Recall that we can add waves so lets take a plane wave traveling in the z
direction and break it into components.
E
⃗
x
=
E
0
x
cos
(
k
z
−
ω
t
)
ı
̂
E
⃗
x
=
E
0
x
cos
(
k
z
−
ω
t
)
ı
̂
E
⃗
y
=
E
0
y
cos
(
k
z
−
ω
t
+
δ
)
̂
E
⃗
y
=
E
0
y
cos
(
k
z
−
ω
t
+
δ
)
̂
Where
δ
δ
is some arbitrary phase between the two components. The electric field for the
wave is
E
⃗
=
E
⃗
x
+
E
⃗
y
E
⃗
=
E
⃗
x
+
E
⃗
y
Now
there are a number of different cases that arise.
If
δ
=
0
,
2
π
,
4
π
,
…
δ
=
0
,
2
π
,
4
π
,
…
Then we can write the field as
E
⃗
=
(
E
0
x
ı
̂
+
E
0
y
̂
)
cos
(
k
z
−
ω
t
)
E
⃗
=
(
E
0
x
ı
̂
+
E
0
y
̂
)
cos
(
k
z
−
ω
t
)
Polarization of light for the
case
E
⃗
=
(
E
0
x
ı
̂
+
E
0
y
̂
)
cos
(
k
z
−
ω
t
)
E
⃗
=
(
E
0
x
ı
̂
+
E
0
y
̂
)
cos
(
k
z
−
ω
t
)
The field is linearly polarized, that is the E field lies along a straight
line.
Likewise If
δ
=
π
,
3
π
,
5
π
,
…
δ
=
π
,
3
π
,
5
π
,
…
Then again it is linearly polarized, but is now "flipped"
E
⃗
=
(
E
0
x
i
̂
−
E
0
y
j
̂
)
cos
(
k
z
−
ω
t
)
E
⃗
=
(
E
0
x
i
̂
−
E
0
y
j
̂
)
cos
(
k
z
−
ω
t
)
Polarization of light for the
case
E
⃗
=
(
E
0
x
ı
̂
−
E
0
y
̂
)
cos
(
k
z
−
ω
t
)
E
⃗
=
(
E
0
x
ı
̂
−
E
0
y
̂
)
cos
(
k
z
−
ω
t
)
Circularly polarized light is a particularly interest example. Let
δ
=
−
π
/
2
δ
=
−
π
/
2
and
E
0
x
=
E
0
y
=
E
0
E
0
x
=
E
0
y
=
E
0
Then
E
⃗
x
=
E
0
cos
(
k
z
−
ω
t
)
ı
̂
E
⃗
x
=
E
0
cos
(
k
z
−
ω
t
)
ı
̂
E
⃗
y
=
E
0
cos
(
k
z
−
ω
t
−
π
/
2
)
̂
E
⃗
y
=
E
0
cos
(
k
z
−
ω
t
−
π
/
2
)
̂
E
⃗
y
=
E
0
sin
(
k
z
−
ω
t
)
̂
E
⃗
y
=
E
0
sin
(
k
z
−
ω
t
)
̂
or
E
⃗
=
E
0
[
cos
(
k
z
−
ω
t
)
ı
̂
+
sin
(
k
z
−
ω
t
)
̂
]
E
⃗
=
E
0
[
cos
(
k
z
−
ω
t
)
ı
̂
+
sin
(
k
z
−
ω
t
)
̂
]
The direction of
E
⃗
E
⃗
is changing with time. For example consider the case at
z
=
0
z
=
0
.
Then
E
⃗
=
E
0
[
cos
(
−
ω
t
)
ı
̂
+
sin
(
−
ω
t
)
̂
]
E
⃗
=
E
0
[
cos
(
−
ω
t
)
ı
̂
+
sin
(
−
ω
t
)
̂
]
E
⃗
=
E
0
[
cos
(
ω
t
)
ı
̂
−
sin
(
ω
t
)
̂
]
E
⃗
=
E
0
[
cos
(
ω
t
)
ı
̂
−
sin
(
ω
t
)
̂
]
In this case the electric field undergoes uniform circular rotation. For
light coming out of the page, it will have the motion shown in the
figure
Polarization of light for the
case
E
⃗
=
E
0
[
cos
(
ω
t
)
ı
̂
−
sin
(
ω
t
)
̂
]
.
E
⃗
=
E
0
[
cos
(
ω
t
)
ı
̂
−
sin
(
ω
t
)
̂
]
.
I, and most physicists would
call the Left Hand Circular polarization (LHC). The thumb points in the
direction of the light ray and the fingers curve in the direction of rotation.
(This is known as the Angular momentum convention, optical scientists will use
the "Optical" convention which is opposite - we will stick to the angular
momentum convention.)
Suppose
δ
=
π
/
2
δ
=
π
/
2
then you get
E
⃗
=
E
0
[
cos
(
k
z
−
ω
t
)
ı
̂
−
sin
(
k
z
−
ω
t
)
̂
]
E
⃗
=
E
0
[
cos
(
k
z
−
ω
t
)
ı
̂
−
sin
(
k
z
−
ω
t
)
̂
]
This
now has the opposite rotation, it is Right Handed.
The most general case of polarization has
δ
δ
arbitrary and
E
0
x
≠
E
0
y
E
0
x
≠
E
0
y
and is elliptically polarized.
Elliptical
polarization
Natural light is emitted with the
E
⃗
E
⃗
field in a mixture of random directions. This in unpolarized light. At any
particular instant
E
⃗
E
⃗
has a particular direction, but that direction changes rapidly and randomly.
A polarizer is a device that takes incident natural light and transmits
polarized light. For example a linear polarizer will take incident light and
select only that component of the light that has its
E
⃗
E
⃗
field lined up along the transmission axis. Suppose there is a linear
polarizer that transmits light along a particular axis. This is followed by a
second linear polarizer that has its transmission axis at a different angle
with
θ
θ
being the angle between the transmission axes. Since
I
∼
E
2
I
∼
E
2
then at the second polarizer
I
(
θ
)
=
I
(
0
)
cos
2
θ
I
(
θ
)
=
I
(
0
)
cos
2
θ
where
I
(
0
)
I
(
0
)
is the Irradiance of the light hitting the second polarizer.
Lets consider unpolarized light hitting a linear polarizer. Then half the
Irradiance will get through. If this is followed by a second polarizer at
90
0
90
0
then no light will pass through the second polarizer. Now what happens if a
third polarizer is placed between them?