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256 Chapter Twelve
in a 20-cm-diameter envelope. Assume that the bulb is painted so
that only a 1-cm square transmits energy, and that the source radiates
one-fiftieth of a watt through this square. (We assume, for conve-
nience, that the radiation intercepted by the painted envelope is thereby
totally removed from consideration.) Now the filament has an area of
2
0.01 cm and is radiating 0.02 W into a solid angle of (approximately)
2
0.01 steradian. Therefore, it has a radiance of 200 W ster 1 cm , but
only within the solid angle subtended by the window! Outside this
angle the radiance is zero. This concept of radiance over a limited
angle becomes important in dealing with the radiance of images and
must be thoroughly understood.
There are several interesting consequences of Lambert’s law that
are worthy of consideration, not only for their own sake but because
they illustrate the basic techniques of radiometric calculations. The
radiance of a surface is conventionally taken with respect to the area of
a surface normal to the direction of radiation. It can be seen that,
although the emitted radiation per steradian falls off with cos according
to Lambert’s law, the “projected” surface area falls off at exactly the
same rate. The result is that the radiance of a Lambertian surface is
constant with respect to . In visual work the quantity corresponding
to radiance is brightness (or luminance), and the above is readily
demonstrated by observing that the brightness of a diffuse source is
the same regardless of the angle from which it is viewed.
12.4 Radiation into a Hemisphere
Let us determine the total power radiated from a flat diffuse source into
2
a hemisphere. If the source has a radiance of N W ster 1 cm , one might
expect that the power radiated into a hemisphere of 2 steradians
2
would be 2 N W/cm . That this is twice too large is readily shown.
With reference to Fig. 12.2, let A represent the area of a small source
Figure 12.2 Geometry of a lambertian
source radiating into a hemisphere.