ON OPTICAL THEORIES. 223 
‘To solve these, put 
1 = p COS w 
C ? 
k = p sin @, 
2 
‘Then 
ea p? cos 20 = F 
cz om 
u A 4 } (25) 
<” = p? sin 2u0= 
om TP Sin Aw G 
Thus the value of &, on which the absorption depends, is proportional 
to y’, the coefficient of dU /dt in the equation, and vanishes if y? is zero; 
that is, if there be no frictional resistance to the matter motion. If k& be 
at all appreciable, the light-disturbance will penetrate but a little way 
into the medium, so that for transparent media we may treat k, and there- 
fore G, as small. 
In this case we have 
1 1G 
are shan peepee : 
e ae oe 
while in the small term we may put for G/F the value 2ke/n. 
In these circumstances, then, 
Las pty? 1 
G= : x ‘ ape 
2 2a u? (n?— 7)? +4 (2? + wo”) a? ( 6) 
where 
(27) 
pr? =a? + ? — 4 /2p 
w= y*/4y? 
Thus, as » changes k/c is a maximum when n=1; if the corresponding 
values of & and c be iy and co, then 
Co (n2— 12)? yas 
B11 + gaan ae iam yf De leat 
If the value of y be zero, then, for n= 1, k is intinite compared with c; all 
the light is absorbed. 
At the same time A is large, and we have, in dealing with the motion 
of the matter particles, to consider the limit of Ae—*:, 
Turning, now, to the refraction, let C be the velocity of light in free 
space, N the refractive index, and suppose that the term 1G?/F may be 
neglected, then 
W= Or ei at p [4A (v? + 2a? — n?) (29) 
2 mn? myn? {(v? — n?)? + 4a2(1? + w)} 
a 
and the maxima and minima values of this expression lead to the limiting 
values of the refractive index. 
These, it is shown, are given approximately by n? — »? = + 2ya, which 
