FresnelEquations

23. Fresnel Equations
• EM Waves at boundaries
• Fresnel Equations:
Reflection and Transmission Coefficients
• Brewster’s Angle
• Total Internal Reflection (TIR)
• Evanescent Waves
• The Complex Refractive Index
• Reflection from Metals
We will derive the Fresnel equations
r : reflection coefficient
rTE
Er cos θ − n2 − sin 2 θ
=
=
E cos θ + n2 − sin 2 θ
rTM
Er − n2 cos θ + n2 − sin 2 θ
=
=
E
n2 cos θ + n2 − sin 2 θ
E
θ
Et
2n cos θ
=
E n2 cos θ + n2 − sin 2 θ
θr
n1
n2
θt
t : transmission coefficient
E
2cos θ
tTE = t =
E cos θ + n2 − sin 2 θ
tTM =
Er
Et
n≡
ntransmitted n2
=
nincident
n1
EM Waves at an Interface
r r
r
r
Ei = Eoi exp ⎡i ki ⋅ r − ωi t ⎤
⎣
⎦
r r
r
r
Reflected beam : Er = Eor exp ⎡i kr ⋅ r − ωr t ⎤
⎣
⎦
r r
r
r
Transmitted beam : Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤
⎣
⎦
(
Incident beam :
)
(
)
(
TE mode
)
r
Eoi
r
Ei
r
ki = n1k0
r
kr = n1k0
r
kt = n2 k0
r
ki
n1
n2
r
Et
r
n2
θi
n1
r
kr
r
Eot
r
kt
TM mode
n2
r
Er
n1
Note the definition of the positive E-field directions in both cases.
r
Eor
EM Waves at an Interface
r r
r
r
Ei = Eoi exp ⎡i ki ⋅ r − ωi t ⎤
⎣
⎦
r r
r
r
Reflected beam : Er = Eor exp ⎡i kr ⋅ r − ωr t ⎤
⎣
⎦
r r
r
r
Transmitted beam : Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤
⎣
⎦
(
Incident beam :
)
(
)
(
)
At the boundary between the two media (the x − y plane), all waves must exist simultaneously,
and the tangential component must be equal on both sides of the interface.
r
Therefore, for all time t and for all boundary points r on the interface,
) r
) r
) r
n × Ei + n × Er = n × Et
r r
r r
r r
) r
) r
) r
⎡
⎤
⎡
⎤
⎡
n × Eoi exp i ki ⋅ r − ωi t + n × Eor exp i kr ⋅ r − ωr t = n × Eot exp i kt ⋅ r − ωt t ⎤
⎣
⎦
⎣
⎦
⎣
⎦
(
)
(
)
(
)
Assuming that the wave amplitudes are constant ,
the only way that this can be true over the entire interface and for all t is if :
r r
r r
r r
⇒ ki ⋅ r − ωi t = kr ⋅ r − ωr t = kt ⋅ r − ωt t : Phase matching at the boundary!
(
) (
) (
)
r
Eoi
r
Ei
r
ki
n1
n2
r
Er
n̂
r
kr
r
Eor
r rr
Eot
r
kt
r
Et
EM Waves at an Interface
Phase matching condition:
r r
r r
r r
ki ⋅ r − ωi t = kr ⋅ r − ωr t = kt ⋅ r − ωt t
(
) (
) (
r
Ei
)
r
At r = 0, this results in
ωi t = ωr t = ωt t
⇒
r
ki
n1
n2
ωi = ωr = ωt
(Frequency does not change at the boundary!)
At t = 0, this results in
r r r r r r
⇒ ki ⋅ r = k r ⋅ r = kt ⋅ r
r
Er
r
kr
r
ki = n1k0
r
kr = n1k0
r
kt = n2 k0
r
r
r
kt
r
Et
n2 k0
n1k0
r
kr
r
ki
(Phases on the boundary does not change!)
r
r
⇒ ki ,r ,t ⋅ r = constant
r
r
→ the equation for a plane perpendicular to ki ,r ,t and r .
r
kt
Normal
r r
r
⇒ ki , kr , and kt are coplanar in the plane of incidence.
