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ELEC425-summer, 2012 1
Assignment 5
Assignment 5.
1.13, 7.1, 7.2, 7.5, 7.11, 7.12, 7.15, 7.21
1.13. TIR and FTIR
a) By considering the electric field component in medium B in Figure 1.20 (b),
explain how you can adjust the amount of transmitted light.
B = Low refractive index
transparent film ( n 2)
Reflected
light
n2
n1
n1
TIR
i > c
A
(a)
Transm itted
Incident
light
Glass prism
n1
Reflected
FTIR
Incident
light
i > c
C
A
(b)
(a) A light incident at t he long face of a glass prism suffers T IR; t he prism deflects the
light.
(b) T wo prisms separated by a thin low refractive index film forming a beam-splitt er cube.
T he incident beam is split into two beams by FT IR.
© 1999 S.O. K asap,Optoelectronics(P rentice H all)
Figure 1.20. from Optoelectronics, S. O. Kasap
b) What is the critical angle at the hypotenuse face of a beam splitter cube
made of glass with n₁ = 1.6 and having a thin film of liquid with n₂ = 1.3.
Can you use 45° prism with normal incidence?
c) Explain how a light beam can propagate along a layer of material between
two different media as shown in Figure 1.34 (a). Explain what the
requirements are for the indices n₁, n₂, n₃. Will there be any losses at the
reflections?
d) Consider the prism coupler arrangement in Figure 1.34 (b). Explain how this
arrangement works for coupling an external light beam from a laser into a
thin layer on the surface of a glass substrate. Light is then propagated inside
the thin layer along the surface of the substrate. What is the purpose of the
adjustable coupling gap?
ELEC425-summer, 2012 2
Assignment 5
Las er light
Air
d = Adjustable coupling gap
n2
n 1 Thin layer
Glass substrate
P rism
Thin layer
n3
Glass substrate
(a)
(b)
(a) Light propagation along an opt ical guide. (b) Coupling of laser light int o a t hin layer optical guide - using a prism. The light prop agates along t he t hin layer.
© 1999 S.O. K asap,Optoelectronics(P rentice H all)
Figure 1.34. from Optoelectronics, S. O. Kasap
Solution.
a) Consider the prism A when the neighboring prism C in Figure 1.20 (b) in far
away. When the light beam in prism A is incident on the A/B interface,
hypotenuse face, it suffers TIR as θi > θc. There is however an evanescent
wave whose field decays exponentially with distance in medium B. When
we bring prism C close to A, the field in B will reach C and consequently
penetrates C. (The tangential field must be continuous from B to C). One
cannot just use the field expression for the evanescent wave because this was
derived for a light beam incident at an interface between two media only; no
third medium. The transmitted light intensity from A to C depends on the
thickness of B.
b) For the prism A in Figure 1.20 (b), n₁ = 1.6 and n₂ = 1.3 so that the critical
angle for TIR at the hypotenuse face is
 n2 
 1.3 
  sin 1 
  54.3
 1.6 
 n1 
 c  sin 1 
in this case 45° prism cannot be employed, Figure 1.20 (a).
c) If the angle of incidence θi at the n₁/n₂ layer is more than the critical angle
θc12 and if angle of incidence θi at the n₁/n₃ layer is more than the critical
angle θ₁₃ then the light ray will travel by TIR zigzagging between the
boundaries as sketched in Figure 1.20 (a). For example, suppose that n₁ = 2
(thin layer), n₂ = 1 (air), n₃ = 1.6 (glass),
ELEC425-summer, 2012 3
Assignment 5
 n2 
1
  sin 1    38.8
2
 n1 
 c12  sin 1 
 n3 
 1.6 
  sin 1 
  53.1
n
2


