696
14 Electromagnetism II
14.95 v g =
dω
dk
=
d(ω ¯
h)
d(k ¯
h)
=
dE
d p
=
d
p 2 /2m
d p
=
1
2m
d p 2
d p
=
2 p
2m
=
mv
m
= v
14.96 By prob. (14.85)
V p − V g
V p
= −nλ
dn
dλ
(1)
where V p =
c
n
(2)
Re-arranging (1) with the aid of (2) and writing μ for n we find
V g = c
1
μ
+ λ
d μ
d λ
(3)
μ = 1.420 +
3.60 × 10 −14
λ 2
(by problem)
(4)
Substituting λ = 500 nm = 5 × 10 −7 m in (4)
μ = 1.564
(5)
Differentiating μ with respect to λ in (4)
dμ
dλ
= −
7.2 × 10 −14
λ 3
or λ
dμ
dλ
= −
7.2 × 10 −14
λ 2
= −0.288
(6)
Substituting (5) and (6) in (3), we find V g = 0.35 c.
14.3.4 Waveguides
14.97 (a) Assuming the dominant mode, for a = 2.5 cm, and b = 2.5 cm, for the
rectangular waveguide, and c = 3× 10 8 m/s, the velocity of electromagnetic waves in free space, the phase velocity is given by
V p =
c
1 − (λ/2a) 2
=
3 × 10 8
1 − (4/5) 2
= 5 × 10
8 m/s
(b) V g = c
1 − (λ/2a) 2 = 3 × 10 8
1 − (4/5) 2 = 1.8 × 10 8 m/s
(c) λ g =
λ
1 − (λ/2a) 2
=
4
1 − (4/5) 2
= 6.67 cm
14 Electromagnetism II
14.95 v g =
dω
dk
=
d(ω ¯
h)
d(k ¯
h)
=
dE
d p
=
d
p 2 /2m
d p
=
1
2m
d p 2
d p
=
2 p
2m
=
mv
m
= v
14.96 By prob. (14.85)
V p − V g
V p
= −nλ
dn
dλ
(1)
where V p =
c
n
(2)
Re-arranging (1) with the aid of (2) and writing μ for n we find
V g = c
1
μ
+ λ
d μ
d λ
(3)
μ = 1.420 +
3.60 × 10 −14
λ 2
(by problem)
(4)
Substituting λ = 500 nm = 5 × 10 −7 m in (4)
μ = 1.564
(5)
Differentiating μ with respect to λ in (4)
dμ
dλ
= −
7.2 × 10 −14
λ 3
or λ
dμ
dλ
= −
7.2 × 10 −14
λ 2
= −0.288
(6)
Substituting (5) and (6) in (3), we find V g = 0.35 c.
14.3.4 Waveguides
14.97 (a) Assuming the dominant mode, for a = 2.5 cm, and b = 2.5 cm, for the
rectangular waveguide, and c = 3× 10 8 m/s, the velocity of electromagnetic waves in free space, the phase velocity is given by
V p =
c
1 − (λ/2a) 2
=
3 × 10 8
1 − (4/5) 2
= 5 × 10
8 m/s
(b) V g = c
1 − (λ/2a) 2 = 3 × 10 8
1 − (4/5) 2 = 1.8 × 10 8 m/s
(c) λ g =
λ
1 − (λ/2a) 2
=
4
1 − (4/5) 2
= 6.67 cm