r
r
x
EM Waves at an Interface
r
Ei
At t = 0,
r r r r r r
ki ⋅ r = kr ⋅ r = kt ⋅ r = constant
r
ki
θι
Considering the relation for the incident and reflected beams,
r r r r
ki ⋅ r = k r ⋅ r
ki r sin θi = kr r sin θ r
⇒
Since the incident and reflected beams are in the same medium,
nω
sin θi = sin θ r
ki = k r = i
θi = θ r : law of reflection
⇒
⇒
c
θr
r
Er
r
kr n
1
θt
x
n2
r
kt
r
Et
n2 k0
n1k0
Considering the relation for the incident and transmitted beams,
r r r r
⇒
ki ⋅ r = kt ⋅ r
ki r sin θi = kt r sin θt
But the incident and transmitted beams are in different media,
nω
nω
ki = i
kt = t
ni sin θi = nt sin θt : law of refraction
⇒
c
c
r
kr
θr
θi
θt
r
ki
r
kt
Normal
r
r
x
Development of the Fresnel Equations
From Maxwell ' s EM field theory,
we have the boundary conditions at the interface
TE-case
for the TE case :
Ei + Er = Et
Bi cos θi − Br cos θ r = Bt cos θ t
The above conditions imply that the tangential
r
r
components of both E and B are equal on
both sides of the interface. We have also
assumed that μi ≅ μt ≅ μ0 , as is true for
most dielectric materials.
For the TM mode :
Ei cos θi + Er cos θ r = Et cos θ t
− Bi + Br = − Bt
TM-case
Development of the Fresnel Equations
⎛c⎞
Recall that E = v B = ⎜ ⎟ B ⇒
⎝n⎠
B=
nE
c
n2
TE-case
Let n1 = refractive index of incident medium
n2 = refractive index of refracting medium
n1
For the TE mode :
Ei + Er = Et
n1 Ei cos θi − n1 Er cos θ r = n2 Et cos θ t
TM-case
n2
For the TM mode :
Ei cos θi + Er cos θ r = Et cos θt
− n1 Ei + n1 Er = − n2 Et
n1
Development of the Fresnel Equations
Eliminating Et from each set of equations
n2
and solving for the reflection coefficient we obtain :
TE case : rTE =
TM case : rTM =
where n =
Er cos θi − n cos θt
=
Ei
cos θi + n cos θt
TE-case
n1
Er −n cos θi + cos θt
=
n cos θi + cos θt
Ei
n2
n1
TM-case
n2
We know that
sin θi = n sin θt
sin 2 θi
n cos θt = n 1 − sin θt = n 1 −
=
2
n
2
n 2 − sin 2 θi
n1
Now we have derived the Fresnel Equations
Substituting we obtain the Fresnel equations for reflection coefficients r :
TE case : rTE
TM case : rTM
cos θi −
E
= r =
Ei
cos θi +
n 2 − sin 2 θi
n2
n − sin θi
2
2
2
2
2
Er − n cos θi + n − sin θi
=
=
Ei
n 2 cos θi + n 2 − sin 2 θi
n≡
TE-case
n2
n1
n1
For the transmission coefficient t :
2 cos θi
E
TE case : tTE = t =
Ei cos θi + n 2 − sin 2 θi
TM case : tTM =
TE :
2n cos θi
Et
=
Ei n 2 cos θi + n 2 − sin 2 θi
tTE = rTE + 1
TM : ntTM = 1 − rTM
TM-case
n2
These just mean the boundary conditions.
For the TE case : Ei + Er = Et
For the TM mode : − Bi + Br = − Bt
n1
Power : Reflectance (R) and Transmittance (T)
The quantities r and t are ratios of electric field amplitudes.
