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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" 
{TEXT 257 125 "copyright 2000 Oregon State University              Cap
stones in Physics: Electromagnetism                  Philip J. Siemens
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 20 "WAV
ES AT A BOUNDARY\n" }}{PARA 0 "" 0 "" {TEXT -1 266 "\nThis worksheet i
s a learning tool to help visualize the field of a linearly polarized \+
plane wave reflected from a dielectric interface.  We will assume that
 the media in which the wave propagates are insulators, so that there \+
is no loss of energy due to conduction." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 81 "Our aim is to display the electric a
nd magnetic fields as a function of position " }{TEXT 262 1 "r" }
{TEXT -1 37 " .  We choose Cartesian coordinates:\n" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 53 "restart: with (linalg): with (plots):\nrvec:=[x,y,
z];\n" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 32 "A LINEARLY POLARIZED PL
ANE WAVE " }}{PARA 0 "" 0 "" {TEXT 258 2 "\n\n" }{TEXT -1 188 "We begi
n by looking at the field of a single plane wave.  This will eventuall
y become the Incident wave.  We will set its wavelength to 1 display u
nit, so the magnitude of the wave vector " }{TEXT 261 1 "k" }{TEXT -1 
4 " is " }{XPPEDIT 18 0 "2*pi" "*&\"\"#\"\"\"%#piGF$" }{TEXT -1 108 ".
  \n\nTo best exploit Maple's built-in display options, we choose the \+
direction of propagation to be in the  " }{TEXT 259 4 "x-y " }{TEXT 
-1 19 "plane, at an angle " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }
{TEXT -1 10 " from the " }{TEXT 260 2 "x " }{TEXT -1 5 "axis:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "kIn
c:=[2*Pi*cos(theta),2*Pi*sin(theta),0];\n\n" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 4 "The " }{TEXT 282 3 "x-y" }{TEXT -1 38 " plane will be th
e plane of incidence." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 53 "The electric and magnetic fields will only depend on \+
" }{TEXT 284 1 "x" }{TEXT -1 5 " and " }{TEXT 285 1 "y" }{TEXT -1 13 "
, but not on " }{TEXT 283 1 "z" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "We will make the arrows c
orresponding to different values of " }{TEXT 263 1 "z" }{TEXT -1 82 " \+
have different colors, to help us see the 3-dimensional perspective in
 the plots." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 57 "CHOOSE \"WINDOW\" FROM THE \"OPTIONS/PLOT DISPLAY\" MENU NOW." 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 118 "We may look at the wave at different
 times.  We will set its period to 1 time unit.  That means its angula
r frequency " }{XPPEDIT 18 0 "omega" "I&omegaG6\"" }{TEXT -1 9 " will \+
be " }{XPPEDIT 18 0 "2*Pi" "*&\"\"#\"\"\"%#PiGF$" }{TEXT -1 4 " : \n" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "omega:=2*Pi;\n" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 72 "\nThe argument of the sines and cosines in the \+
incident wave will then be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 35 "argInc:=dotprod(rvec,kInc)-omega*t;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 333 "\nNow we have to specify the ampl
itude of the incident wave.  We can conveniently choose the magnitude \+
of the incident field as one unit of vector length.  \n\nNext, we have
 to specify the direction of polarization of the wave. \n\nBy conventi
on, the polarization is the direction of the electric field's amplitud
e eInc . \n\nLet's choose  " }{XPPEDIT 18 0 "alpha" "I&alphaG6\"" }
{TEXT -1 60 "  as the angle between the electric field amplitude and t
he " }{XPPEDIT 18 0 "x-y" ",&%\"xG\"\"\"%\"yG!\"\"" }{TEXT -1 18 " pla
ne.\n\nThen the " }{XPPEDIT 18 0 "z" "I\"zG6\"" }{TEXT -1 28 " compone
nt of eInc  is just " }{XPPEDIT 18 0 "sin(alpha)" "-%$sinG6#%&alphaG" 
}{TEXT -1 35 " .  \n\nThe component of eInc in the " }{TEXT 265 3 "x-z
