Since this layer is isothermal and since its absorptivity
a
equals its emis-
sivity
ε
, it follows that it must emit
fl
ux density
F
1
in both the upward
and downward directions. Hence, it loses energy at a rate
2
F
1
.
The loss
due to emission is balanced by the absorption of upwelling radiation from
below. The upwelling radiation from below is
(
F
+
F
1
)
and the fraction
of it that is absorbed in the topmost layer is
a
.
Hence
2
F
1
=
a
(
F
+
F
1
)
(1)
Solving, we obtain
F
1
=
αF
(2
−
α
)
(2)
But from the Stefan Boltzmann law and Kirchho
ff
’s law
F
1
=
aσT
4
1
(3)
where
T
1
is the equivalent blackbody temperature of the topmost layer.
Combining Eqs. (2) and (3) yields
aσT
4
1
=
αF
(2
−
α
)
In the limit, as
a
→
0
,
σT
4
1
=
F
2
=
σT
4
E
2
Solving, we obtain
T
1
= (0
.
5)
1
/
4
T
E
For the Earth;s atmosphere,
T
E
= 255
K and
T
1
, the so-called skin tem-
perature, is 214 K, which is quite comparable to the mean temperature of
the lower stratosphere.
63

4.40
Consider an idealized aerosol consisting of spherical particles of radius
r
with a refractive index of 1.5. Using Fig. 4.13, estimate the smallest radius
for which the particles would impart a bluish cast to transmitted white
light, as in the rarely observed "blue moon".
To impart a bluish (as opposed to a reddish) cast to transmitted light, the
particles would need to exhibit a range of sizes for which the scattering
e
ﬃ
ciency
K
λ
decreases with increasing values of the size parameter, the
reverse of the situation for Rayleigh scattering.
It is evident from Fig.
4.13, that the smallest range of values of
x
for which
dK
λ
/dx <
0
occurs
around
x
=
2
πr
λ
∼
5
For visible light with
λ
∼
0
.
5
µ
m
2
πr
∼
5
×
0
.
5
µ
m
= 2
.
5
µ
m
r
∼
0
.
4
µ
m
4.41
Consider an idealized cloud consisting of spherical droplets with a uniform
radius of
20
µ
m and concentrations of 1 cm
−
3
. How long a path through
such a cloud would be required to deplete a beam of visible radiation by a
factor of
e
due to scattering alone?
(Assume that none of the scattered
radiation is subsequently scattered back into the path of the beam.)
From Eq. (4.16),
dI
λ
=
−
I
λ
K
λ
Nσds
The size parameter
x
= 2
πr/λ
= 2
π
×
20
µ
m
÷
0
.
5
µ
m
>
50
.
Hence, if
absorption is assumed to be negligible, Fig. 4.13 yields a value of
K
λ
= 2
.
The number density
N
= 1
cm
−
3
= 10
6
m
−
3
and the cross-sectional
area
σ
=
π
×
¡
20
×
10
−
6
¢
2
= 1
.
2
×
10
−
9
.
Hence, the volume scattering
coe
ﬃ
cient
K
λ
Nσ
= 2
.
4
×
10
−
3
m
−
1
.
The incident radiation is depleted
by a factor of
e
over a path in which the optical depth
τ
λ
=
R
K
λ
Nσds
is equal to unity. If we assume that conditions are uniform along the ray
path, the path length is given by
s
= (
K
λ
Nσ
)
−
1
=
¡
2
.
4
×
10
−
3
m
−
1
¢
−
1
∼
400
m
4.43
Consider radiation with wavelength
λ
and zero zenith angle passing through
a gas with an volume absorption coe
ﬃ
cient
k
λ
of 0.01 m
2
kg
−
1
. What frac-
tion of the beam is absorbed in passing through a layer containing 1 kg
m
−
2
of the gas?
What mass of gas would the layer have to contain on
order to absorb half the incident radiation?

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- Fall '15
- Thermodynamics, Atmosphere, Remote Sensing, Black body