ECE 455: Optical Electronics
Homework 5 Solutions
Due: Oct. 15, 2009
1
Derivations
(a) Consider the flux of energy accross the
dx

dy
plane in a time
dt
. This energy is:
E
=
Idxdydt
In the time
dt
, the light has moved a distance
dz
=
c
n
dt
. Therefore the energy density is:
ρ
=
E
dxdydt
=
Idxdydt
dxdy
c
n
dt
=
n
c
I
(b) See Gamma Tom’s course notes for the derivation
dI
dz
=
γ
0
I
1 +
I/I
sat
(c) See Gamma Tom’s course notes for the derivation
ln
I
out
I
in
+
I
out

I
in
I
sat
=
γ
0
L
(d) See Gamma Tom’s course notes for the derivation
I
out
=
I
in
e
γ
0
L
(e) See Gamma Tom’s course notes for the derivation
I
out
=
I
in
+
γ
0
I
sat
L
2
2Level System
Verdeyen 7.13
(a)
dN
2
dt
=

σI
p
hν
"
N
2

g
2
g
1
N
1
#

N
2
τ
dN
1
dt
=
σI
p
hν
"
N
2

g
2
g
1
N
1
#
+
N
2
τ
(b) First set
dN
2
/dt
= 0. Solving for
N
2
we get:
N
2
=
g
2
g
1
N
1

N
2
hν
I
p
στ
As
I
p
→ ∞
, clearly
N
2
→
(
g
2
/g
1
)
N
1
.
1
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(c) Again assume a steady state distribution. Set
N
1
= 2
N
2
in the equation we found in part A.
Then we can solve for
I
p
.
I
p
=
hν
τσ
2
g
2
g
1

1
!

1
= 7
.
78
W/cm
2
(d) In writing the rate equations, there are many terms that are neglected becauase they are so
small compared to the optical effects. For example, the rate of excitation from state 1 to state
2 due to high energy inelastic collisions is ignored. When the pump beam is turned off, these
effects are all the only mechanisms that exchange particles between populations 1 and 2. The
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 Fall '08
 Eden,J
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