614 EFFECT OF APERTURE IN PROJECTION [Cn. XIV 



Consider a sphere of one meter radius having this lighted surface at its 

 center. The light received by one square meter of the surface of this sphere 

 will then be B cos e lumens. Only half of the sphere can receive light from 

 this opaque surface and the entire light received by this hemisphere will be : 



sin0 cos9 



7TB 



Now if the reflecting surface is perfectly white there will be no light lost and 

 the entire light received by the hemisphere will equal the light incident upon 

 the reflecting surface, that is ir~B = 1/10,000 and B = i/io,oooir candle- 

 power per square centimeter. In the above example where the incident 

 illumination is 48,000 meter candles, the surface considered as a source of 

 light will have 48,000 candle power per square centimeter. 



IO.OOOIT 



This same formula will apply to the case of opaque projection ( 2/4a) where 

 it is desired to determine the ratio of the light getting through the objective to 

 form the screen image and the light falling on the opaque surface, assuming that 

 this opaque surface is perfectly diffusing and perfectly white. In the case of 

 the objective, light over a certain zone of the hemisphere is used. If the angle 

 which the objective subtends with a point on the object taken as the center is 

 called 26, then the angle between the axis and the edge of the objective is e, 

 and the above formula will apply, i. e., the light flux striking the objective 

 from one square centimeter is 7rB/2 (i cos 26). Also the total light flux 

 reflected from the surface over the entire hemisphere is irB, hence the ratio of 

 the light flux striking the objective to form the screen image to the light flux 

 received by the reflecting surface is i cos 26. This takes no account of losses 



2 



due to reflection and absorption by the objective. 



