1.4 Origin of Refractive Index
9
or
d
2 x
dt 2 + ω
2
0 x = −
qE
m
(1.19)
where −q and m are the charge and the mass of an electron, k 0 is the restoring force
constant, and ω 0 =
√
k 0 /m is the resonant frequency. On solving Eq. 1.18, we get
x = −
qE
m(ω
2
0 − ω 2 )
(1.20)
If the number of dispersion electrons per unit volume is N, then the polarization P
induced in the medium by electric field E is given by
P = −Nqx
(1.21)
=
Nq
2
m(ω
2
0 − ω 2 )
E
(1.22)
= 0 χ E
(1.23)
where
χ =
Nq
2
m 0 (ω
2
0 − ω 2 )
(1.24)
is the electric susceptibility and 0 is the permittivity of free space. Hence, the permittivity of the medium is given by
= 0 + 0 χ
(1.25)
= 0 (1 + χ) = 0 r
(1.26)
where r is the relative permittivity of the medium given by
r = 1 + χ = 1 +
Nq
2
m 0 (ω
2
0 − ω 2 )
(1.27)
It is known that the relative permittivity is square of the refractive index, and hence
Eq. 1.27 can be written as
n
2
= 1 + χ = 1 +
Nq
2
m 0 ω
2
0
1 −
ω
2
ω
2
0
−1
(1.28)
≈ 1 +
Nq
2
m 0 ω
2
0
1 +
ω
2
ω
2
0
(1.29)
≈ 1 +
Nq
2
m 0 ω
2
0
+
Nq
2
m 0 ω
4
0
4π
2 c
2
λ
2
0
(1.30)