Solution 10
1. An electro-optical tunable retarder (Kerr-cell) is able to produce from linear polarized light ( =
5145 Å) circular polarized light by a retardation of ∆ = /4 when being switched from zero to E
= 1200 volts/cm. What switching voltage must be applied in order to use the Kerr-cell as a
shutter to switch on and off the linearly polarized light from an argon laser with  = 5145 Å?
Make sketch of electro-optical arrangement used and the relevant light polarization.
To make a shutter for a linearly polarized light, we need a tunable /2-retarder and a
linear polarizer. The polarization of the linearly polarized light can be changed 90°
by the tunable /2-retarder, and the light is blocked by or transmitted through the
linear polarizer.
For a Kerr cell, the retardation  = d·n = d· o·K·E2. When  = /4, we need E =
1200 volts/cm. Therefore, in order to get  = /2,
we need E = √2 · 1200 volts/cm. = 1697 volts/cm
E= 1697 V/cm
No
Light
Laser
E=0
2. A transparent isotropic material of thickness d = 15 mm changes under uniaxial compression of
100 atm its refractive index n in the following way:
(a) ∆n || (100 atm) = + 8.5  10-5 for light polarized parallel to the compression direction.
(b) ∆n (100 atm) = – 3.0  10-5 for light polarized perpendicular to the compression direction.
How much uniaxial pressure do you have to apply if you want to obtain circular polarized
light from linearly polarized light of  = 6500 Å? Make a sketch of arrangement, and show
what you would do (with the same elements) in order to change between right and left circular
polarized light?
For the stress-birefringent material,
  

n = (∆n || – ∆n) =  S.
Since n = 1.15  10-4 for S = 100 atm, the constant  is  = 1.15  10-6 atm-1.
In order to obtain circular polarized light from linearly polarized light,
we need a /4-retarder for  = 6500 Å.
The retardation is given by  = d · n = d ·  · S , so that
6.5  10-7 (m)


S =
=
=
d·
4·d·
4 · 1.5  10-2 (m) · 1.15  10-6 (atm-1)
= 9.42 atm
With the same arrangement, if you want to change from right to left circular polarized
light, we need to apply more pressure to make  = 3/4, that is, to increase the
pressure S = 28.3 atm.
S= 9.42 atm
Right
Circular
s
f
Laser
Left Circular
under
S = 28.3 atm
3. Two narrow parallel slits illuminated by light from a He-Ne laser ( = 6328Å) produce
interference fringes with a separation of 0.5 mm on a screen 3 meter away. What is the distance
of the two slits? What would change (and by how much?) if the same experiment would be
performed under water (n = 1.33)? What would change in the fringe-pattern if you use instead
of the He-Ne laser an Ar laser operating at  = 4880Å in air and under water ?
Let the distance of the two slits = a. Then, the angular positions of fringes at the
m·
m·
intensity maxima are m = sin-1( a
) ≈ a for the small angles.
The spatial positions of the fringes on the screen 3 meter away are
m·
ym = X· tan m ≈ X· a
for the small angles.
X·
Therefore, the fringe separation is y =
a .
 a=
X·
3 · 6.328  10-7
=
y
5  10-4
(m) = 3.8  10-3 m = 3.8 mm
Under water only the wavelength of light will change to ' = /n = 4758Å. Thus,
y' =
X·'
a
=
y
n = 0.376 mm (fringes are getting closer).
With Ar laser also the wavelength of light changes to " = 4880Å, so that
y" =
Under water,
y" 
X·"
a
y"
n
=
y
n = 0.386 mm
= 0.290 mm
4. Imagine that in a double-slit (separation = a) interference experiment (in which you can neglect
the slit-width) one of the slits becomes covered by a very thin slab of transparent material of
refractive index n and thickness D. Discuss qualitatively (possibly with a sketch) the difference
between interference patterns before and after inserting the slab. Can you derive the expression
for the change in the angular position of the zeroth order interference maximum, as a function
of a, D, and n, caused by inserting the transparent slab?
D
r1
r2
a

Qualitatively,
the optical pathlength r1 will increase by inserting the slab, so that the interference
pattern on the screen will shift to the up-ward direction (i.e., the direction of the
covered slit).
Quantitatively,
before inserting the slab, the optical pathlength difference  between two paths r1 and
r2 is  = r2 – r1 ≈ a· sin. Then, the zeroth-order maximum ( = 0) occurs at  =
0.
After inserting the slab, the optical pathlength r2 is still the same but the optical
pathlength r1 increases by (n–1)·D due to the slower light speed inside the
material. Then, the optical pathlength difference changes to
 = r2 – {r1 + (n–1)·D} = r2 – r1 – (n–1)·D ≈ a· sin – (n–1)·D
For the new zeroth-order maximum,  = 0. Then, sin =
(n–1)·D
a
5. Monochromatic light ( = 600 nm) from a particular source has a frequency bandwidth of  =
5 1011 Hz.
(A) What is the coherence length x and the coherence time t of this light?
The coherence time
The coherence length
1
= 2  10-12 sec

