Homework 6

Photonics
Sheet 6
1. Assume the following set up for a Pockels Cell:
Use Jones Matrices to find the transmission formula T=sin 2 (φ/2), with
φ being the phase angle introduced by the retarder. Hint: T is proportional
to E 2 , and use the following identity:
e
− e
and remember that e 0 =1.
jx
jy
= 2 je
⎛ x−
y ⎞
j⎜
⎟
⎝ 2 ⎠
⎛
⎜ e
⎜
⎜
⎝
⎛ x−
y
j⎜
⎝ 2
⎞
⎟
⎠
− e
2 j
⎛ x−
y
− j⎜
⎝ 2
⎞
⎟
⎠
⎞
⎟
⎟
⎟
⎠
For the following questions assume that the initial phase angle is zero, so that
the transmission becomes T=sin 2 (π/2·V/Vπ), Vπ=λ0/(2r63no 3 )
2. Assume a Pockels Cell with an ordinary KDP crystal with linear
electrooptic coefficient r63 of 10.6 pm / V and a refractive index for the
ordinary ray of 1.51 .
a) What voltage is needed to turn the angle of incident linearly polarized
light with the wave length 632.8 nm by 90 degrees (this voltage is normally
called Vπ)
b) How many percent of incident light is transmitted if half of this voltage is
used
c) By how many percent would the transmission change if the incident light
has the wave length 480 nm
3. Assume a Pockels Cell with an ordinary KDP crystal with linear
electrooptic coefficient r63 of 10.6 pm / V and a refractive index for the
ordinary ray of 1.51. For a wave length of 1.06 µm Vπ is 14.5 kV. Now
assume a signal that has V at the beginning. Then the voltage is increased
in two steps, each time by 3V. If this signal is transformed into a light

signal by the desribed Pockels Cell, what would be the signal change in %
(in relation to the 3V-Intensity) at each step
4. Assume a Pockels Cell with an ordinary KDP crystal with linear
electrooptic coefficient r63 of 10.6 pm / V and a refractive index for the
ordinary ray of 1.51 . For a wave length of 1.06 µm Vπ is 14.5 kV.
Calculate the voltage where the intensity curve shows linear behaviour. In
what voltage range do you find linear behaviour if you assume that a
change in the slope of 10% is still linear
5. Assume a Pockels Cell with an ordinary KDP crystal with linear electrooptic
coefficient r63 of 10.6 pm / V, a refractive index for the ordinary ray of 1.51, and
for a wave length of 1.06 µm Vπ is 14.5 kV.
a) Calculate the voltage for the “point” where the “intensity curve shows
linear behaviour” for the described Pockels Cell (see question 4).
b) Calculate the linear function that describes a tangent line that passes
through the point calculated in question a) and which has the same slope as
the intensity curve in that point.
c) Now it should be found out in what voltage range the described Pockels
Cell can be used as a linear modulator. Because of the symmetry of the
problem it is sufficient to calculate only the lower voltage value of this
range. If we assume that the intensity curve can be considered linear if the
difference between the tangent of question b) and the original intensity curve
is below 0.1, what is then the lower limit of the voltage range
Hint: Because the function f(V) you will obtain in order to answer the question is
not solvable by standard means, use the Newton-Method to find the solution
(i.e. to solve the function f(V) = 0 for V):
V
n+
1
= V
n
−
f ( Vn
)
f ′(
V )
n
Use V = 900 Volts as a start value, and stop when f(Vn) gets below 0.001 (or
after 3 iterations).
The intensity curve’s function values are (in the here interesting range)
greater than the tangent’s function values; the derivative of the function f(V)
you will have to solve has in the here interesting range only negative function
values; if the function values of the derivative of f(V) get below a value of -0.0001,
you can assume that the function values of the derivative do not change
anymore and take -0.0001 as the value for the function value of the derivative,
it is then not necessary anymore to calculate the function values of the
derivative.