Theory of Colour Vision. 415 



(3) If white is specified by 5 = 12*7, P = 210, find the 

 complementary colour to 4659. 



This resolves itself into finding where the straight line 

 through 5 = 12*7, £ 2 = 210, and s= — 12*7, k 2 = 161 cuts the 

 straight line through 5 = 15, #=198, and 5 = 12*4, A 2 = 163. 

 Writing y for s 2 + k' 2 these straight lines are 



,y-371 5-12-7 



371-322 ~ 12-7 -f 12-7' 



y- 423 5-15 



423-317 " 15-12-4* 



Their intersection gives 5 = 13*8, and by the principle of 

 proportional parts \=5748. The last two problems could 

 be solved more easily graphically. 



(4) Light of X = 5842 can be matched by a mixture of 

 X=6063 and 5638 and also by a mixture of A,= 6302 and 

 5451. Show from the energy curves, that in each case we 

 obtain the same value of k 2 . 



Since the shape of the energy-curves is specified only 

 by P, we shall draw them as rectangles in order to save 

 trouble. The upper diagram of fig. 3 represents the first 



Ffe. 3. 



-10 



10 



20 



30 



40 



-10 



10 



20 



30 



40 



mixture, and the lower one the second. In the upper 

 diagram 240 units of X = 5638, i. e. 5 = 12*4, are added to 

 156 units of \=6063, i. e. 5=19*0. The area of the 

 rectangle to the right is consequently 240 units and of the 

 rectangle to the left 156 units, and their centroids are 

 respectively at 5 = 12*4 and 5 = 19*0. It will be found that 

 the centroid of the combination lies on the dotted line at 

 5 = 15*0, i. e. X=5842. The values of k 2 corresponding to 

 5=12*4 and 5=19*0 are 163 and 219. The lengths of half 



2 F 2 