x = c· t = 6  10-4 m = 0.6 mm
t =
(B) What is approximately the largest optical pathlength difference in an interference experiment
with this light, which would still give you observable interference fringes? (Make a schematic
sketch of such an experiment.)
We can observe an interference of two light beams separated from the same light
source only when their optical pathlength difference is smaller than the coherence
length of the light source. Therefore, the largest optical pathlength difference to
observe interference fringes must be the same as the coherence length of the light
source, that is, 0.6 mm.
For example,
Mirror
Mirror
Screen
distance L
Light
Source
Beam
Splitter
Beam
Splitter
you can observe the interference fringes on the screen only when 2L ≤ 0.6 mm.
(C) For a double slit of a = 1.2 mm slit distance, what will be the angular separation of
neighboring interference fringes, and what will be their spatial separation on a screen 2
meters away from the double slit? What will be the approximate angle at which the highest
order interference fringe can be observed?
The angular separation of neighboring interference fringes is
6  10-7 m

 ≈ a =
= 5  10-4 rad = 0.0286 Degree
3
1.2  10 m
The spatial separation on the screen is y ≈ X·  = 1.0  10-3 m = 1 mm
In order to find the highest order of the observable interference fringes,
max = x ≥ mmax  = a· sinmax .
x
 mmax = largest integer but smaller than
≈ 1000

1000 

max ≈ sin-1(
) = 30°
a
No fringes
Highest Order
= x
(D) If you chop the above light into pulses of pulse-duration t = 5  10-13 sec, what changes under
(A), (B), and (C) ? Find the new values if changed.
Since the pulse duration is shorter than the coherence time,
the coherence time changes to the pulse duration, that is, t = 5 10-13 sec.
Then, the coherence length x = 1.5  10-4 m = 0.15 mm = 150 m, and the
bandwidth  = 2 1012 Hz (or, for wavelength,  = c·  / 2 = 2.4 nm) is still
very small compared to the  = 5  1014 Hz.
The largest optical pathlength difference for observable interference fringes is
0.15mm.
Since the wavelength of light does not change, the angular separation (or spatial
separation) of neighboring interference fringes does not change either.
However, the highest order interference fringe will occur at  = 7.18° with the highest
order of mmax ≈ 250 instead of  = 30° with mmax = 1000.
6. What is the diameter of the light spot on the moon which is produced by “parallel” (plane
wave) laser beam (D=1mm) that is pointed to moon.
The 1mm diameter can be considered as the size of an aperture. This leads to the Airy
disc formula with R the distance between Earth and the moon (384,400 km).
q1[m]  1.22

R
3.844 10 8
 1.22
  4.7 1011 [m]  4.7 10 5 [m]
d
103
7.
Discuss how to built an electro-optical modulator using LiNbO3. Try to be as quantitative
as possible. Give directions, magnitudes of voltages,….Discuss different possibilities and
compare them to each other.
This is a question that may have more than one solution.
Let’s first right down the changes to the index ellipsoids
  1  

 n 2 1 
  1  

 0 r22 r13 
r22 E y  r13 E z 
 n 2 2 
r22 E y  r13 E z 
r22 r13 
  1    0
E x 
 2
 0


0 r33 
r33 E z
 n 3 
E 
 
y


  1  
0
r51 0  
r51E y
 2

E

 z 

 n 4   r51
0
0
r51E x




 1   r
0
0


r
E

22
22 x
2
 n 5 
 1  
2
 n 6 
This leads to the following ellipsoid
x 2  y 2  z 2  2
 2   2   2  x r22 E y  r13 E z  y 2 r22 E y  r13 E z  z2 r33 E z  2r51 E y yz  2r51E x xz  2r22 E x xy  1
no  no  ne 
We want to use one of the large coefficient:
(1) r33
we need to apply the field in z-direction
x 2  y 2  z 2  2
 2   2   2  x r13 E z   y 2 r13 E z   z2 r33 E z  1
no  no  ne 
 1


  r E 
 2  1  r13 E z 
y 2   1  r33 E z 
z2  1
13 z x 
2
2
no

no

n 2e

to use r33 we have to have light polarized in z-direction, so let’s assume we propagate in y-direction:
1
 1

 2  r13 E z 
x 2   2  r33 E z z2  1
no

ne

3
no
r31E z
2
n3
nz  ne  e r33 E z
2
nx  n o 
we have a phase difference:


c

n 3o
n 3  
r31  e r33 E z y
 2
2  
nx  nz y  no  n e  

3
 3
E z y  
Problem: intrinsic birefringence. We could have  no  ne y   so we can have no r31  ne r33 
c
in this case V 
n r
3
o 31
d
 n 3e r33 y
Similar for r51, we apply the field along x direction
x 2  y 2  z 2 
 2   2   2  2r51 E x xz  2r22 E x xy  1
no  no  ne 
if we can neglect r22 we rotate in the xz plane:
1
y 2   1

 2  r51E x 
x ' 2  2   2  r51 E x z 2  1
no
 no  ne

Let’s propagate in y-direction
1
 1

 2  r51E x 
x ' 2  2  r51 E x z 2  1
no
 ne


2
2