The ratios R and T are the ratios of reflected and transmitted powers,
respectively, to the incident power :
P
P
R= r
T = t
Pi
Pi
A
From conservation of energy :
⇒ 1= R +T
Pi = Pr + Pt
We can express the power in each of the fields
in terms of the product of an irradiance and area :
Pi = I i Ai
Pr = I r Ar
Pt = I t At
⇒ I i Ai = I r Ar + I t At
I i A cos θi = I r A cos θ r + I t A cos θt
I i cos θi = I r cos θ r + I t cos θt
But I =
⇒
2
I out cos θ out ⎛ nout Eout cos θ out
Power _ ratio =
=⎜
I in cos θin ⎜⎝ nin Ein 2 cos θin
1
1
1
n1ε 0 cE02i cos θi = n1ε 0 cE02r cos θ r + n2ε 0 cE02t cos θ t
2
2
2
2
2
2
⎛ cos θt ⎞ E02t
E0 r n2 E0t cos θt
E0 r
⇒ 1= 2 +
= 2 + n⎜
⎟ 2 = R+T
co
s
E0i n1 E02i cos θi
E0i
θ
i ⎠ E0 i
⎝
1
n ε 0 cE02 ⇒
2
E02r
R = 2 = r2
E0i
⎛ cos θt ⎞ E02t
⎛ cos θt ⎞ 2
T = n⎜
⎟t
⎟ 2 = n⎜
⎝ cos θi ⎠ E0i
⎝ cos θi ⎠
R = rr* = r
2
⎛ cos θ t
T = ⎜⎜ n
⎝ cos θ i
⎛ cos θ t
⎞
⎟⎟tt* = ⎜⎜ n
⎝ cos θ i
⎠
⎞ 2
⎟⎟ t
⎠
⎞
⎟
⎟
⎠
23-2. External and Internal Reflection
rTE =
cos θ i −
n 2 − sin 2 θ i
co s θ i +
n 2 − sin 2 θ i
rTM =
− n 2 cos θ i +
t
n 2 − sin 2 θ i
n 2 c os θ i +
n 2 − sin 2 θ i
tTE ,TM > 0
rTE ,TM > 0
rTM
External Reflection [ n = n2 / n1 > 1 ]
⇒ n2 > n1
⇒ n = n2 / n1 > 1 ⇒ ( n 2 − sin 2 θ ) ≥ 0
rTE
rTE ,TM < 0
n=1.50
⇒ rTE ,TM are always real
⇒ If rTE ,TM > 0 then there are no phase changes after reflection.
⇒ If rTE ,TM < 0 then there are always π (= 180o ) phase changes.
→ rTE ,TM = − rTE ,TM = eiπ rTE ,TM
Note for the TM case :
⇒ rTM (θ = θ p ) = 0 when θ p = tan n
−1
Brewster’s angle (or, polarizing angle)
(No reflection of TM mode)
Internal Reflection [ n = n2 / n1 < 1 ]
rTE =
cos θ i −
n 2 − sin 2 θ i
co s θ i +
n 2 − sin 2 θ i
rTM =
− n 2 cos θ i +
n 2 c os θ i +
rTE ,TM > 0
n 2 − sin 2 θ i
n 2 − sin 2 θ i
TIR region
rTE ,TM < 0
n1 > n2 ⇒ n = n2 / n1 < 1
⇒ ( n 2 − sin 2 θ ) > 0, or , ( n 2 − sin 2 θ ) < 0
⇒ If ( n 2 − sin 2 θ ) > 0, rTE ,TM are always real
→ If rTE ,TM > 0 then there are no phase changes after reflection.
→ If rTE ,TM < 0 then there are π (= 180o ) phase changes.
⇒ If ( n 2 − sin 2 θ ) = 0, rTE ,TM =1
→ sin θ c = n = (n2 / n1 )
critical angle
⇒ If ( n 2 − sin 2 θ ) < 0, rTE ,TM =1, BUT rTE ,TM are complex !
→
rTE ,TM =1
Note Brewster's angle (θ p = tan −1 n )
for the TM case : rTM = 0
Total internal reflection (TIR) when θ > θc
→ rTE ,TM = rTE ,TM eiφ = eiφ
→ φ (-π ~ +π ) phase change may occur after reflection
Derivation of Brewster’s Angle
θc
Brewster's angle θ p ( for polarizing angle) :
rTM (θ p ) =
⇒
− n 2 cos θ p +
n 2 − sin 2 θ p
n cos θ p +
n − sin θ p
2
2
2
θp
=0
R
n cos θ p = n − sin θ p
4
2
2
2
n 4 cos 2 θ p − n 2 + sin 2 θ p
= (n − 1) ⎡⎣ n c os θ p − sin θ p ⎤⎦ = 0
2
2
θp
2
2
TE
TM
internal
reflection
⇒ θ p = tan −1 n
For n = 1.50, θ p = 56.31°
Brewster ‘s angle : tan θ p = n
: n > 1 or n < 1
Æ External & Internal reflections, but TM-polarization only
Critical angle :
sin θ c = n
:n < 1
Æ TE & TM polarizations, but Internal reflection only
external
reflection
Total Internal Reflection
Internal reflection : n =
n2
<1
n1
(TIR)
R
θc
R=1
For θ ≥ θ c = sin −1 n , called total internal reflection(TIR),
⇒ r = 1 and R = rr* = 1 for both (TE and TM) cases.
