Chapter 11. Photoelasticity. Introduction
Sample problems
11-S1
Assuming coordinate axis x-y that are respectively at 45o from a vertical beam of linearly
polarized light incident on a half wave plate with the fast axis making an angle of 450 with the
vertical axis. Get the expression of the emerging wavefronts and the type of polarization.
Solution to 11-S1
Figure P11.1 shows the scheme of the plane polarized light incident on a quarter wave plate. Due
to the birefringence of the plate two wave fronts will be exiting the plate. We know that the
emerging vectors representing the wave fronts are perpendicular to the propagating ray. We also
know as shown in Figure 11.6 that one wave front is faster than the other, in this case the
extraordinary beam because a negative crystal is assumed. One must remember that according to
the definition of index of refraction, the lower index of refraction corresponds to the fastest
wave front in this case the extraordinary beam that is polarized in the direction of the x-axis,
called the fast axis of the plate. At the same time the extraordinary beam is plane polarized in the
direction parallel to the optical axis (look at 11.6 and at P11.1). The ordinary beam that obeys the
Snell’s law is polarized in the plane perpendicular to the fast plane. As it can be seen in Figure
P11.1 the incident plane of the plane polarized beam is at +45o with respect to the x-axis of the
coordinate system.
Figure P11.1. Plane polarized light incident on a quarter–wave plate retarder.
According to sample problem 6-S21 the light exiting the quarter wave plate has the equation
1
J CR  E 0e   . We can see that in sample problem 6-S21 the assumption is that that the beam
i 
along the x-axis is ahead by π/2 with respect to the component in the y axis, as it is depicted in
Figure P11.1. The light emerging from the quarter wave plate is right circularly polarized light.
The value of E0=Ex=Ey=E/ 2 .Returning to the Cartesian notation,


E 

sin(
kz


t
)
i

cos(
kz


t
)
j is the representation of the wavefronts emerging

2

from the wave plate. It is possible to verify that the above results depend on the adopted initial
conditions and the sign adopted for the relative phase. In the case of sample problem 6-S21 we
started with the condition at time zero is that the amplitude is zero, hence the x-component of the
vector is given by a sine function and the y-direction by a cosine function. If the vector rotates
clockwise the x-component is ahead of the y- component, the x-increases and the y-component
decreases, the opposite occurs for the counterclockwise direction.
We can now verify that the instrument matrix can provide the answer to the problem. The
 i 4

0  given by equation (11.25). The beam is given
instrument matrix is (table 11.3) T1  e

i 

4
4
0
e



E( z, t ) 

1
by E( z, t )  E e i  
1
2
 
The resulting beam is
 i 4
E( z, t )  e

0


 4 
1


E
E
E  0
0 
e  
ei   
ei  e   =

i 
1
 4 
2
2
2
4
i 

e 
e 
Since what is important is the relative phase  
   
   
4  4 2
11-S2
A beam of right circular polarization is incident on a quarter wave plate with the fast axis in the
vertical direction. Determine the state of polarization of the emerging beam.
Solution to 11-S2
The Jones vector of right circularly polarized light is J CR 
E  1
e   . The Jones vector of a π/4
2 i 
 i 4
retarder with the fast axis in the vertical direction is (Table 11.3) TV1  e

4
 0
 i 4
E( z, t )  e

 0


0  , hence

i 
e 4
  

0  E ei 1  E ei  e 4 

i 
i 
 4 
2
e 4 2  
 ie 

But the symbol i stands for e 2 hence we have
φ/4.


3
e 2  e 4  e 4 , then taking the common factor -


E i   4  1
e
i 
2

Returning to Cartesian coordinates,

E 
 
 
E( z, t ) 
sin
(
kz


t

)
i

cos(
kz


t

)j
4
4 
2 

E( z, t ) 
The wave front continues as right circularly light but experiences a phase change of π/4.
11-S3
Between two parallel linear polarizers with parallel axis, in the x-direction and y-direction
respectively a quarter wave plate is located with the fast axis making a +450 with the direction of
the x-axis. A non polarized beam is incident normal to the plane of the vertical polarizer.
Describe the beam emerging from this set up.
Solution to 11-S3
Upon impinging the vertical polarizer the ordinary light beam is transformed into a vertically
0 
polarized bean,   .The beam encounters first the quarter wave plate with the axis at +45(Table
1 
1 1 i 
11.3) the instrument matrix is T45 1 

 and the encounters a vertical polarizer with the
2 i 1
4
0 0
instrument matrix Table 11.2 
 . Taking into consideration the non commutative property
0 1
of matrices the whole operation can be represented by,

