*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **instruments. Now we better check that the “good conductor” approximation really holds, i.e., is σ >> ωε ? ωε = (2 π )(134)(81)(8 . 85 × 10-12 ) = 6 . 04 × 10-7 σ = 4 is indeed much bigger than this. Problem 12.11 Phase propagation in the x direction must match on either side of the boundary at z = 0. Hence, k ix = k t x = ⇒ k tx = ω c sin θ i , which is independent of the material. k tz = s ω 2 c 2 ± 1-j 30 4 ¶-k 2 tx = ω c s ± 1-j 30 4 ¶-1 4 = ω c √ 3 2 p 1-j 10 Now we ﬁnd the angle of transmission from θ t = tan-1 • k tx <{ k tz } ‚ = tan-1 " 1 / 2 √ 3 2 <{ √ 1-j 10 } # = tan-1 • 1 √ 3 <{ √ 1-j 10 } ‚ = 13 . 8 ◦ CORNELL UNIVERSITY c ± WES SWARTZ (09/9/15) 7–1...

View
Full
Document