Question

1. [50 pts] Given: Figure 6.23, page 366 of the course textbook. Find: Derive the expression for the transverse magnetic reflection coefficient, Itm for a normally incident wave. Solution: HP FIGURE 6.23 TM polarization Answer: ITM 1/2 cos - cos, 1/2 cos 0+11 cos

Answer 1

Waves at Interfaces - TE and TM Waves & Ho & Ho Transverse Electric (TE) wave H Transverse Magnetic (TM) wave H Z=0 Plane of Incidence: The plane containing the incident wavevector k; and a vector that is normal to the interface is called the plane of incidence in the figure above the x-z plane is the plane of incidence) TE Wave: If the E-field of the wave is perpendicular to the plane of incidence then the wave is called a TE-wave TM Wave: If the H-field of the wave is perpendicular to the plane of incidence then the wave is called a TM-wave

TM Wave - Wavevectors ki E HE Transverse Magnetic E, (TM) wave O, @ @ ki E & Ho & Ho H z=0 ni ki = @ Mo & = 0 с kr = ki = V10 6 = " kt = @ Hot = 0 k; = Kix*+kizź = ki (sin(0) +cos(0) 2] kr = krx*+kr22 =k, (sin(e) & -cos(6) 2] ke = kex & + Kxz2 = ke [sin(0) + cos(0) 2] ni nt с ECE 303 - Fall 2006 - Farhan Rana - Cornell University

TM Wave - First Boundary Condition Transverse Magnetic (TM) wave Er & 14 Bt Ab H 20 H(7)\zco =H; e-jK;.' +ý H, e-jk. k;=k; (sin(0) * + cos(0) 2] k= k, (sin(0) 8-cos(6) 2] H(F),30 = Ỳ Hy e-jker kę = kt (sin(@)X + cos(0) 2] Use boundary conditions: (1) At z = 0 the H-field parallel to the interface must be continuous across the interface for all This gives: H; e-jk; sin(@;)* + H, e-jk, sin(@, )* = Hy e-j kų sin(@.)x ECE 303-Fall 2008 - Farhan Rana-Cornell University

TM Wave - Phase Matching Condition Transverse Magnetic (TM) wave E4 H & Ho H 0 H; e-jk; sin(Ⓡ)* + H, e-jk, sin(Q)* = Hy e-jke sin(0,)x The only way the above boundary condition can be satisfied for all x is if all the x- dependent phase factors are the same this is called "phase matching") k; sin(e)k, sin(@r)kt sin(@) The first equality gives using k;=k): sin(@:)=sin(0) 0; = 0, angle of incidence equals the angle of reflection ECE 303 - Fall 2008 - Farhan Rana - Cornell University

TM Wave - Snell's Law Transverse Magnetic (TM) wave & 14 Bt Ho kix = kx = ktx ki sin(@:)=k, sin(@)=k+ sin(@) The second equality gives: k; sin(@})=ksin(0) w" sin(Q;)= 0"* sin(0) = n; sin(0)=nsin(0) H; e-)ksin(@;)* +H, e-ik, sin(@, )x =H → Hi+H, = HT Snell's Law e-ik sin(@)x (1) ECE 303 - Fall 2006 - Farhan Rana - Cornell University

TM Wave - Second Boundary Condition Transverse Magnetic (TM) wave н 20 (2) At z = 0 the E-field component parallel to the interface must be continuous for all x E(),co --R xºm, H, e-jk. – 6, xỹ), H, e-k. k;=k; (sin(@)X + cos(@:) 2] E(7),50 --(kexý) me He ejer kr = kr (sin(er) - cos(er) 2] kt = kt (sin(@) * + cos(6) 2] * cos(0)n: H; e-/ky sin(( )x – cos(@.), H, e-jk, sin(@.)* = * cos(@.)He ej ke sin(s)x cos(@)(7H; -7 Hx) = cos(6) He (2) ECE 303 - Fall 2006 - Farhan Rana - Cornell University

TM Wave - Reflection and Transmission Coefficients kr H Transverse Magnetic (TM) wave & 10 & Ho H 20 The solution is: 2 & Kiz 21/cos(@) тTM ni/cos(@) не , 1/cos(@) Hicos(@ +1 11t/cos(@) Transmission coefficient Ei kz &t kiz +1 si kz ni/cos(@)-1 8t kiz -1 nicos(@) ne/cos@) +1 n/cos(@) ne/cos(0). nicos(0) nt/cos(@) +1 ni/cos(@) rTM H1/cos(@ Hicos(@) 1/cos(@) +1 &ktz & Kiz +1 Ei kiz Reflection coefficient ECE 303 - Fall 2006 - Farhan Rasa - Cornell University