Page 304 - Modern Optical Engineering The Design of Optical Systems

Principles of Radiometry and Photometry 283

Exercises
1 A point source emits 10 W/steradian toward a 4 in diameter optical system.
How much power is collected by the optical system when its distance from the
source is (a) 10 ft, (b) one mile?
ANSWER: (a) At 10 feet the solid angle subtended by the 4 in aperture is
2
2
area/distance 2 /120 0.00087266 ster. The power equals the angle
2
times the intensity 0.00087266 10 0.0087266 W.
(b) At a mile, the distance is 5280 12 63,360 in and the solid angle
2
2
8
2 /63,360 3.130 10 9 ster. Thus the power collected is 3.130 10 .
2 A 10 candlepower point source illuminates a perfectly diffusing surface
which is tilted at 45 to the line of sight to the source. What is the brightness
of the surface if it is 10 ft from the source?
ANSWER: A 10 candlepower source emits 10 lumens per steradian.
2
At a distance of 10 ft, one square foot subtends 1/10 0.01 steradian.
The illumination on a surface normal to the propogation equals the intesity
times the solid angle 10 lumens 0.01 ster 0.1 lumens per square foot.
Because the surface is tilted at 45 , the area within 0.01 steradian is increased
by 1/cos 45 , and the illumination is 0.0707 lumens per square feet, which
is 0.0707 footcandles. On a perfect diffuser this produces a brightness of
0.0707 foot lamberts.
Using an alternate approach, 0.0707 lumens per square feet will be emitted/
reflected in a lambertian fashion into a hemisphere of 2 ster to produce a
2
1
brightness of 0.0707/ 0.022508 lumens ster ft . Since there are 929 cm 2
1
in a square feet this is 0.24228 10 4 lumen ster cm 2 or 0.24 10 4 stilb.
3 A fluorescent lamp 10 in long and 1 in wide illuminates a slit, parallel to the
lamp and 10 in from the lamp. If the lamp has a brightness of 0.5 candles/cm ,
2
(a) what is the illumination at the center of the slit, and (b) at the ends of the
slit? (Hint: divide the lamp into 10 one-inch-square sources.)
ANSWER: At normal incidence and emission a one inch square of 0.5 cd/cm 2
brightness which is 10 in away will produce an illumination
2
2
2
E B 0.5 cd/cm (1 /10 ) 0.005 lumens/cm 2
0
At an angle from the normal this becomes
4
E cos 0.005 lm/cm 2

At the center of the lamp the two adjacent squares have their centers displaced
0.5 in from the center. The next squares are 1.5 in off center, the next 2.5 in, the
3.5 in, and 4.5 in. These squares make tilt angles from the centerline equal to
arctan (displacement/10 in)
4
and we can tabulate the cos for displacements of 0.5 in to 4.5 in [and to 9.5
in for use in part (b)] as follows: