694
14 Electromagnetism II
(c) Differentiating with D with c constant, ω
dω
dk
= c 2 k
∴ v g =
dω
dk
=
c 2 k
ω
=
c 2
v p
= c
1 −
D 2
ω 2
∴ v p v g = c
2
(5)
(d) Substituting D = 1.2 × 10 11 /s, ω = 2π × 20 × 10 9 Hz and c = 3 ×
10 8 m/s in (3), we find v p = 1.016 × 10 9 m/s
Substituting υ p = 1.016 × 10 9 m/s in (5) we find v g = 8.858 × 10 7 m/s.
It is observed that while v p > c, v g < c.
14.88 V g =
dω
dk
(1)
ω = vk
(2)
∴ V g =
d
dk
(vk) = v + k
dv
dk
(3)
14.89 V g = v + k
dv
dk
(by prob. 14.88)
( 1 )
V = c/n
(2)
Substituting (2) in (1)
V g = c/n + ck
d
dk
1
n
=
c
n
−
ck
n 2
dn
dk
(3)
Now
dn
dk
=
dn
dλ
dλ
dk
=
dn
dλ
d
dk
2π
k
= −
2π
k 2
dn
dλ
(4)
Using (4) in (3)
V g =
c
n
+
λc
n 2
dn
dλ
14.90 v ∝
1
λ
(by problem)
∴ v = Ak (where A = constant)
v g = v + k
dv
dk
= Ak + k
d
dk
(Ak) = Ak + Ak = 2Ak = 2v
14.91 v g = v + k
dv
dk
= v + k
dv
dω
dω
dk
= v + kv g
dv
dω
(1)