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ECE 318
Fundamentals of Optics
Lecture 16 & 17
Light as Waves VI
• Total Internal Reflection
* Topics, questions, or homework problems with an asterisk (*) are at a difficulty level or having
a scope beyond the expectation of this course. Only qualitative understanding is required.
Total Internal Reflection
So far we have considered the cases where there always exists a
propagating transmitted wave. Snell’s law is satisfied with a transmitted
angle θt between 0 and π/2.
Now we are going to consider a special case called Total Internal
Reflection (TIR). Two conditions must be satisfied for TIR to occur:
(1) ni>nt, that is, the incident light wave is in the denser medium
(2) θi >θc , that is, the incident angle is greater than the critical angle
In what follows, we’ll show that, upon TIR, both TE and TM polarizations
are 100% reflected, however, they will experience different phase shifts.
Q16-1: Next time you go for a swim, think about how you could verify first hand the zero
transmission phenomenon when total internal reflection occurs at the water-air
boundary.
2
© Li Qian
Lecture 16 &17
Critical Angle
Critical angle refers to the incident angle when the refracted angle is exactly π/2. This is a
special case that occurs only for internal reflections.
Snell’s Law:
ni
nt
ki
θc
ni sin θ i = nt sin θ t
kr
(1)
Therefore:
π 
ni sin θ c = nt sin  
2
kt
or
(2)
 nt 
θ c ≡ sin  
 ni 
−1
Above: Solving for θc graphically using
k-matching at the boundary.
(3)
Example: For a glass-air interface, the critical angle occurs at 41.8o, for a water-air
interface, this occurs at 48.8o.
3
© Li Qian
Lecture 16 &17
What happens to Snell’s Law beyond
the critical angle?
It can be shown that (see Homework #1), even in the case of total internal reflection, Snell’s
law still holds, except that the sine of the transmitted angle becomes greater than 1, and the
cosine of the transmitted angle becomes imaginary. Though there is no physical interpretation
of the “transmitted angle”, it is still useful as a mathematical tool. The Fresnel reflection
formulae (Lecture 14&15, Eq (24) & (26)) still apply if we continue to use the cosθt term,
despite its imaginary value.
sin θ t =
ni sin θ i
>1
nt
(4)
2
 n sin θ i 
 − 1
cos θ t = i  i
 nt 
(5)
Q16-2: What’s the origin of Snell’s Law from the Wave Optics point of view?
4
© Li Qian
Lecture 16 &17
“Transmitted” wave vector is complex
Let’s analyze the wave vector kt of the “transmitted wave”, assuming it exists. In order to
satisfy the wave equations (Lecture 9, Eq (27) & (28)), we must have:
y
ω
(6)
k t = ktx2 + kty2 = nt
c
ki θi
ni
nt
θr
And in order to satisfy the boundary phasematching condition, we must have
kr
kt=?
ktx = kix = ki sin θ i =
x
Above: Boundary conditions cannot be
satisfied with a real kt .
kty = kt2 − ktx2 = kt2 − (ki sin θ i ) =
2
c
ni sin θ i
(7)
And in order to satisfy both (6) and (7), and
noting that nisinθi > nt, kty must become
imaginary!
ω
c
nt2 − (ni sin θ i )
2
2
 n sin θ i 
 − 1
= ±i nt  i
c
 nt 
ω
(8)
≡ ± iβ
5
ω
© Li Qian
Lecture 16 &17
“Transmitted” wave is evanescent
What does an imaginary kty signify and what is the consequence? If we substitute (7) and
(8) into our usual wave expression for the transmitted wave, we will have:
 ω