" }{TEXT -1 18 " plane has length " }{XPPEDIT 18 0 "cos(alpha)" "-%$co
sG6#%&alphaG" }{TEXT -1 39 ", and has to be perpendicular to kInc:\n" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "eInc:=[-sin(theta)*cos(alpha),cos
(theta)*cos(alpha),sin(alpha)];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
134 "\nNow we are ready to construct the electric field of this plane \+
wave.  It is given by the magnitude times the cosine of the argument:
\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "EInc:=[cos(argInc)*eInc[1],co
s(argInc)*eInc[2],cos(argInc)*eInc[3]];" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 31 "\nWe can plot the field at time " }{XPPEDIT 18 0 "t=0" "/
%\"tG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 48 "First we have to  choose values of the angles.  \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "We'll
 take an incident angle of " }{XPPEDIT 18 0 "pi/6" "*&%#piG\"\"\"\"\"'
!\"\"" }{TEXT -1 59 ", or 30 degrees.  This will allow us to see thing
s clearly." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "theta:=Pi/6;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 98 "We'll want to try various polarizations. \+
 The simplest is perpendicular to the plane of incidence:" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "alpha:=Pi/2;
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
7 "and...\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "fieldplot3d(subs(t=
0,EInc),x=-1..1,y=-1..1,z=-1..1,grid=[9,17,3],orientation=[-120,20],ax
es=boxed,shading=zhue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 54 "Be sure to pick out the wave fronts, perp
endicular to " }{TEXT 264 4 "kInc" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 1 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 125 "Now let's look at the magnetic field of the sa
me plane wave.\n\nAccording to Faraday's Law (see Page 5.3 of the cour
se notes), " }}{PARA 263 "" 0 "" {TEXT -1 2 "\n " }{XPPEDIT 286 0 "H  \+
=(k  cross  E)/(omega*mu)" "/%\"HG**%\"kG\"\"\"%&crossGF&%\"EGF&*&%&om
egaGF&%#muGF&!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 12 "We can take " }
{XPPEDIT 18 0 "mu =u[0]" "/%#muG&%\"uG6#\"\"!" }{TEXT -1 14 ", and inc
lude " }{XPPEDIT 18 0 "mu[0]" "&%#muG6#\"\"!" }{TEXT -1 26 " in choosi
ng the units of " }{XPPEDIT 18 0 "H" "I\"HG6\"" }{TEXT -1 29 " .\nThen
, in a Maple command,\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "HInc:=ev
al(crossprod(kInc,subs(t=0,EInc)/omega));" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 52 "Let's look at the magnetic field of the wave at t=0." }
{MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 53 "In the same perspective as the electric field, it is " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "fi
eldplot3d(HInc,x=-1..1,y=-1..1,z=-1..1,grid=[17,17,3],orientation=[-12
0,20],axes=boxed,shading=zhue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "Agai
n, be sure to pick out the wave fronts perpendicular to " }{XPPEDIT 
18 0 "kInc" "I%kIncG6\"" }{TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 1 " " }}{PARA 0 "" 0 "" {TEXT -1 118 "When you have time, outside \+
of class, try changing the angles to see how the wave looks with diffe
rent polarizations.." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 264 "" 0 "" {TEXT -1 31 "REFLECTED AND
 TRANSMITTED WAVES" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Now we are ready \+
to consider reflection from a boundary in the plane " }{XPPEDIT 18 0 "
x=0" "/%\"xG\"\"!" }{TEXT -1 11 ", i.e. the " }{XPPEDIT 18 0 "y-z" ",&
%\"yG\"\"\"%\"zG!\"\"" }{TEXT -1 7 " plane." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "The wave vector " }{XPPEDIT 18 
0 "kRef" "I%kRefG6\"" }{TEXT -1 64 "  of the reflected wave is like th
e incident wave, but with the " }{XPPEDIT 18 0 "x" "I\"xG6\"" }{TEXT 