0 0
E( z, t )  

0 1
1 i  0 i 
i 1 1  1 ,

   
1 1 i  i 0

 Ee 1 , performing the corresponding operation in order
2 i 1
 
0 0 0
0 1  1 then,

  
Ee i 0 
E( z, t ) 
 
2 1 
The operation produces vertically polarized light but the amplitude of the field has been reduced
by 2

11-S4
Between the set up of the polarizers with orthogonal polarizing axis x,y horizontal and vertical
respectively there is two quarter wave plates. The first quarter wave plate has its fast axis at
45ofrom the x-axis and the second wave plate at -45 . Determine the output of the above
described set up.
Solution to 11-S4
The first polarizer is vertical and ordinary polarized light impinges normal to this polarizer,
Following the procedure of the preceding problem and utilizing Table 11.13,

1 0 1  1  i 1 1 i  i 0
E( z, t )  




 Ee 1
0 0 2  i 1  2 i 1
 
From the preceding problem we know that the first operation produces
1 i  0 i   1  i i 
0 1 0  0 0
i 1 1  1 ,  i 1  1  21 , 0 0  1  0

   
 
  
   
The final result is that no light goes through the system. This is a similar case to the first
arrangement of the circular polariscope in Figure 11.11. In this problem the orientation of the
quarter wave plates is different but the fast axes are also orthogonal as shown in Figure 11.11.
11-S5
In the arrangement of sample problem 11-S1 the second plate axis is set parallel to the axis of the
first polarizers. Determine the output of the system.
Solution to 11-S5
0
From problem 11.8 the output of the second quarter wave plate is   . The instrument matrix of
1
0 0
the second polarizer is 
 , the output is then,
0 1
0 0 0 0
0 1 1  1 .

   
The final output is then,

0
E( z, t )  Ee i   . The plane polarized exits the system.
1
Problems to solve
11.1
Quartz and calcite are uniaxial crystals. For quartz ne=1.5534 and no=1.5433. In calcite
ne=1.4864 and no=1.6584. Represent the wave fronts that will propagate in both media if there is
a point source inside them. Utilize the intersections of the corresponding ellipsoids of revolution
with a principal plane of the ellipsoid.
11.2
In a Wollaston type cube made out of quartz n0=1.544, ne=1.553 a beam of non-polarized light
impinges normal to the phase as shown in figure P11.2. Draw the continuation on the beam
through the rest of the cube and out on the air. Explain the procedure utilized.
Figure P11.2.Non coherent light incident on a Wollaston cube
11.3
A calcite crystal is grounded into a parallel plate with the optical axis parallel to the face of the
crystal. For beams incident normal to the face of the crystal compute the thicknesses of the
crystal such that the relative retardation between the extraordinary and the ordinary beams is
equal to π/2 for light with λ=0. 590. The indices of refraction of the crystal are no=1.568, ne=1.486.
Figure P11.3. Calcite crystal
11.4
Derive the instrument of a 180o retarder. Analyze the effect of the retarder on a polarized beam
that makes 45o with the fast axis of the retarder.
11.5
A plane polarized beam in the y direction of a coordinate system with x-axis horizontal and the
y-axis vertical, goes through a π retarder whose fast axis forms a 600 with the x-axis. Describe
the emerging wave fronts.
11.6
A half wave plate is inserted between two crossed polarizers, the fast axis of the plate is at 45o of
the polarizer. Describe the outgoing beam
11.7
A retarder with a total retardation δ is inserted between two orthogonal plane polarizers. The fast
axis makes an angle θ with the polarizer. Give the equation of the fringes that will be observed.
11.8
Derive the expression of the output that will be observed with the arrangement of Figure 11.11
(a).
11.9
Derive the expression of the output that will be observed with the arrangement of Figure 11.11
(b).
11.10
Derive the expression of the output that will be observed with the arrangement of Figure 11.11
(c).
11.11
Show that by superimposing two circularly polarized beams one right and the other left, elliptical
polarization is produced.
11.12
Show that upon entering a retarder circularly polarized light is divided into two equal orthogonal
components independent of the orientation of the fast and slow axis.