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the area of the tube end times the solid angle subtended at the eye by the far end
of the tube. The fact that things look the same means that the radiance has not
changed as you moved along the ray from the wall to your initial eyepoint.
To the degree that the pixel area on the sensor in a digital camera can be
regarded as infinitesimal (or that the variation of radiance with position on the
sensor can be assumed small) and the solid angle of rays that hit that pixel can
be considered infinitesimal (or the variation with direction assumed small), the
response of the sensor (assuming it responds to total arriving energy) is propor-
tional to radiance. In fact, high-quality cameras used for computer-vision exper-
iments produce images whose individual entries are radiance values. To be more
precise, the values they produce are often integrals, with respect to wavelength,
of spectral radiance multiplied by a response function that characterizes how the
sensor responds to radiance of each wavelength.
R
9
S
r
S
r
V
26.7.3 Two Radiance Computations
n
For a Lambertian emitter, the radiance in all outgoing directions is the same.
Let's suppose that we have a Lambertian emitting sphere S of radius r , emitting
total power Φ . We'll now compute the radiance along each ray leaving the sphere.
The idea (see Figure 26.26) is to surround the emitter with a concentric sphere S
of radius R
P
Figure 26.26: A radiating sphere
inside a large receiving sphere.
We'll
r . All power emitted from S must arrive at S , and the arriving
power density (in Wm 2 )on S is independent of position. If we call this density
D , then
>>
compute
arriving
power
density at P.
R 2 D .
4
π
(26.43)
We'll compute the power density at the point P in terms of the unknown con-
stant emitted radiance L , which will allow us to solve for L in terms of Φ .The
power density at P is
D =
L
| v ·
n ( P )
|
d
(26.44)
v
S + ( P )
=
r
L
| v ·
n ( P )
|
d
(26.45)
v
Ω
V
= L
Ω | v ·
n ( P )
|
d
.
(26.46)
v
n
P
R
The transition between Equations 26.44 and 26.45 is justified by noting that for
v
outside Ω , the radiance at P in direction
is zero, so the integral over the whole
hemisphere can be reduced to an integral over just Ω .
v
For sufficiently large values of R ,
v ·
n ( P ) is very close to 1, so in the limit as
R approaches infinity, we get
D = L
Figure
26.27:
Computing
the
measure of the solid angle Ω .
1 d
= Lm (Ω) .
(26.47)
v
Ω
The sphere of radius R around P (drawn in light gray in Figure 26.27) has
total area 4
R 2 , and subtends a solid angle of 4
R 2 area, an area
π
π
at P ;ofthis4
π
r 2 is occupied by the radiating sphere (i.e., as seen from P ,the
radiating sphere occludes a disk of area
of approximately
π
r 2
R 2 ). The solid angle
π
in the entire 4
π
 
 
 
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