COLOR PHOTOGRAPHY. 181 



But since 



Ui sin a\ = n- 2 sin a 2 , 



t. 2dn 2 2dn t / 1 th 2 • 2 



1) = — — i cos a 2 — __= J 1 — -i. sm 2 a^. 

 A A V w^ 



Generally in observing - the photographic film one looks perpendicu- 

 larly upon it. In this case cos a 2 = 1 and 



^ 2dn 2 



D = ^- 

 The difference in path of the two rays is a wave length when D = 1 

 since D is measured in wave lengths. The color then appears whose 

 wave length is 



l Q = 2dn 2 ; 



hence that which is caused by the stationary wave. 



In the general case however D = 1 for another wave length: 



A = 2dn 2 cos a 2 = A cos a 2 . 



One is thus able to obtain these values by multiplying A by a factor 



, A 



f = cos a 2 = ~. 



The degree of the color change is thus determined, that is the relative 

 wave lengths of the altered and the original color. 



Let these be designated in the case of incidence less than 45° in air 

 with/t in the prism with / p . 

 (1) Then 



(2) 



Let ratio /p//i be designated as/ pl . It determines the color change in 

 the experiment described in this chapter in which a part of the colored 

 image is viewed through air, another part through the prism. It is: 



(3) . fln?-n? 



• /pl ~" V 2w 2 2 -l 



The equations (1) to (3) show also in what ratio the wave lengths of 

 the colors observed must change if the indices of refraction of the 

 photographic film and of the prism are known. They can conversely 

 serve in the case/ and n\ suffer a known change to calculate the index 

 of refraction n 2 of the layer. Thus, the question may be answered, 

 given a value n 2 , how great an index of refraction the prism should 

 have in order that / shall have a value markedly different from 1, so 

 that there shall be a distinct color change. To be sure, one could make 

 use of greater angles of incidence than 45° ; but then the reflection 

 from the upper surface of the film might easily be so strong as to 

 destroy the value of the experiment. 



