Download 12 Fresnel Equations

12 Fresnel Equations
Contents
12.1 Laws of reflection and refraction
12.2 Electric field parallel to the plane of incidence
12.3 Electric field perpendicular to the plane of incidence
Keywords: Snell’s law, Polarisation, Brewster’s law
Ref: J. D. Jackson: Classical Electrodynamics; A. Sommerfeld: Electrodynamics.
12.1 Laws of reflection and refraction
In this section we derive laws of reflection and Snell’s law from Maxwell’s
equations and boundary conditions. These are kinematical properties and
are followed directly from the wave nature of light and the boundary conditions. The dynamical properties such as intensities of reflected and refracted
light, phase changes and polarisations are dictated by the specific nature
of electromagnetic fields and their boundary conditions. Consider a plane
electromagnetic wave (in medium 1, with permittivity and permeability ǫ1
and µ1 respectively) incident obliquely on a medium 2 (with permittivity
and permeability ǫ2 and µ2 respectively) at an incidence angle i to the in-
12 Fresnel Equations
2
terface normal (Fig. 12.1). The angles of reflection and refraction with
respect to the interface normal (n̂) are r′ and r respectively. The interface
is given by the plane z = 0. Let the incident plane wave be represented by
EI exp(ikI · r − iωt) (BI =
√
µ1 ǫ1
k kI
× EI ). The frequency ω will force the
charges(electrons) in the medium 2 to oscillate with the same frequency and
in turn the oscillating charges will radiate with the same frequency. So we
have one reflected ray, ER exp(ikR · r − iωt) (BR =
√
µ1 ǫ1
k kR
× ER ), and one
refracted (or transmitted) ray ET exp(ikT · r − iωt) (BT =
√
µ2 ǫ2
k ′ kT
× ET )
oscillating with the same frequency as the incident one. Wave vectors have
√
√
magnitudes |kI | = |kR | = k = ω µ1 ǫ1 and |kT | = k ′ = ω µ2 ǫ2 . Applying
boundary condition at z = 0,
[EI exp(ikI · r−iωt)+ER exp(ikR · r−iωR t)]|z=0 = ET exp(ikT · r−iωT t)|z=0
(12.1)
kI
i
r’=i
kR
µ1 , 1
∋
µ2 , 2
∋
r
kT
Fig. 12.1: Electromagnetic waves from one medium to the other
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12.2 Electric field parallel to the plane of incidence
3
The condition should be satisfied for all points x, y on the plane z = 0 and
also for all time. This is a stringent condition and would demand,
kI · r|z=0 = kR · r|z=0 = kT · r|z=0 .
(12.2)
The above constraint gives (for all r on z = 0 plane)
(kI · r)x = (kR · r)x = (kT · r)x ,
(12.3)
(kI · r)y = (kR · r)y = (kT · r)y .
(12.4)
Without loss of generality if the incident wave vector kI is taken in the x − z
plane the first expression of the second condition (12.4) is zero and in turn
it would restrict kR and kT in the x − z plane. So that makes all the rays in
the same plane. The remaining condition (12.3) now reads,
k sin i = k sin r′ = k ′ sin r.
(12.5)
The first equality in (12.5) sets i = r′ , that is the angle of incidence is equal
to the angle of reflection. The second equality is nothing but the Snell’s law,
√
µ2 ǫ 2
v1
n2
k′
sin i
=
= .
= =√
sin r
k
µ1 ǫ 1
v2
n1
(12.6)
In the following we will consider polarisation in the incidence plane and perpendicular to the incidence plane separately. The linear combination of the
two can produce any desired polarisation.
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12 Fresnel Equations
4
π/2 − i
π/2 − i
EI
ER
i
µ 1,
∋
1
µ2 ,
∋
2
^
n
r’=i
r
π/2 − r
ET
Fig. 12.2: Electric field parallel to the incident plane
12.2 Electric field parallel to the plane of incidence
For the electric field in the plane of incidence (or polarisation parallel to the
incidence plane, Fig. 12.2, the magnetic field are perpendicular to the plane
of the paper and are coming out of the paper shown by encircling a dot i.e.
the arrow head), we have the following boundary conditions (see the last
chapter),
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12.2 Electric field parallel to the plane of incidence
5
ǫ1 n̂ · (EI + ER ) = ǫ2 n̂ · ET ,
(12.7)
n̂ · (BI + BR ) = n̂ · BT ,
(12.8)
n̂ × (EI + ER ) = n̂ × ET ,
(12.9)
1
1
n̂ × (BI + BR ) = n̂ × BT .
