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Fundamental of Optical Engineering
Lecture 8

A linearly polarized plane wave with Ē vector
described by
Einc  (

) eˆx  eˆy

is incident on an optical element under test.
Describe the state of polarization of the
output wave (linear, elliptrical, or circular) if
the optical element is:

(a) A linear polarizer oriented to transmit
light polarized in the ex direction.

(b) A half-wave plate with birefringence axes
oriented to coincide with ex and ey.

(c) A half-wave plate with birefringence axes
oriented at 45º relative to ex and ey .

(d) A quarter-wave plate with birefringence
axes oriented to coincide with ex and ey .

(e) A quarter-wave plate with birefringence
axes oriented at 45º relative to ex and ey .

(f) A half-wave plate with birefringence axes
oriented at 25º relative to ex and ey .

(g) A quarter-wave plate with birefringence
axes oriented at 25º relative to ex and ey .

A linearly polarized light propagating in the
z-direction with polarization vector in the xdirection is incident on a birefringent crystal.
What is the state of polarization of the light
after passing through the crystal if:
(a) the crystal is a quarter-wave plate with optic axis
in the xy plane oriented at 30º relative to the yaxis?
◦ (b) the crystal is a half-wave plate with optic axis in
the y-direction?
◦ (c) the crystal is a half-wave plate with optic axis in
the xy plane oriented at 11º relative to the x-axis?
◦ (d) the crystal is a quarter-wave plate with optic
axis in the xy plane oriented at 45º relative to the
y-axis?
◦ (e) the crystal is a quarter-wave plate with optic
axis in the z-direction?

For a birefringent median with n0 = 1.654
and nE = 1.485 as shown in the figure. Find
the length L that makes it be (a) a full wave
plate (b) a half wave plate (c) a ¼ -wave plate
if the wavelength is 656 nm.


This makes use of electrooptic effect (applied
electric fields used to change the optical
properties).
There are 2 kinds of electrooptic effect: linear
and quadratic.



The linear electrooptic effect is called
“Pockels effect”.
This refers to the change in the indices of the
ordinary and extranordinary rays proportional
to applied electric field.
This effect exists only in crystals without an
inversion symmetry such as LiNbO3.


For a crystal with an inversion symmetry, the
linear electrooptic effect can not exist, while
the quadratic electrooptic effect known as
“Kerr effect” is observed.
This is where the induced index change is
proportional to the square of applied electric
field.


V (pi-voltage) or half-wave voltage is the
applied voltage that makes the relative phase
shift be  in a cube of material.
In general,  
2

 n
x

 n y L
nx and ny = refractive index changes
produced by applied voltage.
V L
 

V h


It is preferable to design L >> h to have a low
applied voltage V.
After applying a voltage, indices are changed
as
nE  nE  nE (V )
nO  nO  nO (V )

V for the material is 2,700 V, L = 2 cm, and h = 0.5 mm.
Find applied voltage V to have Δ =  (complete extinction).

An electrooptic crystal has dimensions of 2x2x3 along the
x,y, and z axes with nE = 1.487 and nO = 1.536. An input
wave propagating in the z-direction at λ = 0.63 μm is linearly
polarized at a 45º angle relative to the x- and y- axes. A
voltage V applied across the crystal in the x-direction. The
voltage is increased from V = 0 until, when V = 245 V, the
output polarization from the crystal is the same as that
observed for V = 0. Assume that optic axis is along the yaxis.
◦ (a) What is the total phase retardation, in rad, for V = 0?
◦ (b) What is pi-voltage for the material?
◦ (c) What is the refractive index change Δnx produced by the
applied voltage of 245 V, assuming Δny =0?

Recall: interference eq.
  A1ei  A2 ei
1
2
PA  A12  A22  2 A1 A2 cos 1  2 

Assume that
n n
n1  n2
 1 and 2 3  1
n1  n2
n2  n3
 n2  n3  i
 n1  n2 
Er  Ei 
e
  Ei 
 n1  n2 
 n2  n3 
4 n2 t2

 round trip phase shift


Er  Ei A12  A23 ei
ni  n j

amplitude of reflected wave
Aij 

ni  n j
amplitude of incident wave

Reflectance
Er
R
Ei
2
 A12  A23 e
i
2
R  A122  A232  2 A12 A23 cos 

We can consider R into 3 cases:
◦ n1 = n3.
◦ n1 < n2, n2>n3 , n1 ≠ n3.
◦ n1 < n2 < n3.

Case 1: n1 = n3
R  A122  A212  2 A12 A21 cos 

From a definition: Aij = Aji
R  2 A122 1  cos  

Max in R for    2 N  1  ;
N  0,1, 2,...
Rmax  4 A122

Min in R for   2 N ;
Rmin  0
N  0,1, 2,...

n1 = 1.5, n2 = 1.6, and λ = 0.63 μm. Find t2 for Rmax
and Rmin.

Case 2: n1 < n2, n2>n3 , n1 ≠ n3.
A12 A23  0

Max in R for    2 N  1  ;
N  0,1, 2,...
Rmax   A12  A23 

Min in R for   2 N  ;
2
N  0,1, 2,...
Rmin   A12  A23 
2

n1 = 1.5, n2 = 1.6, n3 = 1.4, λ = 0.63 μm. Find t2 for
Rmax and Rmin.

Case 3: n1 < n2 < n3
A12  0
A23  0

Max in R for   2 N  ;
N  0,1, 2,...
Rmax   A12  A23 

Min in R for    2 N  1  ;
Rmin   A12  A23 
2
N  0,1, 2,...
2

n1 = 1.5, n2 = 1.6, n3 = 1.7, λ = 0.63 μm. Find t2 for
Rmax and Rmin.