ILLCMl \.\Tln.\ .\M> PHOTOMETRY 



417 



Example. A light has an intensity of 25 candlepowei downward in a 

 tl direction. What is the illumination in foot-candles on a hori- 

 zontal table 4 ft. below this light. 



E = 3 



I'.') 



26 

 L6 



1.56 foot-candles. Arw. 



183. Law of Inverse Squares. Figure 378 shows that portion 

 of the light emitted by a certain source which is included within 

 a given solid angle. Let A! be a perfectly transparent surface 

 at a distance DI from the source. Let A 2 be a similar surface at 

 a distance D 2 from the source. By geometry, the areas A^ and 

 .1. are proportional to the squares of their distances from the 

 apex of the cone or pyramid. That is 



Source 

 FlO. 378. Variation of linht intensity with distance from source. 



Tho light flux passing through A\ is equal to the light flux 

 passing through A*, as none of the light flux passes out through 

 the sides of the solid angle. If the light flux passing through A\ 



and .1, is tin- same, then the density of the light flux or the 

 lumens [KM s(|iiare foot must be inversely as the areas. 7 

 the intensiti/ of ill H mi mil ion from (i point source varies ///nr.sr/// 

 as the square of the distance from tin' source. 



Let A' : he the illumination on surface A\ and E* the illumina- 

 tion on surface A*. Then 



The above law of inverse squares is strictly true only when 

 the light snum is a point. It is impossible to ol.tain a point 



sourer in practice. With the u-ual liizht Bourdee, oof 



is introduced in a.-sumini: a point source, unless the illuminated 

 27 