µ1
µ2
(12.10)
Now since B is always perpendicular to the plane of incidence and hence n̂,
the second condition (eqn. (12.6))is trivially satisfied (each term being zero).
Simplifying the rest of the conditions using r′ = i,
ǫ1 (EI + ER ) sin i = ǫ2 ET sin r,
(12.11)
(EI − ER ) cos i = ET cos r,
(12.12)
1
1
(BI + BR ) = BT .
µ1
µ2
(12.13)
Now using BI = kI × EI /ω, BR = kR × ER /ω and BT = kT × ET /ω with
Snell’s law one finds the first and the third conditions above are identical.
Problem 1: Prove the previous statement.
So now from the first two independent conditions one finds,
ER
ǫ2 sin r cos i − ǫ1 sin i cos r
=
,
EI
ǫ2 sin r cos i + ǫ1 sin i cos r
2ǫ1 sin i cos i
ET
=
.
EI
ǫ2 sin r cos i + ǫ1 sin i cos r
(12.14)
(12.15)
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6
Problem 2: Show that there is no reflected ray for tan i = n2 /n1 , use approximation µ1 ≡ µ2 .
The angle, θW = tan−1 (n2 /n1 ) is known as Brewster angle. So for this angle
of incidence there is no inplane polarisation component in the reflected light
and hence the reflected light is completely plane polarised with the polarisation perpendicular to the plane of incidence. Typical values of reflected and
refracted amplitudes are shown in Fig (12.3) for the air-glass interface.
1.0
E
___
T
EI
0.5
0.5
1.0
1.5
Angle of incidence (in radians)
E
R
___
EI
-0.5
-1.0
For air−glass interface (n2 /n1 = 1.5)
Fig. 12.3: Reflected and transmitted amplitudes as a function of angle of incidence when polarisation of the electric field is in the plane of incidence.
12.3 Electric field perpendicular to the plane of incidence
Now we consider the case where the electric field is perpendicular to the plane
of incidence (Fig. 12.4).
Since the electric in this case the electric field is normal to the plane of
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12.3 Electric field perpendicular to the plane of incidence
π/2
7
π/2
i
BI
i
BR
i ^n r’=i
µ 1,
∋
1
µ 2,
∋
2
r
π/2
r
BT
Fig. 12.4: Electric field perpendicular to the incident plane
incidence the first condition, (12.5), vanishes identically, and the rest of the
conditions are written as,
(BI + BR ) sin i = BT sin r
(12.16)
(EI + ER ) = ET
(12.17)
1
1
(BI − BR ) cos i = BT cos r.
µ1
µ2
(12.18)
Among the above three the first two conditions are identical.
Problem 3: Like the previous case show that the first two conditions are the
same.
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12 Fresnel Equations
8
The last condition, in terms of electric field, is written as,
(EI − ER ) cos i =
r
ǫ 2 µ1
ET cos r.
ǫ 1 µ2
(12.19)
Using the independent conditions we have,
√
√
ǫ1 µ2 cos i − ǫ2 µ1 cos r
ER
=√
,
√
EI
ǫ1 µ2 cos i + ǫ2 µ1 cos r
√
2 ǫ1 µ2 cos i
ET
=√
.
√
EI
ǫ1 µ2 cos i + ǫ2 µ1 cos r
(12.20)
(12.21)
Fig. 12.5 shows reflected and refracted amplitudes for air-glass interface when
the polarisation of the electric field is perpendicular to the plane of incidence.
1.0
E
___
T
EI
0.5
Angle of incidence (in radians)
0.5
1.0
E
R
___
EI
-0.5
-1.0
1.5
For air−glass interface (n2 /n1 = 1.5)
Fig. 12.5: Reflected and transmitted amplitudes as a function of angle of incidence when polarisation of the electric field is perpendicular to the plane of incidence.
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