1
Physics 316
Solution for homework 7
Spring 2005
Cornell University
I.
EXERCISE 1
How can a device be built that outputs horizontally polarized light of an intensity that is independent of the polar
ization direction of an incoming linearly polarized beam?
One way to get a linearly polarized wave independent of the initial polarization is to let light go trough a
λ/
4plate,
and then use a polarizer making an angle of 45
◦
with the fast and the slow axis of the
λ/
4plate. Let’s show what
happens mathematically. The initial beam can be written as:
~
E
=
E
cos(
kz

ωt
)(
~e
x
cos
θ
+
~e
y
sin
θ
)
,
(1.1)
where
θ
is the angle of polarization (unknown).
The
λ/
4plate introduces a phase diFerence of
π/
2 between the
x
component and the
y
component. Thus, the beam state at the output is:
~
E
1
=
E~e
x
cos
θ
cos(
kz

ωt
) +
E~e
y
sin
θ
sin(
kz

ωt
)
.
(1.2)
Now, let this wave go through the polarizer. The amplitude of the
x
component going through the polarizer will be
reduced by cos
π/
4 = 1
/
√
2. The amplitude of the
y
component going trough the polarizer will be reduced by the
same amount since sin
π/
4 = 1
/
√
2. Thus, the ±nal output, in the direction of polarization, is:
E
2
=
E
√
2
cos
θ
cos(
kz

ωt
) +
E
√
2
sin
θ
sin(
kz

ωt
) =
E
√
2
cos(
kz

ωt

θ
)
.
(1.3)
II.
EXERCISE 2
Cross two linear polarizers with an angle of
90
◦
so that no light is transmitted. Insert a third polarizer in between
these two with an angle
θ
to the Frst polarizer. What is the transmission as a function of
θ
?
θ
FIG. 1: A polarizer (
B
) is inserted between two orthogonal polarizers (
A
and
C
).
De±ne the polarizers as in ±gure 1. Assume
I
A
is the intensity of the light at the output of polarizer
A
. Then,
I
B
=
I
A
cos
2
θ
. The angle between polarizers
B
and
C
is
π/
2

θ
. Therefore,
I
C
=
I
B
cos
2
(
π/
2

θ
) =
I
B
sin
2
θ
.
Hence,
I
C
=
I
A
cos
2
θ
sin
2
θ
=
I
A
4
sin
2
2
θ.
(2.1)
In general,
I
C
6
= 0.
2
III.
EXERCISE 3
A.
Given a horizontally polarized beam with
~
E
=
~e
x
E
x
cos(
kz

ωt
+
φ
x
)
and
ω
=
ck
. Verify that

~
E
(
z,t
)

2
depends
on
z
and
t
. Compute the time average
<

~
E
(
z,t
)

2
>
t
. Does it depend on
z
?
Let’s ±rst calculate

~
E
(
z,t
)

2
:

~
E
(
z,t
)

2
=
~
E
·
~
E
=
E
2
x
cos
2
(
kz

ωt
+
φ
x
)
.
(3.1)
We then compute the time average,
D

~
E
(
z,t
)

2
E
t
. To do so, we need to integrate over one period of oscillation of

~
E
(
z,t
)

2
. At any given
z
, the period of oscillation is the smallest
T >
0 such that:

~
E
(
z,t
)

2
=

~
E
(
z,t
+
T
)

2
.
(3.2)
Thus
cos
2
(
kz

ωt
+
φ
x
) = cos
2
(
kz

ωt
+
φ
x

ωT
)
⇒
ωT
=
π,
(3.3)
that is,
T
=
π/ω
. Let’s calculate the time average:
D

~
E
(
z,t
)

2
E
t
=
1
T
Z
t
+
T
t

~
E
(
z,t
)

2
dt
(3.4)
=
ω
π
Z
kz

ωt
+
φ
x

π
kz

ωt
+
φ
x

E
2
x
ω
cos
2
θdθ
(with
θ
=
kz

ωt
+
φ
x
)
=
E
2
x
π
Z
kz

ωt
+
φ
x
kz

ωt
+
φ
x

π
1
2
(1 + cos 2
θ
)
dθ
=
E
2
x
π
θ
2
f
f
f
f
kz

ωt
+
φ
x
kz

ωt
+
φ
x