Et = Eto exp i (ktx x + kty y − ωt + φ ) = Eto e βy exp i ni sin θ i x − ωt + φ 

c
[
]
Amplitude
exponentially
decays with –y.
(9)
A wave propagating in the x
direction with vφ=c/(nisinθi)
Note 1. In deriving (9), only kty=–iβ is taken, since kty=+iβ leads to an unphysical result.
Note 2. For finite incident beam size, the x-direction propagation of the evanescent wave is sustained
only within the extent of the incident beam.
The wave expressed by (9) is a peculiar wave: it differs from the regular plane wave in that
its amplitude exponentially decays in the directions orthogonal to its propagation direction.
It is called an evanescent wave.
Q16-3: Does this evanescent wave have planar wavefronts?
Q16-4: In an absorbing medium, the E field also decays exponentially (see Lecture 12&13,
Eq(15)). How is the wave propagating in an absorbing medium different from the
evanescent wave described here?
6
© Li Qian
Lecture 16 &17
Penetration Depths
From Eq (9), we can draw three conclusions regarding the amplitude of the transmitted E
field:
(1) E field is finite in the nt medium, that is, there is field penetration into the
transmitted medium. The field does not stop abruptly at the interface.
(2) E field is constant in the planes parallel to the interface, i.e., for y = constant.
(3) E field decays exponentially with the distance away from the interface.
We can define two penetration depths:
(1) The distance from the interface at which the E field decays to its 1/e value
Field penetration depth = 1/β
(10)
(2) The distance from the interface at which the intensity decays to its 1/e value
(11)
Intensity penetration depth = 1/2β
Q16-5: Penetration depth is a function of incident angle. At what angle is penetration
depth the greatest?
Q16-6: For optical waves, what’s the approximate length scale of the penetration depths
for large incident angles?
7
© Li Qian
Lecture 16 &17
“Transmitted” wave is non-transverse*
It can be shown that the evanescent wave is non-transverse, that is, it has an E or an H
component in the direction of propagation (x-direction). For example, if the incident
polarization is TE, then the transmitted wave can be written as:
Et = Etoe βy exp[i (kix x − ωt + φ )]zˆ
(12)
Now if we calculate the transmitted H field from the E field using one of the Maxwell’s
Equations:
∂B
∂H
∇×E = −
= −µ
(13)
∂t
∂t
We obtain:
∂H x
∂Et
β
⇒ βEt = iµωH x ⇒ H x = −i
= −µ
Eto e βy exp[i (ktx − ωt + φ )] (14)
∂t
µω
∂y
∂H y
∂E
k
− t = −µ
⇒ − iktx Et = iµωH y ⇒ H y = − tx Eto e βy exp[i (ktx − ωt + φ )] (15)
∂x
∂t
µω
Eq (14) indicates that the H field has a finite x-component. Note that Hx is π/2 out-of-phase
with Ez . This is another peculiar feature of the evanescent wave. (Recall that, for a “normal”
propagating wave, E and H are in phase.)
8
© Li Qian
Lecture 16 &17
“Transmitted” wave is not transmitting
energy across the boundary*
We can also show that there is no net energy transport across the boundary. Again, take
TE polarization as an example, from (12), (14) & (15), we can write the E and H fields
using real representations:
Et = Eto e βy cos(ktx x − ωt + φ ) zˆ
(16)
 β

k
H t = Etoe βy 
sin (ktx x − ωt + φ )xˆ − tx cos(ktx x − ωt + φ )yˆ 
µω
 µω

(17)
Poynting vector of the “transmitted” wave becomes:
Et × H t =
β 2 2 βy
k
Etoe cos(ktx x − ωt + φ )sin (ktx x − ωt + φ )yˆ + tx Eto2 e2 βy cos2 (ktx x − ωt + φ )xˆ
µω
µω
(18)
The time-average of the y-component of the Poynting vector is zero, as shown below. This
signifies that, on average, no net energy flow in the y-direction.
Et × H t ⋅ yˆ =
=
β 2 2 βy
Eto e cos(ktx x − ωt + φ )sin (ktx x − ωt + φ )
ωµ
β
Eto2 e 2 βy sin (2ktx x − 2ωt + 2φ ) = 0
2ωµ
(19)
Q16-7: How do you reconcile the fact that there is finite field penetration into the nt
medium, but no net energy transport?
9
© Li Qian
Lecture 16 &17
Complex Fresnel Reflection Coefficients
As mentioned before, the Fresnel reflection coefficients can still be applied to the TIR case,
as long as we incorporate the imaginary cosθt term, expressed by Eq (5).
TE-case:
rTE =
TM-case:
ni cos θ i − int
ni cos θ i + int
(ni sin θ i
(ni sin θ i
nt )2 −1
nt ) −1
2
∴ rTE = eiφTE
φTE