 1
 c13  sin 1 
So that θi > 53.1° will satisfy TIR. There is no loss in TIR as the magnitude
of the amplitude of the reflected way is the same as that of the incident
wave.
Note: There is an additional requirement that the waves entering the thin
film interfere constructively, otherwise the waves will interfere destructively
to cancel each other. Thus there will be an additional requirement, called the
waveguide condition (Chapter 2).
d) The light ray entering the prism is deflected towards the base of the prism.
There is a small gap between the prism and the thin layer. Although the light
arriving at the prism base/gap interface is reflected, because of the close
proximity of the thin layer, some light is coupled into the thin layer as it was
discussed in part (a) due to frustrated TIR. This arrangement is a much more
efficient way to couple the light into the thin layer because the incident light
is received by the large hypotenuse face compared with coupling the light
directly into the thin layer.
7.1. Polarization.
Suppose that we write the Ex and Ey components of a light wave generally as:
Ex  Ex 0 cost  kz and E y  E y 0 cost  kz   
Show that at any instant Ex and Ey satisfy the ellipse equation on the Ey vs. Ex
coordinate system:
2
2
 Ex   E y 

 E 
  2 Ex  y  cos   sin 2 

  





 Ex 0   E y 0 
 Ex 0  E y 0 
Sketch schematics what this ellipse look like assuming Ex0 = 2Ey0. When would
this ellipse form an (a) ellipse with its major axis on the x-axis, (b) a linearly
polarized light at 45° to the x-axis, (c) right and left circularly polarized light?
ELEC425-summer, 2012 4
Assignment 5
Solution.
Consider the LHS of equation
2
2
 Ex   E y 

 E 
  2 E x  y  cos  

  





 Ex 0   E y 0 
 E x 0  E y 0 
 E cos t  kz   E y 0 cos t  kz    

 E cos t  kz    
  2 E x 0 cos t  kz  y 0
 cos  
  
  x 0





Ex0
Ey0
E
E
x
0
y
0

 





2
2
 cos t  kz  cos t  kz     2 cos t  kz cos t  kz    cos  
2
2
 cos 2 t  kz  cos t  kz cos   sin t  kzsin   
 2 cos t  kzcos t  kz cos   sin t  kzsin  cos  
2
 cos 2 t  kz  cos 2 t  kz cos 2   sin 2 t  kzsin 2  
 2 cos t  kzsin t  kz cos  sin  
 2 cos 2 t  kz cos 2   2 cos t  kzsin t  kz cos  sin  
 cos 2 t  kz  cos 2 t  kz cos 2   sin 2 t  kzsin 2  


 cos 2 t  kz 1  cos 2   sin 2 t  kzsin 2  
 cos 2 t  kzsin 2   sin 2 t  kzsin 2   [cos 2 t  kz  sin 2 t  kz] sin 2   sin 2 
The equation is proved.
Consider the general expression
2
2
 Ex   E y 

 E 
  2 Ex  y  cos   sin 2 

  
 E  E 


 Ex 0   E y 0 
 x 0  y 0 
This is a quadratic equation in Ey (or Ex),
aE y2  bE y  c  0
where
2
 E  1 


1
 cos  ; c   Ex   sin 2 
a  2 ; b  2 x 




E yo
 Ex 0  E y 0 
 Ex 0 
Thus, for a given ϕ and a given Ex0 and Ey0, the graph Ey vs. Ex can be sketched
where Ex determines b and c and Ey is given by
ELEC425-summer, 2012 5
Assignment 5
Ey  
b  b 2  4ac
2a
If Ex0 = 2Ey0 then
a) for ϕ = π/2, Ey0 = 1 and Ex0 = 2 we will obtain an ellipse with its major axis
on the x-axis
ELEC425-summer, 2012 6
Assignment 5
b) For ϕ = 0, Ey0 = 1 and Ex0 = 1 we obtain the line (linear) polarization at an
angle π/4 to the x-axis
c) For ϕ = π/2, Ey0 = 1 and Ex0 = 1we obtain a circle, if ϕ= π/2 right circular
polarization; shift ϕ = -π/2 results in left circular polarization
ELEC425-summer, 2012 7
Assignment 5
7.2. Linear and circular polarization
Show that a linearly polarized light wave can be represented by two circularly
polarized light waves with opposite rotations. Consider the simplest case of a wave
linearly polarized along the y-axis. What is your conclusion?
Solution.
if the wave is polarized along the y-direction then circularly polarized light has
retarded electric field in x-direction and E₀ = Ex0 = Ey0 = 1

for right circularly polarized light: ERx  E0 cos t  kz   , ERy  E0 cost  kz

2

for left circularly polarized light: ELx  E0 cos t  kz   , ELx  E0 cost  kz

Total x and y components of two polarizations are
2
ELEC425-summer, 2012 8
Assignment 5




