( 883 ) 
n the real index of refraction, x the index of extinetion, while » 
means the frequency (i.e. the number of vibrations in the time 22) 
of the kind of light we are considering. The meaning of 9, y, and rr 
zh 
is clear from the formula. That V4 aa r, is the frequency of the 
“free vibration” of the electron becomes apparent, if in equation (1) we 
put 4=0 and VY =O, which means that no damping and no external 
electric force are supposed to exist. Each kind of electrons contributes 
a term in the sum. 
Separating the real from the imaginary terms in (2) leads to the 
relations 
an | OD =P") 
te a et ee 
Pm Ui) lc a 
| (3) 
ovv 
ORR | 
(ery ees 
from which » and x have to be solved. This problem becomes much 
simpler, and yet scarcely less general, if we confine our attention 
to the surroundings of each of the characteristic frequencies r, separately. 
In equation (2) we therefore put apart the term relating to the 
selected »,, and designate the other terms of the sum by a variable 
index /: 
On oO 
: ERN ee 
2 RADAR 
? Po WV 
Hee he 
Pi Hive 
decause we only consider such values of r as differ little from 
> we are allowed to replace v by rv, in the terms of the sum- 
mation >, and then to neglect 7zr',r, relatively to ps — »,* (the 
damping connected with rv, being imperceptible near rv). Writing 
yp — v, — wu, and therefore (u being small) »* — r,° = 2v, uw, we may 
put, instead of (4): 
1 
Qh Q 
ob 
Dh Plu) 
or 
3 : Q 5) 
ZEN —-— nt ae EL LE alst Come ee o 
Y,(2u ip’) ( 
where 2, represents the value which, in the small spectral region 
we are considering, the index of refraction would show if there 
were no electrons having the proper frequency r,. 
One might proceed to the separation of the real from the imaginary 
part of this simplified equation, and then solve 7 and x; but as the 
result would not yet be very simple, we shall first consider the 
special case that the modulus of the complex second term is small 