-1 20 " component reversed:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 43 "kRef:=[-2*Pi*cos(theta),2*Pi*sin(theta),0
];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "The frequency of the reflected \+
wave is the same as the incident wave, " }}{PARA 0 "" 0 "" {TEXT -1 
68 "So we can find the argument function (phase) for the reflected wav
e:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
35 "argRef:=dotprod(rvec,kRef)-omega*t;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 98 "\n\nThe wave vector of the transmitted wave is determined
 by the ratio of the indices of refraction " }{XPPEDIT 18 0 "n21=nInci
dent/nTransmitted" "/%$n21G*&%*nIncidentG\"\"\"%-nTransmittedG!\"\"" }
{TEXT -1 36 " \n.\n\nThis is a good point to choose " }{XPPEDIT 18 0 "
n21" "I$n21G6\"" }{TEXT -1 5 " .  \n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 9 "n21:=1.4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "\n\nThe angle o
f the transmitted wave is given by Snell's law, in terms of " }
{XPPEDIT 18 0 "n21" "I$n21G6\"" }{TEXT -1 5 "  as\n" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 41 "thetaTran:=evalf(arcsin(sin(theta)/n21));" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "\nThe magnitude of the transmitte
d wave vector is inversely proportional to the index of refraction.\n
\n" }}{PARA 0 "" 0 "" {TEXT -1 19 "So the wave vector " }{XPPEDIT 18 
0 "kTran" "I&kTranG6\"" }{TEXT -1 29 "  of the transmitted wave is " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "kT
ran:=[2*Pi*cos(thetaTran)/n21,2*Pi*sin(thetaTran)/n21,0];" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 72 "The frequency of the transmitted wave is the sa
me as the incident wave, " }}{PARA 0 "" 0 "" {TEXT -1 70 "So we can fi
nd the argument function (phase) for the transmitted wave:" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "argTran:=do
tprod(rvec,kTran)-omega*t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 267 "" 0 "" 
{TEXT -1 25 "REFLECTION COEFFICIENTS\n\n" }}{PARA 0 "" 0 "" {TEXT -1 
55 "The reflected wave's amplitude will change by a factor " }
{XPPEDIT 18 0 "CRef" "I%CRefG6\"" }{TEXT -1 246 " , which is given by \+
one of Fresnel's formulas.  The perpendicular part of the field, and t
he component of the field in the plane of incidence,each have their ow
n factors.\n\nThe factors are given in terms of the ratio of the indic
es of refraction " }{XPPEDIT 18 0 "n21=nIncident/nTransmitted" "/%$n21
G*&%*nIncidentG\"\"\"%-nTransmittedG!\"\"" }{TEXT -1 4 " .\n\n" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "CRefPerp:=(cos(theta)-n21*cos(theta
Tran))/(cos(theta)+n21*cos(thetaTran));" }{TEXT -1 1 "\n" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 74 "CRefPar:=(n21*cos(theta)-cos(thetaTran))/(n21
*cos(theta)+cos(thetaTran));\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "C
RefPerp:=evalf(CRefPerp);CRefPar:=evalf(CRefPar);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 90 "\nThe part of the reflected wave that is perpendic
ular to the plane of incidence, i.e. the " }{XPPEDIT 18 0 "z" "I\"zG6
\"" }{TEXT -1 82 " component, is proportional to the perpendicular par
t of the incident wave, \n\nThe " }{XPPEDIT 18 0 "z" "I\"zG6\"" }
{TEXT -1 13 " component of" }{TEXT 269 4 " eIn" }{TEXT -1 10 "c is jus
t " }{XPPEDIT 18 0 "sin(alpha)" "-%$sinG6#%&alphaG" }{TEXT -1 11 " .  \+
So the " }{XPPEDIT 18 0 "z" "I\"zG6\"" }{TEXT -1 14 " component of " }
{XPPEDIT 18 0 "eRef" "I%eRefG6\"" }{TEXT -1 4 "  is" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "eR
ef[z]=sin(alpha)*CRefPerp" "/&%%eRefG6#%\"zG*&-%$sinG6#%&alphaG\"\"\"%
)CRefPerpGF," }{TEXT -1 1 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "eRe
fPerp:=CRefPerp*[0,0,sin(alpha)];" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "\nThe component of " }{TEXT 267 4 
"eRef" }{TEXT -1 8 " in the " }{TEXT 268 3 "x-z" }{TEXT -1 18 " plane \+
