Page 305 - Modern Optical Engineering The Design of Optical Systems

284 Chapter Twelve

4
displacement cos
0.5 in 0.955019
1.5 in 0.956474
2.5 in 0.885813
3.5 in 0.800620
4.5 in 0.651560
5.5 in 0.589447
6.5 in 0.494192
7.5 in 0.409600
8.5 in 0.337040
9.5 in 0.276281

4
At the center E 2 0.005 cos where the summation is from displace-
ment 0.5 in to 4.5 in
2 0.005 4.329476 0.04329 lumens/cm 2
which is equal to 929 0.04329 40.22 footcandles
(b) At the end of the slit we remove the factor of 2 and sum from 0.5 in to 9.5 in,
2
which gives us 0.03218 lumens/cm , which equals 29.90 footcandles.
4 A 16-mm movie projector uses a 2 in f/1.6 projection lens and a lamp with
a filament brightness of 3000 candles/cm2. If the condenser fills the pupil of
the lens with the filament image, what is the illumination produced on a
screen 20 ft from the lens? (Assume transmission of 95 percent for the lens and
85 percent for the condenser.)
ANSWER: Using Eq. 12.11 in photometric symbols
2
E t B sin
The diameter of the aperture for a 2 in f/1.6 lens is 2/1.6 1.25 in. At a dis-
tance of 20 ft 240 in the sine of the half-angle subtended by the lens aper-
2
1
ture is / 1.25/240 0.0026404 and sin 0.000006782
2
E 0.95 0.85 3000 0.000006782
0.051612 lumens/cm 2
2
929 0.051612 47.9 lumens/ft 47.9 footcandles
5 (a) What is the spectral radiant emittance of a 1000-K blackbody in the
region of 2000 nm wavelength? What is the radiance? (b) If an idealized band-
pass filter, transmitting 100 percent between 1950 and 2020 nm, is used, what
2
is the total power falling on a 1 cm detector placed one meter from a 1 cm 2
1000K blackbody? (Use Eq. 12.7)
ANSWER: (a) Per Eq. 12.16 the wavelength for peak emittance is 2897.8/T m
5
2.8978 m and per Eq. 12.17 the peak emittance is 1.286 T /10 15
2
1.286 W/cm m. Our wavelength as a fraction of the peak is 2.0/2.8978 0.69,
and from Fig. 12.6, at this fractional wavelength the emittance equals 0.7 of
2
the peak, or 0.9 W/cm m.