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Unformatted text preview: Physics 505 Fall 2007 Homework Assignment #11 — Solutions Textbook problems: Ch. 7: 7.3, 7.4, 7.6, 7.8 7.3 Two plane semiinfinite slabs of the same uniform, isotropic, nonpermeable, lossless dielectric with index of refraction n are parallel and separated by an air gap ( n = 1) of width d . A plane electromagnetic wave of frequency ω is indicent on the gap from one of the slabs with angle of indicence i . For linear polarization both parallel to and perpendicular to the plane of incidence, a ) calculate the ratio of power transmitted into the second slab to the incident power and the ratio of reflected to incident power; We introduce (complex) electric field vectors of the form ~ E i e i ~ k · ~x and ~ E r e i ~ k · ~x on the incident side, ~ E + e i ~ k · ~x and ~ E e i ~ k · ~x in the air gap, and ~ E t e i ~ k · ( ~x ~ d ) on the transmitted side. (We have removed an unimportant phase from the transmitted side by shifting ~x by the vector ~ d pointing from the incident to the transmitted side of the air gap). d k k’ k’ k k i n n n =1 (air) r If i is the incident angle, then the angle r from the normal in the air gap is given by Snell’s law, n sin i = sin r , and the transmitted angle is also i (because it is the same dielectric). We see that cos r = p 1 sin 2 r = p 1 n 2 sin 2 i and that cos r is purely imaginary in the event that i is greater than the critical angle for total internal reflection. To obtain E t and E r in terms of E i , we may match the parallel components of ~ E as well as the parallel components of ~ H . We consider two cases. For ~ E perpendicular to the plane of incidence, the matching becomes first interface second interface E k : E i + E r = E + + E , E + e iφ + E e iφ = E t H k : n ( E i E r )cos i = ( E + E )cos r, ( E + e iφ E e iφ )cos r = nE t cos i where we have introduced the phase φ = ~ k · ~ d = k d cos r = ωd cos r c The matching conditions at the first interface may be written as E + = 1 2 E i (1 + α ) + 1 2 E r (1 α ) E = 1 2 E i (1 α ) + 1 2 E r (1 + α ) (1) where we have defined α = n cos i cos r = n cos i p 1 n 2 sin 2 i Similarly, the matching conditions at the second interface yield E + = 1 2 e iφ E t (1 + α ) E = 1 2 e iφ E t (1 α ) (2) Equating (1) and (2) allows us to solve for the ratios E t E i = 4 α (1 + α ) 2 e iφ (1 α ) 2 e iφ = 2 α 2 α cos φ i (1 + α 2 )sin φ E r E i = (1 α 2 )( e iφ e iφ ) (1 + α ) 2 e iφ (1 α ) 2 e iφ = i (1 α 2 )sin φ 2 α cos φ i (1 + α 2 )sin φ (3) where α = n cos i p 1 n 2 sin 2 i , φ = ωd cos r c = ωd p 1 n 2 sin 2 i c So long as i is below the critical angle, both α and φ are real. In this case, the transmission and reflection coefficients are T = E t E i 2 = 4 α 2 4 α 2 cos 2 φ + (1 + α 2 ) 2 sin 2 φ = 4 α 2 4 α 2 + (1 α 2 ) 2 sin 2 φ R = E r E i 2 = (1 α 2 ) 2 sin 2 φ 4 α 2 cos 2 φ + (1 + α 2 ) 2 sin 2 φ = (1 α 2 ) 2 sin 2 φ 4 α 2 + (1 α 2 ) 2 sin 2 φ (4)...
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This note was uploaded on 04/11/2010 for the course PHYSICS PHYS taught by Professor Tony during the Spring '07 term at University of MichiganDearborn.
 Spring '07
 Tony
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