has length " }{XPPEDIT 18 0 "cos(alpha)*CRefPar" "*&-%$cosG6#%&alphaG
\"\"\"%(CRefParGF'" }{TEXT -1 34 " , and has to be perpendicular to " 
}{TEXT 266 4 "kRef" }{TEXT -1 2 ":\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 100 "eRefPar:=[-sin(theta)*cos(alpha)*CRefPar,-cos(theta)*cos(alpha)
*CRefPar,0];\n\neRef:=eRefPerp+eRefPar;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 137 "\nNow we are ready to construct the electric field of th
e reflected wave.  It is given by the magnitude times the cosine of th
e argument:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "ERef:=[cos(argRef)
*eRef[1],cos(argRef)*eRef[2],cos(argRef)*eRef[3]];" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 31 "\nWe can plot the field at time " }{XPPEDIT 18 0 
"t=0" "/%\"tG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "fi
eldplot3d(subs(t=0,ERef),x=-1..1,y=-1..1,z=-1..1,grid=[17,17,3],orient
ation=[-120,20],axes=boxed,shading=zhue);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 54 "Be sure to pick out the wave fronts, perpendicular to " }
{TEXT 270 4 "kRef" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 60 "Now let's look at the magnetic field of the same plane wave." }
}{PARA 0 "" 0 "" {TEXT -1 64 "\nAccording to Faraday's Law (see Page 5
.3 of the course notes), " }}{PARA 265 "" 0 "" {TEXT -1 2 "\n " }
{XPPEDIT 18 0 "H  =(k  cross  E)/(omega*mu)" "/%\"HG**%\"kG\"\"\"%&cro
ssGF&%\"EGF&*&%&omegaGF&%#muGF&!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 12 "W
e can take " }{XPPEDIT 18 0 "mu =u[0]" "/%#muG&%\"uG6#\"\"!" }{TEXT 
-1 14 ", and include " }{XPPEDIT 18 0 "mu[0]" "&%#muG6#\"\"!" }{TEXT 
-1 26 " in choosing the units of " }{XPPEDIT 18 0 "H" "I\"HG6\"" }
{TEXT -1 29 " .\nThen, in a Maple command,\n" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "HRef:=crossprod(kRef,subs(t=0,ERef)/omega);" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 53 "In the same perspective as the electric f
ield, it is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 103 "fieldplot3d(HRef,x=-1..1,y=-1..1,z=-1..1,grid=[17,17
,3],orientation=[-120,20],axes=boxed,shading=zhue);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Once again, identify
 the wavefronts.\n\n" }}{PARA 12 "" 0 "" {TEXT -1 0 "" }}{PARA 268 "" 
0 "" {TEXT -1 45 "SUPERPOSITION OF INCIDENT AND REFLECTED WAVES" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 98 "\n\nWe can now superpose the incident and reflected waves, in t
he region where they are valid, i.e. " }{XPPEDIT 18 0 "x<0" "2%\"xG\"
\"!" }{TEXT -1 2 "\n\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 205 "E1:=[Hea
viside(-x)*(EInc[1]+ERef[1]),Heaviside(-x)*(EInc[2]+ERef[2]),Heaviside
(-x)*(EInc[3]+ERef[3])];\nH1:=[Heaviside(-x)*(HInc[1]+HRef[1]),Heavisi
de(-x)*(HInc[2]+HRef[2]),Heaviside(-x)*(HInc[3]+HRef[3])];" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 11 "\nPlotting,\n" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 214 "fieldplot3d(subs(t=0,E1),x=-1..1,y=-1..1,z=-1..1,gri
d=[17,17,3],orientation=[-120,20],axes=boxed,shading=zhue);\n\nfieldpl
ot3d(H1,x=-1..1,y=-1..1,z=-1..1,grid=[17,17,3],orientation=[-120,20],a
xes=boxed,shading=zhue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 110 "From this perspective you can see the wa
ves churning as the reflected wave interferes with the incident wave. \+
" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 120 "The reflection at the interface is cause
d by the need to match the fields there with the fields of the transmi
tted wave." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "To see what is happening in the
 plane of the interface, let's turn the cube to another perspective\n
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
214 "fieldplot3d(subs(t=0,E1),x=-1..1,y=-1..1,z=-1..1,grid=[17,17,3],o
rientation=[-30,70],axes=boxed,shading=zhue);\n\nfieldplot3d(H1,x=-1..