Ex  E0 cos t  kz    E0 cos t  kz    0
2
2


E y  2E0 cost  kz
There is only one component which proves the linear polarization along y-axis
In case of the linear polarization in x-axis, the retarded component will be in y
direction ( ERy  E0 cos t  kz   , Ex  E0 cost  kz )
2

=
+
7.5. Jones Matrices
When we represent the state of polarization of a light wave using a matrix, called a
Jones matrix (or vector) then various operations on the polarization state
correspond to multiplying this matrix with another matrix that represents the
optical operation. Consider a light wave travelling along z with field components
Ex and Ey along x and have a phase difference ϕ between them. If we use the
exponential notation then


Ex  Ex 0 exp  j t  kz  x  and E y  E y 0 exp j t  kz   y 
Jones matrix is a column matrix whose elements are Ex and Ey without the
common exp[j(ωt-kz)] factor
ELEC425-summer, 2012 9
Assignment 5
 Ex   Ex 0 exp  jx 
E 

 E y   E y 0 exp  j y 
(1)
Usually Eq. (1) is normalized by dividing by the total amplitude E0  Ex20  E y20  .
1/ 2
We can also factor out exp(jϕx) to further simplify to obtain the Jones matrix:
J
1
E0
Ex 0


 E exp  j 
 y0

(2)
where ϕ=ϕy – ϕx.
a) Table 7.3 shows Jones vectors for various polarizations. Identify the state of
polarization for each matrix.
b) Passing a wave of given Jones vector Jin through an optical device is
represented by multiplying Jin by the transmission matrix T of the device.
If Jout is the Jones vector for the output light through the device, then Jout = T
Jin. Given
1 0
T 

0 j 
(3)
Determine the polarization state of the output wave given the Jones vectors
1
in Table 7.3, and the optical operation represented by T. Hint: Use   as
1

input for determining T.
Table 7.3 Jones vectors
Jones vector 1 
0 
Jin
 
Polarization ?
Transmission 1 0
0 0 
matrix T


Optical
operation
?
1 1

2 1
?
 e j

0
?
0

e j 
cos  
 sin  


?
1 0 
0 j 


?
Solution.
a)
Polarization state for each matrix
1 1 
 
2  j
1 1
 
2  j 
?
1 0 
0 1


?
?
1 0 
0 e  j 


?
ELEC425-summer, 2012 10
Assignment 5
Jones vectors
Jones vector
Jin
Polarization
1 
0 
 
Linear;
horizontal E
cos  
1 1
1 1 
1
 sin  
 
2
2  j


Linear; E at Linear; E at Right
45° to x-axis θ to x-axis
circularly
polarized
1 1
 
2  j 
Left
circularly
polarized
b) Optical operation represented by T is
Jones vectors
Jones vector
Jin
Polarization
Transmission
matrix T
Optical
operation
1 
0 
 
Linear,
horizontal E
1 0 
0 0 


Linear
polarizer;
horizontal
transmission
axis
cos  
1 1
1 1 
1
 sin  
 
2
2  j


Linear; E at Linear; E at Right
45° to x-axis θ to x-axis
circularly
polarized
j
1 0 
1 0 
e
0




0 1
j
0 j 


0 e 
1 1
 
2  j 
Left
circularly
polarized
1 0 
0 e  j 


Isotropic
Quarterphase
wave plate
changer or
phase
retarder
Wave
retarder; fast
axis along x
Half-wave
plate
The polarization state of the output waves from quarter-wave plate are
Jones vectors
Jones vector
Jin
1 
0 
 
