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PHYSICS 218
SOLUTION TO HW 7
Created: October 23, 2004 8:15am
Last updated: November 1, 2004 8:44pm
1. (a) Let
E
0
=

E
0

e
iϕ
E
and
B
0
=

B
0

e
iϕ
M
. By comparing to
E
0
=
E
r
+
iE
i
and
B
0
=
B
r
+
iB
i
, it is easy
to verify that

E
0

=
q
E
2
r
+
E
2
i
,
B
0
=
q
B
2
r
+
B
2
i
,
ϕ
E
= arctan(
E
i
/E
r
) and
ϕ
M
= arctan(
B
i
/B
r
).
Therefore, we have
E
x
=

E
0

e
i
(
kz
−
ωt
+
ϕ
E
)
,B
y
=

B
0

e
i
(
kz
−
+
ϕ
M
)
.
Taking the real part of
E
x
and
B
y
, we obtain
<
E
x
=

E
0

cos(
kz
−
+
ϕ
E
)
,
<
B
y
=

B
0

cos(
−
+
ϕ
M
)
,
from which we see that
a
E
=

E
0

and
a
M
=

B
0

.
(b) Using the expressions
u
E
=
1
2
²
0
E
2
x
and
u
M
=
1
2
µ
0
B
2
y
for the energy densities of electric and magnetic
Felds respectively, we have
u
E
=
1
2
²a
2
E
cos
2
(
−
+
ϕ
E
)
,u
M
=
1
2
µ
a
2
M
cos
2
(
−
+
ϕ
M
)
.
(c) Averaging over one cycle
T
=2
π/ω
, we Fnd that
h
cos
2
(
−
+
ϕ
E
)
i
=
1
T
Z
T
0
cos
2
(
−
+
ϕ
E
)d
t
=
1
2
.
Thus, we have
h
u
E
i
=
1
4
2
E
and
h
u
M
i
=
1
4
µ
a
2
M
.
(d) We note that for a complex number

z

2
can always be written as

z

2
=
zz
∗
=
z
∗
z
. Sowehave
h
u
E
i
=
1
4
²E
0
E
∗
0
and
h
u
M
i
=
1
4
µ
B
0
B
∗
0
.
(e) The Poynting vector is given by
s
=
1
µ
(
<
E
x
)
×
(
<
B
y
)=
1
µ
(
<
E
x
)(
<
B
y
)
ˆ
k
. The
z
component of the
Poynting vector is thus
s
(
z, t
1
µ
(
<
E
x
<
B
y
1
µ
a
E
a
M
cos(
−
+
ϕ
E
) cos(
−
+
ϕ
M
).
(f) The average of
s
(
)is
h
s
i
=
1
µ
a
E
a
M
h
cos(
−
+
ϕ
E
) cos(
−
+
ϕ
M
)
i
. Using the trigono
metric identity 2 cos
A
cos
B
= cos(
A
+
B
) + cos(
A
−
B
), we Fnd
h
cos(
−
+
ϕ
E
) cos(
−
+
ϕ
M
)
i
=
1
2
cos(
ϕ
E
−
ϕ
M
)
.
(1)
The average of
s
(
) over one cycle is thus
h
s
i
=
1
2
µ
a
E
a
M
cos(
ϕ
E
−
ϕ
M
).
(g) We can rewrite
h
s
i
=
1
2
µ
(
a
E
cos
ϕ
E
a
M
cos
ϕ
M
+
a
E
sin
ϕ
E
a
M
sin
ϕ
M
1
2
µ
(
E
r
B
r
+
E
i
B
i
). On
the other hand, we observe that
<
(
E
0
B
∗
0
<
(
E
∗
0
B
0
E
r
B
r
+
E
i
B
i
, we then have
h
s
i
=
1
2
µ
<
(
E
0
B
∗
0
1
2
µ
<
(
E
∗
0
B
0
).
2. (a) The reﬂection and transmission coeﬃcients with the electric Feld perpendicular to the plane of
incidence and
κ
M
1
=
κ
M
2
= 1 given by Elmore and Heald (page 274) are
R
⊥
=
−
sin(
ϑ
1
−
ϑ
2
)
sin(
ϑ
1
+
ϑ
2
)
,T
⊥
=
2cos
ϑ
1
sin
ϑ
2
sin(
ϑ
1
+
ϑ
2
)
.
Using Snell’s law
n
1
sin
ϑ
1
=
n
2
sin
ϑ
2
,wehave
R
⊥
=
−
n
2
cos
ϑ
2
−
n
1
cos
ϑ
1
n
2
cos
ϑ
2
+
n
1
cos
ϑ
1
⊥
=
2
n
1
cos
ϑ
1
n
2
cos
ϑ
2
+
n
1
cos
ϑ
1
.
At normal incidence,
ϑ
1
=
ϑ
2
= 0, so we recover
R
⊥
=
n
1
−
n
2
n
1
+
n
2
⊥
=
2
n
1
n
1
+
n
2
.
1
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View Full DocumentPHYSICS 218
SOLUTION TO HW 7
(b) When the electric feld is parallel to the plane oF incidence, Elmore and Heald (page 275) gives
R
k
=
tan(
ϑ
1
−
ϑ
2
)
tan(
ϑ
1
+
ϑ
2
)
,T
k
=
2 cos
ϑ
1
sin
ϑ
2
sin(
ϑ
1
+
ϑ
2
) cos(
ϑ
1
+
ϑ
2
)
.
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 Fall '04
 POLLACK
 Physics, Magnetism

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