1,y=-1..1,z=-1..1,grid=[17,17,3],orientation=[-30,70],axes=boxed,shadi
ng=zhue);\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "In the plane " }
{XPPEDIT 18 0 "x=0" "/%\"xG\"\"!" }{TEXT -1 81 "  , the transmitted wa
ve has to match the interference pattern we are looking at." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "Since there ar
e no free currents in insulators, and since we have assumed that both \+
media have permeability " }{XPPEDIT 18 0 "mu[0]" "&%#muG6#\"\"!" }
{TEXT -1 40 " , all components of the magnetic field " }{XPPEDIT 18 0 
"H" "I\"HG6\"" }{TEXT -1 49 "  must match on both sides of the boundar
y plane." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
51 "Only the parallel components of the electric field " }{XPPEDIT 18 
0 "E" "I\"EG6\"" }{TEXT -1 96 "  have to match, because the surface po
larization charge can change the perpendicular component." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {TEXT -1 26 "FIELDS OF TRANSMITTED
 WAVE" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 78 "\nThe transmitted wave that does this
 trick is given by the Fresnel relations\n\n" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 58 "CTranPerp:=(2*cos(theta))/(cos(theta)+n21*cos(thetaTr
an));" }{TEXT -1 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "CTranPar:
=(2*cos(theta))/(n21*cos(theta)+cos(thetaTran));\n" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 55 "CTranPerp:=evalf(CTranPerp);CTranPar:=evalf(CTranPa
r);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "\nThe part of the transm
itted wave that is perpendicular to the plane of incidence, i.e. the \+
" }{XPPEDIT 18 0 "z" "I\"zG6\"" }{TEXT -1 82 " component, is proportio
nal to the perpendicular part of the incident wave, \n\nThe " }
{XPPEDIT 18 0 "z" "I\"zG6\"" }{TEXT -1 13 " component of" }{TEXT 274 
4 " eIn" }{TEXT -1 10 "c is just " }{XPPEDIT 18 0 "sin(alpha)" "-%$sin
G6#%&alphaG" }{TEXT -1 11 " .  So the " }{XPPEDIT 18 0 "z" "I\"zG6\"" 
}{TEXT -1 14 " component of " }{XPPEDIT 18 0 "eTran" "I&eTranG6\"" }
{TEXT -1 4 "  is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 271 "" 0 "" 
{TEXT -1 2 "  " }{XPPEDIT 18 0 "eTran[z]=sin(alpha)*CTranPerp" "/&%&eT
ranG6#%\"zG*&-%$sinG6#%&alphaG\"\"\"%*CTranPerpGF," }{TEXT -1 1 " " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eTranPerp:=CTranPerp*[0,0,sin(alpha
)];" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 18 "\nThe component of " }{TEXT 272 5 "eTran" }{TEXT -1 8 " in the \+
" }{TEXT 273 3 "x-z" }{TEXT -1 18 " plane has length " }{XPPEDIT 18 0 
"CTranPar*cos(aplha)" "*&%)CTranParG\"\"\"-%$cosG6#%&aplhaGF$" }{TEXT 
-1 35 "  , and has to be perpendicular to " }{TEXT 271 5 "kTran" }
{TEXT -1 2 ":\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "eTranPar:=[-sin
(thetaTran)*cos(alpha)*CTranPar,cos(thetaTran)*cos(alpha)*CTranPar,0];
\n\neTran:=eTranPerp+eTranPar;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
139 "\nNow we are ready to construct the electric field of the transmi
tted wave.  It is given by the magnitude times the cosine of the argum
ent:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "ETran:=[cos(argTran)*eTr
an[1]*Heaviside(x),cos(argTran)*eTran[2]*Heaviside(x),cos(argTran)*eTr
an[3]*Heaviside(x)];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 118 "We have included a Heaviside function, because
 this expression for the field only applies in the second medium, wher
e " }{XPPEDIT 18 0 "x>0" "2\"\"!%\"xG" }{TEXT -1 2 " ." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "We can plot the tran
smitted field at time " }{XPPEDIT 18 0 "t=0" "/%\"tG\"\"!" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "fieldplot3d(subs(t=0,ETran),x=-1
..1,y=-1..1,z=-1..1,grid=[17,17,3],orientation=[-135,30],axes=boxed,sh
ading=zhue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "Be sure to pick out the
 wave fronts, perpendicular to " }{TEXT 275 5 "kTran" }{TEXT -1 74 ", \+
which are easiest to see on the top surface of the displayed cube wher
e " }{XPPEDIT 18 0 "z = constant" "/%\"zG%)constantG" }{TEXT -1 23 " a
nd the dependence on " }{TEXT 277 1 "x" }{TEXT -1 5 " and " }{TEXT 
276 1 "y" }{TEXT -1 10 " are seen." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 49 "Then look for the pattern of fields in th
e plane " }{XPPEDIT 18 0 "x=0" "/%\"xG\"\"!" }{TEXT -1 109 ", on the l
ower left of this view.  Will they match the fields in the other mediu
