Physics Assignment: Laser Polarization and Antireflection Analysis

This physics assignment addresses two key tasks: laser polarization and antireflection. Task one involves analyzing laser polarization using Fresnel equations and the Brewster angle, calculating transmission coefficients for parallel and perpendicular polarizations, and determining the number of round trips required for a wave to become 99% polarized in a laser system with Brewster-angle windows. Task two focuses on antireflection coatings, analyzing the transmission of monochromatic light through a dielectric medium with a thin layer of a specific refractive index, and demonstrating that the transmission coefficient is 100% under these conditions, relating the findings to the use of MgF2 coatings on lenses. The solution provides detailed calculations and explanations for each task.

INSTITUTIONAL AFFILIATION
FACULTY OR DEPARTMENT
COURSE ID & COURSE NAME
TITLE: PHYSICS TASKS
STUDENT NAME
STUDENT ID NUMBER
PROFESSOR (TUTOR)
DATE OF SUBMISSION
JUNE 8, 2018

TASK ONE: LASER POLARIZATION
(a) For plane waves polarized perpendicular to the plane of incidence, the Fresnel equations
are ~
E¿
~Eoi
= 1−αβ
1+αβ
~
E¿
~
Eoi
= 2
1+ αβ
Suppose that a beam of eight strikes an interface at the Brewster-angle. Find the ratio of
transmission coefficients for polarizations parallel and perpendicular to the plane of
incidence Tii
Ti , assuming μ1=μ2 and show that it is always greater than 1 for η1 ≠ η2
SOLUTION
Using the sell Meier equation,
n ( λ ) =
[ 1+∑
i
Bi λ2
λ2−Ci ]
2
Let the incident wave be given as,
~
Eoi= ^x E0
i e− j k1 z
k1 =ω √ ϵ1 μ1
η1= √ μ1
ϵ1
The transmitted laser signal is given as,
~
Eot = ^x E0
t e− j k2 z
k1 =ω √ ϵ2 μ2
η1= √ μ2
ϵ2
Defining the reflection coefficient,
r = Eo
r
Eo
i
1

Defining the transmission coefficient,
t= Eo
t
Eo
i
The reflection coefficient is given as,
r = Eo
r
Eo
i =η1−η2
η1+ η2
The transmission coefficient is given as,
t= Eo
t
Eo
i = 2 η1
η1 +η2
Let η1=1 , η2=0
t= 2 η1
η1 +η2
=2
1 =1
The transmitted energy is given as,
R=|r|2 … fraction of incident power reflected
T =1−R , R=reflected energy
(b) In a laser, light is reflected back and forth through an active medium with optical gain,
located between two mirrors. The active medium is commonly enclosed in a cell with
Brewster-angle windows at each end. Suppose the windows are glass with η=1.5,
separated by pass with η=1. Neglect the multiple reflections within and between
windows assuming the initial light beam contains random polarizations, how many round
trips must the light make to produce a wave that is 99 percent polarized?
SOLUTION
window glass , η=1.5
The electric field at any given point is always perpendicular to the direction of travel of
the waves but changes directions randomly for the randomly polarized source. The
Brewster-angle is given as,
θB +θR=900
2

η1 sin ( θB )=η2 sin ( θR )
η1 sin ( θB )=η2 sin ( 90−θB ) =η2 cos ( θB )
η1
η2
= sinθB
cos θB
=tan θB
θB=tan−1
( η2
η1 )=tan−1 1.5
1 =56.310
θB=56.310 … Brewster Angle
For total internal reflection of the laser light, it is important to find out what is the
maximum acceptable angle of the cone of light ray as fed into the optical core,
θc=sin−1
( η2
η1 )
sin−1
( 1
1.5 )=¿ 41.80 ¿
¿ 1 full round trip
TASK TWO: ANTIREFLECTING
Analyze the case of monochromatic light of frequency w passing from a dielectric
medium with index of reflection n, to a medium with n3. The two media are separated by
a thin layer of medium2, which has been specially designed to have an index of refraction
η2= √ η1 η3. The thickness of this layer is adjusted to the λ /4 where λ is the wavelength of
the light in medium 2. Show that the transmission coefficient at the normal incidence is
100 percent under these conditions.
Notes: lenses are commonly coated with MgF2 which has an average index of refraction
of 1.38, which is close to the geometric mean of air and dense flint glass (η ≈ 1.9)
Solution
For such a layered media,
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Ei e− j k1 z …incident
Er e+ j k1 z … reflected
Ef e− j k2 z … forward
Eb e+ j k2 ( z− L ) …backward
Et e− j k3 ( z−L ) …transmitted
Omitting e− jωt
H±=± E±
η
k ≡ ω √ϵμ
η= √ ( μ
ϵ )
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Anti-reflection coating has a huge impact on the reflectance,
The transmitted wave is given as,
Et = Ei t21 t12 e−2 j k2 L
1−r21 r21 e−2 j k2 L
The Fabry-Perot resonance is given as,
t= t21 t12 e−2 j k2 L
1−r21 r21 e−2 j k2 L
At maximum transmission, the Fabry-Perot resonance is given as,
e−2 j k2 L=1
Hence transmission is 100 % at this state.
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