Page 306 - Modern Optical Engineering The Design of Optical Systems

Principles of Radiometry and Photometry 285

2
(b) The total emittance in a 100 nm band is thus 0.1 0.9 0.09 W/cm emitted
2
into 2 steradians. The radiance is 0.09/ 0.0286 W/cm ster. The irradiance
2
2
2
equals N 0.0286 (1 /100 ) 2.86 10 6 W/cm . The flux equals area
irradiance 1 2.86 10 6 W.
Alternately we can consider the blackbody as a point source emitting 1.0 cm
2
2
0.0286 W/ster cm 0.0286 W/ster. With the detector subtending 1 /100
2
2
6
0.0001 ster, it intercepts 0.0001 0.0286 2.86 10 W.
6 Show that, for long projection distance, the maximum lumen output of a
projector is given by
2
F ABT/4(f/#) lumens
where A is the area of the film gate, B is the source brightness, T is the trans-
mission of the system, and (f/#) is the relative aperture of the projection lens.
ANSWER: Illumination E T B sin
2
sin (d/2)/D d/2D [D is the projection distance and d is the lens clear
aperture]
2
E T Bd /4D 2
d efl/(f/#) where efl is the focal length of the projection lens
2
2
E T B efl /(f/#) 4D 2
For long throws the magnification of the film gate to the screen is approxi-
mately D/efl, and thus the area of the screen is A (D/efl) 2
The total lumens on the screen is the product of the illumination E and the
area, or
2
2
2
Flux T B efl /(f/#) 4D times A (D/efl) 2
ABT/4(f/#) 2
QED