Neither of these difficulties occur in the use of the diffraction spectrum, where the pure colours are apportioned by Professor Rood from A to H in the manner shown in table on next page.
Professor Rood further divides the spectrum from A to H into 100 equal divisions, allotting 20 unit divisions of 72,716 wave lengths to the space between each two colour lines. This allots a space of 3,635 W.L. to each unit division, as shown in Table III.
TABLE III.
| Wave Length Position. | No. of Wave Lengths from Colour between each. | Division | W.L.K. per Division | ||
| 760,400 A. | Red | 760,400 —— | 72,717 | == 20 | 3,635 |
| Orange | 687,683 —— | 72,716 | == 20 | 3,635 | |
| Yellow | 614,967 —— | 72,716 | == 20 | 3,635 | |
| Green | 542,251 —— | 72,716 | == 20 | 3,635 | |
| Blue | 469,535 —— | 72,716 | == 20 | 3,635 | |
| 396,819 H. | Violet | 396,819 | |||
| 363,581 | Total W.L. between A. & H. | 363,581 | 100 | ||
Having provided equal wave length positions for the six pure colours, the intermediate colours are necessarily binaries in definite proportions, accounted for by a regular overlapping of two bounding colours in opposite directions from zero to 20, as shown in the following table from Red to Orange, representing the space between these two pure colours.
| Red W.L 760,400 | 20 | 19 | 18 | 17 | 16 | 15 | 14 | 13 | 12 | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 | W.L 687,683 Orange. |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | ||
| 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 | 20 |
It follows, that apart from the six monochromes, all spectrum complex colours in a single wave length must be binaries, whose united values equal 20.
On comparing Professor Rood’s scales of divisions with those of the tintometrical scales already described, they appear to coincide in several particulars, for instance:—
The monochromes correspond both in number and in name.