44 Dr. K. S. Clay. On the Application oj 



This will be the part of the spectrum absorbed by the pink ink. 

 From X 59 to the end of the spectrum they were 



Violet, 0. Green, 15-6. Ked, 65-8. 



These add up to 



Violet, 148-2. Green, 148 -4. Ked, 148 '3. 



Thus the ray compositions of the inks are 



Yellow ink. Blue ink. Pink ink. 



Violet 25-8 148-2 122-4 



Green 143-6 132-8 20-4 



Ked 145-3 82-5 68-8 



The blue ink matches A 50 -4 and white, and the yellow will match 

 X 58 with white. 



I now took the ray composition of some spectrum colour, and found 

 by successive approximation the area to be covered by the inks to match 

 it. Thus X 44 has a ray composition of 



Red 1-47 equivalent to Ked 1-27 



Green 0-20 ' Green 0-0 



Violet 76-68 Violet 76-48 



White 0-2 



Then if 66 per cent, of the area is free from yellow ink, 66 per 

 cent, of the light the yellow would absorb will be transmitted, 

 namely 



Red, 0. Green, 1-2. Violet, 76-6. White, 2. 



So if 5 per cent, of the area is left free from blue ink it will reflect 

 in addition 



Red, 3-2. Green, 0-78. 



The pink ink must be printed all over. Then we shall have 

 left- 

 Red 3-2 equivalent to Red....- 1-22 



Green 1'98 Green O'O 



Violet 78-6 Violet 76-46 



White 2-0 White 3-98 



which matches the spectrum colour except for an excess of 3'7& 

 white. 



In the same way I worked out the percentage areas to be left free 

 from colour to match each of the wave-lengths 38, 40, 42, &c., up to 

 70, and have found in each case the excess of white. 



