Lecture 9. Transmission and Reflection. Reflection at a Boundary. Specific Boundary. Reflection at a Boundary

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1 Lecture 9 Reflection at a Boundary Transmission and Reflection A boundary is defined as a place where something is discontinuous Half the work is sorting out what is continuous and what is discontinuous at the boundary At an optical boundary the refractive index n changes Leads to a wave returning (reflected) and a wave ongoing (transmitted) Reflection at a Boundary pecific Boundary Continuous across the boundary frequency Continuous at the boundary spatial variation For a dielectric tangential E field tangential H field Consider a boundary at z=0 with a plane wave incident on it angle of incidence 2 wrt to z-axis (normal) electric vector in the x-z plane No y variation We must have in the incident waves 1 k x = 1 ksin2 1 1 k z = 1 kcos2 1

2 pecific Boundary At the Boundary At the interface k 1. r - Tt = k 1-. r - Tt = k 2. r - Tt no y components no z component at interface (z=0) 1+ k x x = 1- k x x = 2 k x x For the reflected wave sin2 1 = sin Law of reflection n 1 sin2 1 = n 2 sin2 2 - nell s Law (n sin2 is conserved) Complex Refractive Index If n 2 is complex (conductor, etc) then life gets fun!! sin2 2 = n 1 sin2 1 / ( n 2 + i6 2 ) Leads to the following expression for the propagation T/c [ x n 1 sin2 1 + z p (n 2 cos q sin q) + izp(6 2 cos q + n 2 sin q)] p, q are functions of n 1, n 2, 6 2, 2 1 but not x, z wave propagates spatially as exp(i k 2. r) amplitude given by imaginary part phase given by real part Inhomogeneous Waves Wave propagates as T/c [ x n 1 sin2 1 + z p (n 2 cos q sin q) + izp(6 2 cos q + n 2 sin q)] amplitude given by complex part of equation surfaces of constant amplitude given by z=constant phase given by real part surfaces of constant phase given by x n 1 sin2 1 + z p (n 2 cos q sin q) = constant surfaces of constant phase not parallel to surfaces of constant amplitude inhomogeneous waves

3 Boundary Conditions Boundary Conditions Boundary conditions at the interface Boundary conditions for dielectrics Tangential E and H fields are continuous ( 1+ E + 1- E ) x n = 2 E x n ( 1+ H + 1- H ) x n = 2 H x n ( 1+ k x 1+ E) x n +( 1- k x 1- E ) x n = ( 2 k x 2 E) x n 1 st Case - Electric field parallel to interface transverse electric (TE) s-polarised 2 nd Case - Magnetic field parallel to interface transverse magnetic field p-polarised - (magnets have poles) TE Wave Reflection/Transmission ubstitute in equations ( assuming that µ 1 = µ 2 ) 1+ E E 0 = 2 E 0 1+ E 1+ 0 k cos E 1-0 k cos2 1 = 2 E 2 0 k cos2 2 ubstitute for k = 2B/8 0 n 1- E 0 / 1+ E 0 = [n 1 cos2 1 - n 2 cos2 2 ]/[n 1 cos2 1 + n 2 cos2 2 ] 2 E 0 / 1+ E 0 = [2 n 1 cos2 1 ]/[n 1 cos2 1 + n 2 cos2 2 ] ubstitute from nell s Law 2 E 0 / 1+ E 0 = 2 cos2 1 sin2 2 /sin( ) Fresnel Relations At Normal Incidence 2 E 0 / 1+ E 0 = 2 cos2 1 sin2 2 /sin( ) At normal incidence - expressions are zero!! or are they? as 2 -> 0 1- E 0 / 1+ E 0 -> ( )/( ) -> (n 1 - n 2 )/(n 2 + n 1 ) 2 E 0 / 1+ E 0 -> /( ) -> 2n 1 /(n 2 + n 1 )