⇒ r is a complex number
rTE
rTM
internal
reflection
2
2
Er cos θi − i sin θi − n
=
=
Ei
cos θi + i sin 2 θi − n 2
2
2
2
Er − n cos θi + i sin θi − n
=
=
Ei
n 2 cos θi + i sin 2 θi − n 2
r
Complex value
23-3. Phase changes on reflection
External reflection
tTE ,TM > 0
Phase shift after External Reflection
rTE ,TM > 0
rTM
rTE ,TM is always a real number for external reflection,
then the phase shift is 0° for rTE ,TM > 0,
and the phase shift is 180°(= π ) for rTE ,TM < 0.
rTE ,TM < 0
rTE
n=1.50
External Reflection
TE
For TE case, π phase shift for all incident angles
External Reflection
TM
For TM case, π phase shift for θ < θp
No phase shift for θ > θp
Phase shift after Internal Reflection
Internal reflection
⇒ rTE > 0 for θ < θ c = sin −1 n
⇒ rTE is complex in TIR region where θ > θ c
Complex value
→ rTE = rTE eiφTE = eiφTE
In TIR region
⇒ rTM > 0 for θ < θ p = tan −1 n
: θ > θc
⇒ rTM < 0 for θ p < θ < θ c
→ rTM = − rTM = eiφTM rTM → φTM = π
⇒ rTM is complex in TIR region where θ > θ c
→ rTM = rTM eiφTM = eiφTM
TIR
For TE case, no phase shift for θ < θc
φTE(θ) phase shift for θ > θc
TIR
For TM case, no phase shift for θ < θp
π phase shift for θp < θ < θc
φTM(θ) phase shift for θ > θc
Phase shifts on total Internal Reflection for both TE- and TM-cases
When θ ≥ θ c (TIR case) then r is complex and for both the TE and TM cases has the form :
a − ib cos α − i sin α e − iα
sin α
b
r=
=
= + iα = e − i 2α = eiφ
⇒ tan α =
=
φ = − 2α
a + ib cos α + i sin α e
a
cos α
φ is the phase shift on total internal reflection(TIR ).
TE case : rTE
a = cos θi
⇒
φTE
2
2
Er cos θi − i sin θ i − n
=
=
Ei
cos θi + i sin 2 θi − n 2
b = sin 2 θ i − n 2
Internal reflection
sin 2 θi − n 2
⎛ φTE ⎞
tan α = tan ⎜ −
⎟=
cos θ
⎝ 2 ⎠
⎛ sin 2 θ − n 2
i
= − 2 tan ⎜
⎜
cos θi
⎝
−1
⎞
⎟
⎟
⎠
: θi > θ c
A similar analysis for the TM case gives :
φTM
⎛ sin 2 θ − n 2
i
= π − 2 tan ⎜
2
⎜ n cos θ i
⎝
−1
⎞
⎟
⎟
⎠
: θi > θ c
TIR
(Complex r )
Therefore,
rTE ,TM
after TIR is ………..
Internal reflection
rTE ,TM
For TIR case ( θ incident > θ c )
rTE
rTM
2
2
Er cos θi − i sin θi − n
=
=
Ei
cos θi + i sin 2 θi − n 2
Complex value
2
2
2
Er −n cos θi + i sin θi − n
=
=
Ei
n 2 cos θi + i sin 2 θi − n 2
φTM
⎛ sin 2 θ − n 2
i
= π − 2 tan ⎜
2
⎜ n cos θ
⎝
φTE
⎛ sin 2 θ − n 2
i
= − 2 tan ⎜
⎜
cos θ
⎝
−1
−1
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
φTE ,TM
Summary of Phase Shifts on Internal Reflection
φTM
φTE
⎧
0o
⎪
⎪
⎪⎪
= ⎨π (= 180o )
⎪
⎛ sin 2 θ − n 2
⎪
i
−1
⎪ π − 2 tan ⎜⎜ n 2 cos θ
⎝
⎩⎪
⎧ 0o
⎪⎪
⎛ sin 2 θ − n 2
=⎨
i
−1
⎪−2 tan ⎜⎜
cos θ
⎪⎩
⎝
Δφ = φTM − φTE
⎧ = 0o
⎪
⎨= π
⎪
o
0
>
⎩
θ <θ p'
Internal reflection
TIR
(Complex r )
θ p' < θ <θ c
⎞
⎟
⎟
⎠
θ <θ c
θ <θ c
⎞
⎟
⎟
⎠
θ >θ c
θ <θ p
θ p < θ <θ c
θc < θ
φTM
φTE
Δφ
Fresnel Rhomb
3π
near θ i = 53o when n = 1.5
4
→ After two consequentive TIRs,
Note φTM − φTE =
→ φTM − φTE =
3π
2
→ Δφ = φTM − φTE =
π
φTE
2
→ Quarter − wave retarder
Linearly polarized light (45o)
Circularly
Polarized
light
φTM
Δφ
Quarter-wave retardation after TIR
Note φTM − φTE =
π
2
φTM
near θ i = 69 when n = ???
