14.3 Solutions
697
14.98 λ g =
λ
1 − (λ/2a) 2
(1)
λ g = 3λ (by problem)
(2)
Combining (1) and (2) and solving for λ with a = 3 cm we find the freespace wavelength λ = 4
√
2 cm.
14.99 (a) λ g =
λ
1 − (λ/2a)
2
=
8
1 − (8/10)
2
= 13.33 cm
(b) The cut-off wavelength λ c = 2a = 2 × 5 = 10 cm.
14.100 N =
8πν 2 dνV
c 3
=
8π dλ · V
λ 4
Mean λ = 5500Å = 5.5 × 10 −7 cm
dλ = (6000 − 5000)Å = 10
−7 cm
V = (0.5)
3 cm
3
N =
8π × 10 −7 × (0.5) 3
(5.5 × 10 −7 ) 4
= 3.43 × 10
18
14.101 The cut-off frequency of the TM mn or TE mn mode is
ω mn = c
πm
a
2 +
πn
b
2
1/2
The lowest value we can have for ν mn is for the choice m = 1, a = 15 and
n = 0, that is, for TM 10 wave.
∴ ν 10 =
c
2
×
1
15
=
3 × 10 8
30
= 10
7 Hz = 10
4 kHz
This is much above the range of AM waves (530–1600 kHz). Hence AM
waves cannot propagate in the tunnel.
14.102 The cut-off frequency will be least for TE 10 waves. Of course TM 10 waves
do not exist.
ν 10 =
c
2π
π
a
=
c
2a
=
3 × 10 8
2 × 0.05
= 3 × 10
9 Hz = 3 GHz
Note that we take the higher dimension (5 cm) for the lower value of cut-off
frequency.