4 Energy Transmission The energy reflection R, is given by... resolve Poynting vector along normal - n.< 1- >/n.< 1+ > = (cos2 1 n 1 1- E 0 2 )/(cos2 1 n 1 1+ E 0 2 ) = 1- E 0 2 / 1+ E 0 2 = sin 2 ( )/sin 2 ( ) The energy transmission T, is given by... resolve Poynting vector along normal n.< 2 >/n.< 1+ > = (cos2 2 n 2 2 E 0 2 )/(cos2 1 n 1 1+ E 0 2 ) = sin22 2 sin22 1 / sin 2 ( ) Notice that R + T = 1 R = (n 2 - n 1 ) 2 /(n 2 + n 1 ) 2 T = 4n 1 n 2 /(n 2 + n 1 ) 2 R + T = 1 At Normal Incidence ummary for a TE wave For a dielectric interface there is no phase change (0,B) 2 E 0 / 1+ E 0 = 2 cos2 1 sin2 2 /sin( ) At normal incidence 1- E 0 / 1+ E 0 -> ( )/( ) -> (n 1 - n 2 )/(n 2 + n 1 ) 2 E 0 / 1+ E 0 -> /( ) -> 2n 1 /(n 2 + n 1 ) The reflected power ratio R = sin 2 ( )/sin 2 ( ) The trasmitted power ratio T = sin22 2 sin22 1 / sin 2 ( ) At normal incidence R = (n 2 - n 1 ) 2 /(n 2 + n 1 ) 2, T = 4n 1 n 2 /(n 2 + n 1 ) 2 The Other Case (TM Waves) Maxwell s Equations are (nearly) symmetrical in E, H if there are no free charges Any solution for E,H can be written for H,E if we interchange -µ,, at the same time o if for the TE case we have ( assuming that µ 1 = µ 2 ) 1+ E E 0 = 2 E 0 1+ E 1+ 0 k cos E 1-0 k cos2 1 = 2 E 2 0 k cos2 2 for the TM case it is k 1+ E k 1- E 0 = 2 k 2 E 0 1+ E 0 cos E 0 cos2 1 = 2 E 0 cos2 2 And the solution proceeds...

5 ummary for a TM wave Critical Angle For a dielectric interface there is no phase change (0,B) 1- E 0 / 1+ E 0 = tan( )/tan( ) 2 E 0 / 1+ E 0 = 2 cos2 1 sin2 2 /sin( )/cos( ) At normal incidence 1- E 0 / 1+ E 0 -> ( )/( ) -> (n 2 - n 1 )/(n 2 + n 1 ) 2 E 0 / 1+ E 0 -> /( ) -> 2n 1 /(n 2 + n 1 ) The reflected power ratio R = tan 2 ( )/tan 2 ( ) The trasmitted power ratio T = sin22 2 sin22 1 / sin 2 ( ) At normal incidence R = (n 1 - n 2 ) 2 /(n 2 + n 1 ) 2, T = 4n 1 n 2 /(n 2 + n 1 ) 2 n 1 sin2 1 = n 2 sin2 2 - nell s Law (n sin2 is conserved) If (n 1 /n 2 )sin2 1 > 1 then 2 2 does not exist? There is no transmitted ray reflection must be perfect!! Brewster s Angle For a TM wave the reflected power ratio R = tan 2 ( )/tan 2 ( ) If ( ) = B/2 then R = 0 Perfect Transmission (TM only) For a TE wave the reflected power ratio R = sin 2 ( )/sin 2 ( ) no minimum Phase Changes For TE waves 2 E 0 / 1+ E 0 = 2 cos2 1 sin2 2 /sin( ) Phase change on transmission is 0 Phase change on reflection is 0 if 2 2 > 2 1, B if 2 2 < 2 1 For TM waves 1- E 0 / 1+ E 0 = tan( )/tan( ) 2 E 0 / 1+ E 0 = 2 cos2 1 sin2 2 /sin( )/cos( ) Phase change on transmission is 0 Phase change on reflection 0 if ( ) < B/2 (less than Brewster s) B if ( ) > B/2 (greater than Brewster s)