414 Dr. R. A. Houstoun on a 



the same ns if we had used Dieteriei's R and G. For the 

 purpose o£ illustration I have added four examples : — 



(1) What proportion of light of wave-length 6559 should 

 be added to light of wave-length 5451 to give the same hue 

 as wave-length 5842 ? 



Let the proportion be x of 6559 to 1 of 5451. Then if 

 we use the ordinary method, since the three wave-lengths in 

 question contain only R and G, we may consider R alone. 

 We thus have 



#112 + 6=0+1)41, 



which gives # = '49, 



Jf we use my method, since we are dealing only with the 

 straight part of the curve, it is not necessary to consider k 2 

 at all. For the wave-lengths in question s has the values 

 23% 11, and 15. Hence 



#23*2 + 11 =(# + 1)1 5, 



which gives t r = *49. 



(2) Red of wave-length 6063 is added to blue of wave- 

 length 4659, and it is found that the same hue and intensity 

 can be obtained by adding green of wave-length 5281 to 

 violet of wave-length 4488. Determine the proportion of 

 each. 



Let x oi 6063 be added to 1 of 4659, and let y of 5281 

 be added to z of 4488. 



Then by the ordinary method 



# + l = ?/ + c. 



Considering the proportion of R present 



#74-22=^29, 



and considering the proportion of G present, 



#26 + 51 =#100 -*48. 



These equations give #='531, z/ = *935, and ^ = -596. 

 By my method we have 



x + l=y + z, 



^19-12-7= ( yl0-3-^21-l, 

 and 



#(19 2 + 219) -f (12-7 2 + 161) = ?/(10-3 2 + 128) + z{ 21-1 2 4 243). 



These equations give # = '519, ?/ = *931, and z = '588. 

 The small differences are due to experimental error in taking 

 s and k 2 graphically from fig. 2. 