Polarization Linear
horizontal
polarization,
horizontal E
Transmission 1 0 
0 j 
matrix T


Jones vector 1 
0 
Jout
 
Linear
horizontal
polarization,
horizontal E
cos  
1 1
1 1 
1
 sin  
 
2
2  j


Linear; E at Linear; E at Right
45° to x-axis θ to x-axis
circularly
polarized
1 0 
0 j 


1 1 
 
2  j
Right
circularly
polarized
1 0 
0 j 


 cos  
 j sin  


Right
circularly
polarized
1 1
 
2  j 
Left
circularly
polarized
1 0 
1 0 
0 j 
0 j 




1 1
1 1
1

2 1
2 
Linear; E at Linear; E at
(2π-π/4) to π/4 to x-axis
x-axis
ELEC425-summer, 2012 11
Assignment 5
The polarization state of the output waves are
Jones vectors
Jones vector
Jin
Polarization
1 
0 
 
Linear,
horizontal E
Transmission
matrix T
1 0 
0 0 


Linear
Optical
polarizer;
operation
horizontal
transmission
axis
Jones vector 1 
0 
Jout
 
Linear,
horizontal
polarization
cos  
1 1
1 1 
1
 sin  
 
2
2  j


Linear; E at Linear; E at Right
45° to x-axis θ to x-axis
circularly
polarized
j
1 0 
1 0 
e
0




0 1
j
0 j 


0 e 
1 1
 
2  j 
Left
circularly
polarized
1 0 
0 e  j 


Isotropic
phase
changer or
phase
retarder
e j 1

2 1
Linear, E at
45° to x-axis
with
Ex
leading Ey
Quarterwave plate
Half-wave
plate
Wave
retarder; fast
axis along x
 cos  
 j sin  


Right
elliptical
polarization
1 1
 
2  j 
Left
circular
polarized
1  1 
  j 
2 e 
Linear,
retarded ycomponent,
fast
axis
along x
7.11. Glan-Foucault prism
Figure 7.37 shows the cross section of a Glan-Foucault prism which is made of two
right angle calcite prisms with a prism angle of 38.5°. Both have their optic axes
parallel to each other and to the block faces as in the figure. Explain the operation
of the prisms and show that the o-wave does indeed experience total internal
reflection.
ELEC425-summer, 2012 12
Assignment 5
Solution.
Calcite is optically anisotropic material, birefringent. This is negative uniaxial
crystal since two of their principle indices the same (n₁ = n₂) and n₃ < n₁.
The light has two orthogonally polarized components in uniaxial crystal, (o)
ordinary and (e) extraordinary waves. The refractive indices (from Table 7.1) are
no = 1.658 and ne = 1.486.
The critical angles for TIR in e- and o-waves are
 c o  wave  arcsin 1 / no   37.09
 c e  wave  arcsin 1 / ne   42.3
If the angle of incidence is θ at calcite/air interface then from Figure 7.37
90°+(90°-θ)+38.5° = 180° → θ = 38.5° > θc (o-wave)
θ = 38.5° < θc (e-wave)
Thus, the o-wave suffers TIR while the e-wave does not. Hence the beam that
emerges is the e-wave, with a field Ee along optic axis.
ELEC425-summer, 2012 13
Assignment 5
A bs orbe r
o -ray
38.5
θ
e -ray
C alcite
O p tic axis
A ir-ga p
Th e Glan -Fo uc au lt p rism prov ide s lin ea rly po larize d lig ht
© 1999 S .O. K asap, O ptoel ect roni cs (P renti ce H al l)
7.12. Faraday Effect.
Application of a magnetic field along the direction of propagation of a linearly
polarized light wave through a medium results in the rotation of the plane of
polarization. The amount of rotation θ is given by
  BL
where B is the magnetic field (flux density), L is the length of the medium, and ϑ is
the so-called Verdet constant. It depends on the material and the wavelength. In
contrast to optical activity, sense of rotation of the plane of polarization is
independent of the direction of light propagation. Given that glass and ZnS have
Verdet constants of about 3 and 22 minutes of arc Gauss⁻¹meter⁻¹ at 589 nm
respectively, calculate the necessary magnetic field for a rotation of 1° over a
length 10 mm. What is the rotation per unit magnetic field for a medium of length
1 m? (Note: 60 minutes of arc = 1° and 10⁴ Gauss = 1Tesla).
Solution.
Rotation per unit magnetic field
For glass: B 
For ZnS: B 

