracted ray must lie in the plane of incidence. Thus ¢1 = ¢2. (2) kn ; kt2
7 27mg 27mlsz'n6 —
A 1— ,\ M7102 Since n1 sinﬂl = 112 sinﬁz, then it follows that kg = km. ? kzl = k$2
2 2
7:11 cosgb sinéll ; 71;” cosqﬁ sin62 Since 711 sin01 2 n2 sin02, then it follows that kml = 19,32. (4) kzl ; kz2 27rn16080 l 271'712
A 1‘ ,\ 00362 ?
n1 c0301 = 722 00362 This is not in agreement with m sin61 = n2 327102; thus it follows that kzl aé 1922. k:le kzz2 — 2 knl = 7:1 (cosd) sinﬂl a“: + 00361 2) — 2W . A .
[cm1 = — (cosdJ n1 317191 m + n1 c0301 2) A — 2
kmz = 7:12 (cosqﬁ sin62 (f: + 00362 2) — 2
kng = —7r (c03¢ n2 sin02 5: + 712 c0302 2) A Since n1 52'an = n2 sin02, the 3: components are the same. However, the 2 components are obviously different. Thus kml 75 [92:52. . 7 .
(6) nlszné‘nl = 712517102352 , 0 cos¢ sin61
szn ml = ———.————
(003sz 3171261 + 003201)1/2
n1 cosqﬁ sin61 7 n2 c0545 sin02 (cos2qb 3272261 + cos2t91)1/2 — (c052¢ 3271292 + 008202)1/2
Using n1 sinél = n2 sin62, then
7 (0032(1) sin201 + 005201) é (c052¢ sin202 + c03262) This is obviously not true. Thus n1 sinem 74 ngsin6zzg. ...
View
Full
Document
This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Gaylord

Click to edit the document details