→ Δφ = φTM − φTE =
o
π
2
→ Quarter − wave retarder
Linearly
polarized light
(45o)
φTE
Circularly
Polarized
light
n
Δφ
23-5. Evanescent Waves at an Interface
r r
r
r
Ei = Eoi exp ⎡i ki ⋅ r − ωi t ⎤
⎣
⎦
r r
r
r
Reflected beam : Er = Eor exp ⎡i kr ⋅ r − ωr t ⎤
⎣
⎦
r r
r
r
Transmitted beam : Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤
⎣
⎦
(
Incident beam :
)
(
)
(
)
For the transmitted beam :
r r
Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤
⎣
⎦
(
)
r r
)
)
)
)
kt ⋅ r = ( kt sin θt x + kt cos θt z ) ⋅ ( x x + zz )
= kt ( x sin θ t + z cos θt )
sin 2 θi
But , cos θt = 1 − sin θt = 1 −
n
2
When sin θi > n ( total internal reflection), then :
sin 2 θ i
cos θt = i
−1
n
⇒ a purely imaginary number
Evanescent Waves at an Interface
For the transmitted beam with an TIR condition ( sin θi > n ) ,
we can write the phase factor as :
⎛ sin θ
r r
t
+ iz
k t ⋅ r = kt ⎜ x
⎜
n
⎝
⎡ ⎛ k x sin θt
⎞⎤
Et = E0t exp ⎢i ⎜ t
− ωt ⎟ ⎥ exp ( −α z )
n
⎠⎦
⎣ ⎝
⎞
sin θi
−1 ⎟
⎟
n
⎠
2
Defining the coefficient α :
α = kt
sin θi
2π
−1 =
λt
n
2
z
sin θ i
−1
n
2
n2
n1 > n2
h
We can write the transmitted wave as :
⎡ ⎛ k x sin θt
⎞⎤
− ωt ⎟ ⎥ exp ( −α z )
Et = E0t exp ⎢i ⎜ t
n
⎠⎦
⎣ ⎝
n1
The evanescent wave amplitude will decay rapidly
as it penetrates into the lower refractive index medium.
⎛ ⎞
Penetration depth: Et = ⎜ e ⎟ Eot ⇒ h = α =
⎝ ⎠
1
1
Note that the incident and reflection waves
form a standing wave in x direction
λ
sin 2 θ i
−1
2π
n2
x
Frustrated TIR
d
Tp = fraction of intensity
transmitted across gap
n1=n2=1.517
1.65
Zhu et al., “Variable Transmission Output
Coupler and Tuner for Ring Laser Systems,”
Appl. Opt. 24, 3610-3614 (1985).
d/λ
Frustrated Total Internal Reflectance
Pellin-Broca prism
d
Zhu et al., “Variable Transmission Output
Coupler and Tuner for Ring Laser Systems,”
Appl. Opt. 24, 3610-3614 (1985).
d = 1 ~ λ: changing the reflectance
Rotation: changing the wavelength resonant at θB
23-6. Complex Refractive Index
⎛ σ ⎞
For a material with conductivity (σ ) : n% = 1 + i ⎜
⎟ = nR + i nI
⎝ ε0 ω ⎠
⎛ σ ⎞
2
2
n% 2 = 1 + i ⎜
⎟ = nR − nI + i 2nR nI
⎝ ε0 ω ⎠
Solving for the real and imaginary components we obtain :
nR2 − nI2 = 1
2nR nI =
σ
ε0 ω
⇒
⇒
⎛ σ ⎞
nI4 − nI2 − ⎜
⎟ =0
2
ε
ω
⎝ 0 ⎠
2
⇒
⎛ σ
⎞
2
⎜
⎟ − nI = 1
⎝ 2 nI ε 0 ω ⎠
nR =
σ
2 nI ε 0 ω
2
From the quadratic solution we obtain :
2
⎛ σ ⎞
⎛ σ ⎞
+
+
1± 1+ 4⎜
1
1
4
⎜
⎟
⎟
2
2
ω
ε
ω
ε
⎝ 0 ⎠
⎝ 0 ⎠
⇒
nI2 =
nI2 =
2
2
We need to take the positive root because nI is a real number.