60'

 2000 Gauss or 0.2 Tesla
1
L 3' Gauss meter 110 103 m

60'

 273 Gauss or 0.27 Tesla
1
L 22' Gauss meter 110 103 m
ELEC425-summer, 2012 14
Assignment 5
7.15 Transverse Pockels cell with LiNbO₃
Suppose that instead of the configuration in Figure 7.20, the field is applied along
the z-axis of the crystal, the light propagates along the y-axis. The x-axis is the
polarization of the ordinary wave and z-axis that of the extraordinary wave. Light
propagates through as o- and e-waves. Given that Ea = V/d, where d is the crystal
length along z, the indices are
1
no,  no  no3r13Ea
2
1
ne,  ne  ne3r33Ea
2
Show that the phase difference between the o- and e-waves emerging from the
crystal is,
  e  o 
2L

ne  no   2L 1 ne3r33  no3r13 V
 2
d
where L is the crystal length along the y-axis.
Explain the first and second terms. How would you use two such Pockels cells to
cancel the first terms in the total phase shift for the two cells?
If the light beam entering the crystal is linearly polarized in the z-direction, show
that
 
2ne L

2L ne3r33  V


2 d
Consider a nearly monochromatic light beam of the free-space wavelength λ = 500
nm and polarization along z-axis. Calculate the voltage Vπ needed to change the
output phase ∆ϕ by π given a LiNbO₃ crystal with d/L = 0.01 (see Table 7.2).
Solution.
Consider the phase change between the two electric field components
ELEC425-summer, 2012 15
Assignment 5
  e  o 
2L

ne  no   2L 1 ne3r33  no3r13 V
 2
d
The first term is the natural birefringence of the crystal and occurs all the time,
even without applied electric field. The second term is the Pockels effect, applied
field inducing a charge in the refractive indices.
L
L
x
z
Ea
d
z
Ea
Light
d
light
∆ϕ
y
y
x
Two transverse Pockels cell phase modulators together cancel the natural
birefringence in each crystal.
If the light is linearly polarized with its field along z, we only need to consider the
extraordinary ray (ϕ₀ = 0).
  e 
2ne L


2L no3r13 V
 2 d
The first term does not depend on the voltage. The voltage V that changes the
output phase by π is
2L no3r13 V

 2 d
Or V 
d 
500 109

0
.
01
 15.5 V
L no3r13
2.1873 30.8 1012


7.21. Optical Kerr effect
Consider a material in which the polarization does not have the second order term:
P   0 1E   0 3 E 3 or P /  0 E   1  3 E 2
ELEC425-summer, 2012 16
Assignment 5
The first term with the electric susceptibility χ₁ corresponds to the relative
permittivity εr and hence to the refractive index no of the medium in the absence of
the third order term, i.e. under low fields. The E² term represents the irradiance I of
the beam. Thus, the refractive index depends on the intensity of the light beam, a
phenomenon called the optical Kerr effect:
n  no  n2 I and n2 
3 3
4no2
And η=(μ₀/ε₀)1/2 = 120π = 377 Ω, is the impedance of the free space.
a) Typically, for many glasses, χ₃≈ 10⁻²¹ m²/W; for many doped glasses, χ₃ ≈
10⁻¹⁸ m²/W; for many organic substances, χ₃ ≈ 10⁻¹⁷ m²/W; for
semiconductors, χ₃ ≈ 10⁻¹⁴ m²/W. Calculate n₂ and the intensity of light
needed to change n by 10⁻³ for each case.
b) The phase ϕ at a point z is given by
  0t 
2n

z  0t 
2 n0  n2 I 

z
It is clear that the phase depends on the light intensity I and the change in the
phase along ∆z due to light intensity alone is
  0t 
2n2 I