−1  nt
= −2 tan


(ni sin θi
nt )2 −1 

ni cosθi

(20)
rTM =
nt cos θ i + ini
(ni sin θ i
(ni sin θ i
∴ rTM = eiφTM
(21)
(22)
− nt cos θ i + ini
φTM

−1  ni
= −2 tan


nt ) − 1
2
nt ) − 1
2
(23)
(24)
(ni sin θi
nt )2 −1 
+ π (25)

nt cosθi

Important conclusions: (1) The wave is 100% reflected; (2) The reflected wave is not in phase
with the incident wave; (3) The additional phase incurred in the reflected wave is different from
TE and TM polarized waves.
10
© Li Qian
Lecture 16 &17
Phase Shifts upon Total Internal
Reflection
The figure below shows φTE, φTM, and the relative phase difference (φTM – φTE) as a function
of incident angle for a glass-to-air interface.
π
∆φ=φΤΜ−φΤΕ
Phase
π/2
φΤΜ
0
0
θc
π/2
θi
φΤΕ
−π/2
−π
Q16-8: Do the wavefronts of the incident, reflected and transmitted wave match at the
boundary upon total internal reflection?
11
© Li Qian
Lecture 16 &17
Reflectance and Transmittance
From (21) and (24), rTE and rTM both have a magnitude of unity in the region of Total
Internal Reflection (left plot). Therefore, the reflectances of both TE and TM polarizations
are unity as well, and the transmittances are zero (right plot).
1
1
rTE
0.5
0
0
θp
rTM
-0.5
-1
12
ni=1.5, nt=1
Transmittance or Reflectance
Fresnel Reflection Coefficients
ni=1.5, nt=1
Region of
θc
Total
Internal
Reflection
TTM
0.8
TTE
Incident Angle (θi)
π/2
Region of
0.6
Total
Internal
0.4
Reflection
RTE
RTM
0.2
0
0
RTE=RTM=1
00
TTE=TTM=0
θp θc
π/2
Incident Angle (θi)
© Li Qian
Lecture 16 &17
Frustrated Total Internal Reflection
Even though the evanescent wave does not transport any energy to the nt medium, the finite
E field in the nt medium can still polarize the atoms in the medium. If a higher index medium
with index nt2 is placed near the ni/nt interface, and as long as the boundary conditions can be
satisfied by a propagating wave in the nt2 medium, a propagating transmitted wave will result
(see graph below). Consequently, a finite amount of energy will be transmitted to the nt2
medium, and we say that the initial total internal reflection is frustrated.
y
FTIR has been used in numerous applications,
including:
ni
ki
θi θr
kr
x
nt
nt2
kt2
13
- Prism coupling: coupling light into and out of
waveguides;
- Optical directional couplers with variable ratios;
- Coupling light into the “Whisper-Gallery” mode of
a high Q resonator
With multiple dielectric layers separated by air gaps,
one can create a resonant tunneling condition and
make highly selective wavelength filters.
© Li Qian
Lecture 16 &17
Example: Fresnel Rhomb
The Fresnel rhomb can convert linear polarization into circular polarization, or vice versa. It is
made of a transparent, homogenous, isotropic material, such as glass. Its cross-section is a
parallelogram with an apex angle θ, as shown. Incoming light enters the rhomb normal to its
input facet, experiences two total internal reflections, and exits normal to its output facet. You
are asked to calculate the apex angle. The refractive index of the rhomb is 1.5.
If the linearly polarized input has equal TE and TM
components, say, with an amplitude of Eo, then the input
can be expressed by:
Ei
input
facet
Ei θ
output
facet
14
(26)
After two total internal reflections, and ignoring the facet
