14.3 Solutions
695
Now v = c/n
(2)
∴
dv
dω
=
dv
dn
dn
dω
= −
c
n 2
dn
dω
(3)
Substituting (2) and (3) in (1) and using ω = kv and rearranging we get
v g =
c
n + ω(dn/dω)
14.92 V g =
distance
time
=
50
1 × 10 −6 = 5 × 10
7 m/s
V g = c
1 − (λ/2a) 2
∴ 5 × 10
7
= 3 × 10
8
1 − (λ/5) 2
Solving for λ, we find the free-space wavelength λ = 4.93 cm.
V p =
c 2
V g
=
(3 × 10 8 ) 2
5 × 10
7
= 1.8 × 10
9 m/s
14.93 V g = dω/dk
(1)
Rewriting (1), 1/V g = dk/dω
∴
1
V g
=
d
dω
ω
V p
=
1
V p
−
ω
V 2
p
dV p
dω
(2)
Substituting V p = c/n in (2)
1
V g
=
1
V p
−
ωn 2 c
c 2
−
1
n 2
dn
dω
=
1
V p
+
ω
c
dn
dω
(3)
14.94 V g =
∂ω
∂k
∴
1
V g
=
∂k
∂ω
=
∂ (2π/λ)
∂(2πν)
=
∂ (1/λ)
∂ν
But n =
c
v p
=
c
νλ
→
1
λ
=
nν
c
∴ V g =
∂ν
∂
1
λ
=
∂ν
∂ (nν/c)
=
c∂ν
∂(n ν)