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µ
1
=
µ
,weobtain:
by value
+
(1 −
µ
1
)
(1 −
µ
=
µ
1
χ
χ
φ
2
) cos
,
=
ϕ
+ arccos
⎛
⎞
(2.10)
χ
µ
µ
1
−
ϕ
:
⎝
⎠
.
µ
2
1−
1−
µ
1
It is possible to attribute an evident meaning of the reflection probability to
the albedo in the description of the interaction with the surface: the reflection
occurs if
β
≤
A
and the opposite case corresponds to the photon absorption
bythesurfaceandtotheendofthephotontrajectory.Duetothereflection
orthotropness all possible directions of the photon are uniformly distributed
because:
− cos
2
β
,
µ
=
ϕ
=
πβ
τ
=
τ
0
.
2
,
(2.11)
After the repeating of the above reasoning about the photon free path to
calculate the desired irradiances we are revealing that it is enough to count
the number of the photons passing level
τ
(
z
) to the interested direction, i. e.
µ
µ
>
0) for
F
↓
(
z
)andupwardfor(
<
0) for
F
↑
(
z
). For that we
downward (
ξ
1
(
z
)and
ξ
1
(
z
) with the zeroth initial values. Let
τ
1
be the
introduce functions
τ
2
=
τ
1
+
∆τ
be
photon coordinate before simulating of the free path (2.7) and
the coordinate after the simulation. Then:
ξ
1
(
z
):
=
ξ
1
(
z
)+1, provided
τ
1
≤
τ
≤
τ
2
,
(
z
)
(2.12)
ξ
1
(
z
):
=
ξ
1
(
z
)+1, provided
τ
2
≤
τ
≤
τ
1
.
(
z
)
Equation (2.12) is also used in the case of the photon going out of the at-
mosphere (
τ
2
≥
τ
0
). After the simulation of
a certain number
K
of trajectories the desired irradiances are found as a num-
ber of the counting photons multiplied by energy of the single photon, which
is equal to
F
0
τ
2
<
0) or reaching the surface (
µ
0
|
K
(see above). Thus, we are inferring:
1
1
K
ξ
1
(
z
)
F
0
K
ξ
1
(
z
)
F
0
F
↑
(
z
)
=
µ
F
↓
(
z
)
=
µ
0
,
0
.
(2.13)
ξ
1
(
z
)willbecalledfurther
the counters
and the expressions
analogous to (2.12) will be treated as
writing to the counters
.
Equation (2.12) seems to contradict the formula of a link of the radiance and
irradiance (1.4) because they don't account for the cosine of the photon zenith
angle. However, the photon is a carrier of the very irradiance and not of the
radiance. It is easy to understand from the physical meaning: the real quantum
of light as a photon has energy independent of the direction. The formal proof
of the correctness of (2.12) is elementary. Consider the trajectory of a single
photon (
K
ξ
1
(
z
)and
Va l u e s
=
1).Letthephotongooutfromthetopoftheatmosphereafterthe
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