m, just across the interface?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "Now let's look at the magnetic f
ield of the same plane wave." }}{PARA 0 "" 0 "" {TEXT -1 64 "\nAccordi
ng to Faraday's Law (see Page 5.3 of the course notes), " }}{PARA 270 
"" 0 "" {TEXT -1 2 "\n " }{XPPEDIT 18 0 "H  =(k  cross  E)/(omega*mu)
" "/%\"HG**%\"kG\"\"\"%&crossGF&%\"EGF&*&%&omegaGF&%#muGF&!\"\"" }}
{PARA 0 "" 0 "" {TEXT -1 3 "We " }}{PARA 0 "" 0 "" {TEXT -1 5 "take " 
}{XPPEDIT 18 0 "mu =u[0]" "/%#muG&%\"uG6#\"\"!" }{TEXT -1 14 ", and in
clude " }{XPPEDIT 18 0 "mu[0]" "&%#muG6#\"\"!" }{TEXT -1 26 " in choos
ing the units of " }{XPPEDIT 18 0 "H" "I\"HG6\"" }{TEXT -1 29 " .\nThe
n, in a Maple command,\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "HTran:=
crossprod(kTran,subs(t=0,ETran)/omega);" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
53 "In the same perspective as the electric field, it is " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "fieldplot3d
(HTran,x=-1..1,y=-1..1,z=-1..1,grid=[17,17,3],orientation=[-135,30],ax
es=boxed,shading=zhue);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 54 "Once again, identify the wavefronts, perp
endicular to " }{TEXT 278 5 "kTran" }{TEXT -1 74 ", which are easiest \+
to see on the top surface of the displayed cube where " }{XPPEDIT 18 
0 "z = constant" "/%\"zG%)constantG" }{TEXT -1 23 " and the dependence
 on " }{TEXT 280 1 "x" }{TEXT -1 5 " and " }{TEXT 279 1 "y" }{TEXT -1 
10 " are seen." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 49 "Then look for the pattern of fields in the plane " }
{XPPEDIT 18 0 "x=0" "/%\"xG\"\"!" }{TEXT -1 109 ", on the lower left o
f this view.  Will they match the fields in the other medium, just acr
oss the interface?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 272 "" 0 "" {TEXT -1 17 "THE WHOLE
 PICTURE" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Let's put the fields together so w
e can plot them in the same picture." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "Etot:=E1+ETran; Htot:=[H1[1]+HTran[
1],H1[2]+HTran[2],H1[3]+HTran[3]];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 
"In the same perspective as the previous views of the transmitted wave
s,, they are " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 113 "fieldplot3d(subs(t=0,Etot),x=-1..1,y=-1..1,z=-1..1,g
rid=[17,17,3],orientation=[-135,30],axes=boxed,shading=zhue);" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "fi
eldplot3d(Htot,x=-1..1,y=-1..1,z=-1..1,grid=[17,17,3],orientation=[-13
5,30],axes=boxed,shading=zhue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 51 "Notice especially what is happening \+
near the plane " }{XPPEDIT 18 0 "x=0" "/%\"xG\"\"!" }{TEXT -1 2 " ." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 69 "First, you should see how the fields chan
ge when you change the time." }}{PARA 0 "" 0 "" {TEXT -1 91 "There is \+
another version of this worksheet, \"Electromotion \", which will make
 this easier. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Next, you should restart \+
this worksheet and change the polarization angle " }{XPPEDIT 18 0 "alp
ha" "I&alphaG6\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 60 "Then, you should try different values of \+
the incident angle " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 
30 ", and the index of refraction " }{TEXT 281 4 "n12." }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 273 "" 0 "" {TEXT -1 28 "HOMEWOR
K FOR THIS MAPLESHEET" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 36 "For an arbitrary angle of incidence " }{XPPEDIT 18 0 "
theta" "I&thetaG6\"" }{TEXT -1 24 " and polarization angle " }
{XPPEDIT 18 0 "alpha" "I&alphaG6\"" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "-  Find the polarizatio
n charge density on the interface, " }{XPPEDIT 18 0 "sigma(y,z,t)" "-%
&sigmaG6%%\"yG%\"zG%\"tG" }{TEXT -1 4 " = ?" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "Give your answer in terms of th
e angles, the frequency, and the amplitude of the incident wave." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "Extra cr
edit: Obviously some charge is moving around.  Why doesn't this moving
 charge cause a change in the magnetic field across the surface?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 262 "" 0 "" {TEXT -1 24 "version 22 February 2000" }}}}{MARK "
0 7 0" 52 }{VIEWOPTS 1 1 0 1 1 1803 }