reflections, the output wave becomes:
θ
cross-section
view
Ei = Eo (uˆ TE + uˆ TM ) exp[i (k ⋅ r − ωt + φ )]
Eout
θ
Eout
E out = Eo (rTE rTE uˆ TE + rTM rTM uˆ TM ) exp[i (k ⋅ r − ωt + φ )]
(
= E (uˆ
)
)exp[i(k ⋅ r − ωt + φ + 2φ )]
= Eo ei 2φTE uˆ TE + e i 2φTM uˆ TM exp[i (k ⋅ r − ωt + φ )]
o
TE
+ ei 2(φTM −φTE )uˆ TM
© Li Qian
TE
(27)
Lecture 16 &17
Example: Fresnel Rhomb (Cont’d)
In order to convert to circular polarization, one requires:
2(φTM − φTE ) = π 2 + mπ
(28)
where m is an integer. From the plot shown on Page 11, the most appropriate m value is 1.
Substitute (22) and (25) into (28), and using ni=1.5, nt=1.0, we can solve for θi:
θ = θ i ≈ 54.6o
(29)
Q16-9: Given a Fresnel rhomb with a proper apex angle, would linearly polarized input
always be converted to circularly polarized light at the output? What is the other condition
that must be satisfied?
Q16-10: Can one use the Fresnel rhomb to convert circularly polarized light into linearly
polarized light? Into elliptically polarized light?
15
© Li Qian
Lecture 16 &17
Example: Right-angle Prism
front
facet
The right-angle glass prism is often used in optical
experiments as a retro-reflector / delay stage (top
figure). A circularly polarized collimated laser beam
(bottom graph) enters the prism through the front facet
at normal incidence. The refractive index of the prism
is 1.5.
45o
d
y’ ecte
fl
re
z’
t
n
e
x’ y
id
c
z in
x
Ein
y
x
16
(a) Find the reflectance of the prism.
(b) Calculate the phase difference between the TE
and TM polarizations of the output beam.
(c) Plot the time-evolution of the output E field vector
using the co-ordinate system specified in the top
figure.
(d) Can the TE-TM phase difference due to the two
reflections be compensated exactly by another
identical prism. Why or why not?
© Li Qian
Lecture 16 &17
Example: Right-angle Prism (Cont’d)
Solution:
(a)
Find the reflectance of the prism (i.e., the ratio of output power to input power)
n=1.5
front
facet
45o
45o
45o
17
As shown on the left, the laser beam is refracted and
reflected at 4 interfaces:
1) Normal incidence (air-to-glass) at the front facet.
Let t1 and T1 denote the Fresnel transmission
coefficient and transmittance, respectively.
2) Internal reflection (glass-to-air) at 45o angle. Let r2
and R2 denote the Fresnel reflection coefficient
and reflectance, respectively.
3) Another internal reflection (glass-to-air) at 45o
angle. Use r3 and R3 for notation.
4) Normal incidence (glass-to-air) at the front facet.
Use t4 and T4 for notation.
© Li Qian
Lecture 16 &17
Example: Right-angle Prism (Cont’d)
Using Fresnel Equations, one can easily find
2
 n −1 
T1 = T4 = 1 − 
 = 0.96
 n +1
(30)
For R2 and R3, the incident angle is 45o, and since
1
45o > sin −1   = 41.8o
n
(31)
The incidence angle is greater than the critical angle. Therefore, total internal
reflection occurs at these two interfaces. Hence,
R2 = R3 = 1
(32)
The reflectance of the prism becomes:
R prism ≡
18
output power
= T1 R2 R3T4 = 92.2%
input power
© Li Qian
(33)
Lecture 16 &17
Example: Right-angle Prism (Cont’d)
(b)
Calculate the phase difference between the TE and TM polarizations of the output beam.
Using the co-ordinate system specified, we can write the expression for the incident
polarization:
π
i