2
Complex Refractive Index
Substituting our expression for the complex refractive index back into
our expression for the electric field we obtain
r r
r r
E = E0 exp ⎡i k ⋅ r − ωt ⎤
⎣
⎦
r
r
ω
⎧ ⎡
⎤⎫
= E0 exp ⎨i ⎢( nR + i nI ) ( uˆk ⋅ r ) − ωt ⎥ ⎬
c
⎦⎭
⎩ ⎣
r
r
r ⎤
⎧ ⎡n
⎡ n ω
⎤⎫
= E0 exp ⎨i ω ⎢ R ( uˆk ⋅ r ) − t ⎥ ⎬ exp ⎢ − I ( uˆk ⋅ r ) ⎥
c
⎦⎭
⎣
⎦
⎩ ⎣ c
(
)
The first exponential term is oscillatory.
The EM wave propagates with a velocity of nR / c.
The second exponential has a real argument (absorbed).
Complex Refractive Index
r r
r
r ⎤
⎧ ⎡n
⎤⎫
⎡ n ω
E = E0 exp ⎨i ω ⎢ R ( uˆk ⋅ r ) − t ⎥ ⎬ exp ⎢ − I ( uˆk ⋅ r ) ⎥
c
⎣
⎦
⎦⎭
⎩ ⎣ c
The second term leads to absorption of the beam in metals due to inducing
a current in the medium. This causes the irradiance to decrease as the wave
propagates through the medium.
r
r r* r r *
⎡ 2 nI ω ( uˆk ⋅ r ) ⎤
I ≡ EE = E0 E0 exp ⎢ −
⎥
c
⎣
⎦
r
⎡ 2 nI ω ( uˆk ⋅ r ) ⎤
r
ˆ
=
I
−
u
⋅
r
α
exp
I = I 0 exp ⎢ −
⎡
( k )⎦⎤
⎥
0
⎣
c
⎣
⎦
The absorption coefficient is defined : α =
2 nI ω
4π nI
=
c
λ
23-7. Reflection from Metals
Reflection from metals is analyzed
by substituting the complex refractive index n% in the Fresnel equations :
TE case : rTE
cos θi −
E
= r =
Ei
cos θi +
TM case : rTM =
n% 2 − sin 2 θi
n% 2 − sin 2 θi
−n% 2 cos θ i +
Er
=
Ei
n% 2 cos θi +
Substituting n% = nR + i nI
Reflectance
n% 2 − sin 2 θi
n% 2 − sin 2 θi
we obtain :
Er cos θi −
TE case : r =
=
Ei
cos θi +
(n
(n
2
R
− nI2 − sin 2 θi ) + i ( 2nR nI )
2
R
− nI2 − sin 2 θi ) + i ( 2nR nI )
2
2
Er − ⎡⎣( nR − nI ) + i ( 2nR nI ) ⎤⎦ cos θi +
TM case : r =
=
Ei
⎡( nR2 − nI2 ) + i ( 2nR nI ) ⎤ cos θi +
⎣
⎦
(n
2
R
(n
2
R
θi
− nI2 − sin 2 θi ) + i ( 2nR nI )
− nI2 − sin 2 θi ) + i ( 2nR nI )
Reflection from Metals at normal incidence (θi=0)
At normal incidence, θi = 0° :
rTE =
rTM =
cos θi −
n% 2 − sin 2 θi
cos θi +
n% 2 − sin 2 θi
At normal incidence
(from Hecht, page 113)
=
1 − n%
1 + n%
− n% 2 cos θi +
n% 2 − sin 2 θi
n% 2 cos θi +
n% 2 − sin 2 θi
∴ r=
1 − ( nR − i nI
1 + ( nR − i nI
=
1 − n%
1 + n%
)
)
The power reflectance R is given by
R = r r*
⎡1 − ( nR − i nI ) ⎤ ⎡1 − ( nR + i nI ) ⎤ ⎛ 1 − 2nR + nR2 + nI2 ⎞
=⎢
⎥⎢
⎥=⎜
2
2 ⎟
+
−
+
+
+
+
+
n
i
n
n
i
n
n
n
n
1
1
1
2
(
)
(
)
⎝
R
I ⎦⎣
R
I ⎦
R
R
I ⎠
⎣
( n − 1)
R= R
2
( nR + 1)
2
+ nI2
+ nI2
visible
λ