z
As the light intensity modulates the phase, this is called self-phase
modulation. Obviously light is controlling light.
When the light intensity is small n₂I ≪ n₀, obviously the instantaneous
frequency
   / t  0
Suppose we have an intense beam and the intensity I is time dependent
I=I(t). Consider a pulse of light traveling along the z-direction and the light
intensity vs. t shape is a “Gaussian” (this is approximately so when a light
pulse propagate in an optical fiber, for example). Find the instantaneous
frequency ω. Is this still ω₀? How does the frequency change with “time”, or
across the light pulse? The change in the frequency over the pulse is called
chirping. Self-phase modulation therefore changes the frequency spectrum
of the light pulse during propagation. What is the significance of this result?
ELEC425-summer, 2012 17
Assignment 5
c) Consider a Gaussian beam in which intensity across the beam cross section
falls with radial distance in a Gaussian fashion. Suppose that the beam is
made to pass through a plate of nonlinear medium. Explain how the beam
can become self-focused? Can you envisage a situation where diffraction
effects trying to impose divergence are just balanced by self-focusing
effects?
Solution.
a)
The fractional change in the refractive index is
n 
n  n0 n2 I 3 3 I


n0
n0
4n03
Thus I 
 n 4n03
3 3
If  n  103 or 0.1%
For glasses: n₀ ≈ 1.5, χ₃ = 10⁻²¹ m²/W, I = 1.2·10¹⁶ W/m²
for doped glasses: n₀ ≈ 1.5, χ₃ = 10⁻¹⁸ m²/W, I = 1.2·10¹³ W/m²
for organics: n₀ ≈ 1.5, χ₃ = 10⁻¹⁷ m²/W, I = 1.2·10¹² W/m²
for semiconductors: n₀ ≈ 2.5, χ₃ = 10⁻¹⁴ m²/W, I = 5.5·10⁹ W/m²
large intensities.
b)
the phase ϕ at a point z is given by
  0t 
2n

z  0t 
2 n0  n2 I 

z
So that the instantaneous frequency is
 t  

2n2 z I
 0 
t
 t
ELEC425-summer, 2012 18
Assignment 5
When the light intensity is small n₂I≪n₀, ω = ∂ϕ/∂t=ω₀.
We have an intense beam with time dependent intensity I(t). Consider a pulse of
light traveling along the z-direction and the light intensity vs. t shape is a
“Gaussian”. Then the instantaneous frequency is,
 t  

2n2 z I
 0 
t
 t
So that the frequency changes with “time”, or across the light pulse. Since ∂I/∂t is
rising at the leading edge and falling in the trailing edge, the two ends of the pulse
contain different frequencies. The change in the frequency over the pulse is called
chirping. Self-phase modulation therefore changes the frequency spectrum of the
light pulse during propagation. The second term above results in the broadening of
the frequency spectrum and hence leads to more dispersion for a Gaussian pulse
propagating in an optical fiber.
c)
If an intense Gaussian optical beam is transmitted through a plate of nonlinear
medium, it will change the refractive index in the medium with the maximum
change of the refractive index in the center of the beam. Therefore, the plate will
act as a graded-index medium and change the curvature of the wavefront. Under
certain conditions the plate can act as a lens with a power-dependent focal length,
producing a co-called self-focusing of the beam. Similarly, if an intense Gaussian
beam propagates through a nonlinear medium, the medium can act as a gradedindex waveguide. In this case, under certain conditions, the self-focusing effect can
compensate the divergence of the beam due to diffraction, and the beam will be
confined to its self-created waveguide. Such self-guided beams are called spatial
solutions.
ELEC425-summer, 2012 19
Assignment 5
Figure 1. propagation of the light through nonlinear medium.
Ref. http://www.absoluteastronomy.com/topics/Gradient_index_optics
The intensity variation across the beam cross section leads to a similar refractive
index variation in the nonlinear medium. Thus, the medium resembles a graded
index guide or a GRIN rod and can focus the beam.