Electromagnetism - Lecture 14
Waves at Boundaries
• Normal Incidence
• Energy Transport
• Reflection & Refraction
• E or B in Plane of Incidence
• Brewster Angle
• Total Internal Reflection
1
Plane Waves at Normal Incidence
The incident wave in medium 1 with 1 has:
EI = E0I ei(ωt−k1 z) x̂
BI = B0I ei(ωt−k1 z) ŷ
The reflected wave in medium 1 with 1 has:
ER = E0R ei(ωt+k1 z) x̂
BR = −B0R ei(ωt+k1 z) ŷ
Note - reversals of signs from direction of propagation
The transmitted wave in medium 2 with 2 has:
ET = E0T ei(ωt−k2 z) x̂
BT = B0T ei(ωt−k2 z) ŷ
E and B are always tangential to the surface at normal incidence!
2
Amplitudes at Normal Incidence
Applying boundary condition on E|| at z = 0:
EI − ER = ET
E0I − E0R = E0T
Applying condition on H|| with no surface currents and µr = 1:
1
1
(BI + BR ) =
BT
µ0
µ0
(B0I + B0R ) = B0T
Ratio of amplitudes of E and B is always c/n:
c
c
c
B0I
E0R =
B0R
E0T =
B0T
E0I =
n1
n1
n2
Combining all of these gives the ratios of amplitudes:
E0T
2Z2
2n1
=
=
E0I
(n1 + n2 )
(Z2 + Z1 )
(Z2 − Z1 )
(n1 − n2 )
E0R
=
=
E0I
(n1 + n2 )
(Z2 + Z1 )
3
Notes:
Diagrams:
4
Energy Transport at Boundary
Energy transport is given by Poynting vector:
E02
ẑ
N=E×H=
Z
Fraction of reflected energy is reflection coefficient:
2
(n1 − n2 )2
4n1 n2
E0R
=1−
R= 2 =
E0I
(n1 + n2 )2
(n1 + n2 )2
Fraction of transmitted energy is transmission coefficient:
2
4n1 n2
n2 E0T
=
T =
2
n1 E0I
(n1 + n2 )2
Energy is conserved at normal incidence to boundary: T + R = 1
5
Law of Reflection
One plane polarization state has EI in the plane of incidence and
BI parallel to the boundary
The incident plane wave is:
EI = E0I ei(ωt−k1 .r) r̂⊥
r = sin θI x + cos θI z
where EI is perpendicular to the the direction of propagation r
The reflected wave is:
ER =
i(ωt−k1 .r0 ) 0
E0R e
r̂⊥
r0 = sin θR x − cos θR z
At z = 0 the incident and reflected waves must be in phase:
k1 .r = k1 .r0
sin θI x = sin θR x
The angles of incidence and reflection are equal: θI = θR
6
Law of Refraction
The transmitted wave is:
ET =
i(ωt−k2 .r00 ) 00
E0T e
r̂⊥
r00 = sin θT x + cos θT z
At z = 0 the incident and transmitted waves must be in phase:
k1 .r = k2 .r00
k1 sin θI = k2 sin θT
This leads to Snell’s law of refraction:
n1 sin θI = n2 sin θT
7
Notes:
Diagrams:
8
Amplitudes with E in Plane of Incidence
Applying boundary condition on E|| at z = 0:
(E0I + E0R ) cos θI = E0T cos θT
Applying boundary condition on D⊥ at z = 0:
1 (E0I − E0R ) sin θI = 2 E0T sin θT
with the help of n1 sin θI = n2 sin θT and n2 = this becomes:
n1 (E0I − E0R ) = n2 E0T
Combining the two conditions gives the ratios of amplitudes:
E0T
2n1 cos θI
=
E0I
(n1 cos θT + n2 cos θI )
(n1 cos θT − n2 cos θI )
E0R
=
E0I
(n1 cos θT + n2 cos θI )
9
Notes:
Diagrams:
10
Amplitudes with B in Plane of Incidence
The other polarization state has BI in the plane of incidence and
EI parallel to the boundary.
We assume that µr = 1 for both materials
Applying boundary condition on H|| at z = 0:
(B0I − B0R ) cos θI = B0T cos θT
Applying boundary condition on B⊥ at z = 0:
(B0I + B0R ) sin θI = B0T sin θT
Combining these and converting to the electric field amplitudes
B0R
E0R
=−
E0I
B0I
n1 B0T
E0T
=
E0I
n2 B0I
E0T
2n1 cos θI
=
E0I
(n2 cos θT + n1 cos θI )
(n2 cos θT − n1 cos θI )
E0R
=
E0I
(n2 cos θT + n1 cos θI )
11
Notes:
Diagrams:
12
Fresnel Equations
Snell’s Law and some trig identities can be used to simplify the
amplitude ratios and remove n1 and n2
The results are known as Fresnel’s equations
For E in plane of incidence:
sin 2θT − sin 2θI
E0R
=
E0I
sin 2θT + sin 2θI
4 cos θI sin θT
E0T
=
E0I
sin 2θT + sin 2θI
For B in plane of incidence:
E0T
2 cos θI sin θT
=
E0I
sin(θI + θT )
sin(θI − θT )
E0R
=
E0I
sin(θI + θT )
13
Brewster’s Angle
When E is in the plane of incidence the reflected amplitude is zero
at an angle incidence known as Brewster’s angle θB :
2θT = 180◦ − 2θB
n2 cos θB = n1 cos θT
cos θT = sin θB
tan θB =
n2
n1
There is no angle of incidence for which the reflected amplitude is
zero if B is in the plane of incidence
An unpolarised beam incident at Brewster’s angle becomes plane
polarised on reflection. The polarisation state has E|| to the surface.
14
A Couple More Things about Reflection
From Snell’s law there is no transmitted wave if:
n2
θT = 90◦
sin θI =
n1
Total internal reflection occurs for large angles of incidence
when n1 > n2
If n1 > n2 the signs of EI and ER are the same
If n2 > n1 the signs of EI and ER are opposite
There is a phase change of 180◦ when a reflection takes place off
a medium of higher refractive index
There is no phase change for total internal reflection!
15
Notes:
Diagrams:
16