Page:Proceedings of the Royal Society of London Vol 69.djvu/52

44 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.