{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4500pp07 - Projection of Refracted Wavevector An...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Projection of Refracted Wavevector An electromagnetic plane wave with wavevector E1 is propagating in region 1, a homogeneous isotropic dielectric of refractive index 711 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 n2. The incident ray is refracted into region 2 where it now has a wavevector 79-2. The refraction of this wave is described by phase matching at the boundary. That is, the tangential component of the wavevector along the interface is the same for the incident wave and for the refracted wave. Thus, sin61 : sinfig, Where = 27m1/A, = 27rn2/A, and A is the freespace wavelength. Thus n1 sin 01 = . n2 sin 02. This result occurs when phase matching is applied to the plane of incidence (the plane containing E and the normal to the interface, However, phase matching applies to any plane normal to the interface. Using the above information, derive, showing all work, an expression for 02152 as a function of 02:51, 01, and 62 only. Eliminate kzrl and kzm2 and all other parameters from your final result. Put your final answer in the space provided. 021:2 — Projection of Refracted Wavevector km = km 372110le 2 k1 sin 01 cos 9251 191:2 Z 15232 Sin 02:52 = ’62 sin 92 COS ¢2 kzl : kzzcl COS 021:1 = ’91 003 01 kzz = [922:2 cos 02x1 : k2 C08 02 Dividing kle Sin 02x1 _ k1 Sin 01 COS ¢1 kzzzl COS 021:1 191 cos 01 tan 9211 = tan 01 cos 431 and since ch 2 962, also ‘ tan 0sz = tan 92 cos o1 Dividing again tan 6 tan 02:52 = tan 02,51 2 tan 01 and so ‘ tan 0 021,2 2 tan’1[tan0zzl 2 tan 01 1 ...
View Full Document

{[ snackBarMessage ]}