(34)
E = E  xˆ + e 2 yˆ 




Note we have omitted the exp[i (kz − ωt + φ )] component in (34), as it is not relevant to
our polarization analysis. In fact, (34) is simply the Jones vector for Ein. The output
polarization can be expressed as:
π
i


E out = Eo  t1_ TM r2 _ TM r3 _ TM t 4 _ TM xˆ '+t1_ TE r2 _ TE r3 _ TE t 4 _ TE e 2 yˆ ' 
(35)


For normal incidence, there is no polarization dependence, and therefore,
2
(36)
t1_ TE = t1_ TM ≡ t1 =
n +1
2n
t 4 _ TE = t 4 _ TM ≡ t 4 =
(37)
n +1
in
o
Note that both t1 and t4 are real and positive, which means no phase change after
transmission.
19
© Li Qian
Lecture 16 &17
Example: Right-angle Prism (Cont’d)
For the two total internal reflections, however, there are phase changes, and
furthermore, the TE and TM components experience different phase changes.
r2 _ TE = r3 _ TE = eiφTE
r2 _ TM = r3 _ TM = e iφTM
where
φTE

−1 
= −2 tan 


(n sin 45 )
− 1 
o
=
−
0
.
6435
rad
=
−
36
.
9
o

n cos 45

Eout
20
(39)
o 2
(
)
 n n sin 45o 2 − 1 
 + π = 1.85rad = 106.3o
φTM = −2 tan 

cos 45o



Substitute (36)-(39) into (35), we get the output polarization expression:
−1
(38)
π

 i 2φ
i  2φTE +  
2 
yˆ '
= Eot1t4  e TM xˆ '+e 




© Li Qian
(40)
(41)
(42)
Lecture 16 &17
Example: Right-angle Prism (Cont’d)
Therefore, the phase difference between TE and TM polarizations at the output is:
∆φ = 2φTE +
(c)
π
2
(43)
Plot the time-evolution of the output E field vector.
Eout
y’
t = (Φ o + π 2 ) ω
x’
t = Φo ω
21
− 2φTM = −3.43rad = −196.3o
First, we can define the “rectangular box” that confines
the time-evolution trace. From (42), we notice that the
x’ and y’ components of Eout have equal magnitude,
and therefore the “box” is a square. Second, we need
to determine qualitatively the orientation of the ellipse
and the handedness of the rotation. To do so, we use
the real representations to express the variation of the
two components with time:
Eout _ x ' = E 'o cos(Φ 0 − ωt )
Eout _ y ' = E 'o cos(Φ 0 − ωt + ∆φ )
where Φo is not time varying, and ∆φ is shown in (43).
© Li Qian
Lecture 16 &17
Example: Right-angle Prism (Cont’d)
(d)
Can the TE-TM phase difference due to the two reflections be compensated exactly
by another identical prism. Why or why not?.
Yes. It can be compensated by another identical prism. One needs to switch the role
of TE and TM after the first prism, so that the original x-polarized and y-polarized
components experience the same total phase shift (2φTE+2φTM) after going through
two prisms. This is achieved by rotating the second prism by 90o, so that the plane of
incidence is perpendicular to each other, as shown:
22
© Li Qian
Lecture 16 &17
More Questions
Q16-11: Consider the polarization state change upon reflection at an interface between
two media of refractive indices ni and nt. List all the conditions under which
each of the following scenario occurs:
(a)
Polarization state unchanged
(b)
Linear polarization (incident) changes to an elliptical polarization (reflected)
(c)
Linear polarization (incident) changes to a different linear polarization
(reflected)
(d)
Random polarization (incident) changes to linear polarization (reflected)
Q16-12: Can we apply Fresnel’s Equations in the case of frustrated total internal
reflection? Why or why not?
23
© Li Qian
Lecture 16 &17
Homework
1.
24
*From Maxwell’s Equations and boundary conditions, derive the Fresnel
reflection coefficients in the case of total internal reflection (Eq (20) and
(23)), and therefore show that one can use Eq (5) for cosθt, even though an
imaginary cosθt is not physical.
© Li Qian
Lecture 16 &17