8.2 A plane wave traveling in medium 1 with εr1 = 2.25 is normally incident upon
medium 2 with εr2 = 4. Both media are made of nonmagnetic, non-conducting
materials. If the electric field of the incident wave is given by
Ei = ŷ8 cos(6π × 109 t − 30π x)
(V/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each of
the two media.
(b) Determine the average power densities of the incident, reflected and transmitted waves.
Solution:
(a)
Ei = ŷ 8 cos(6π × 109t − 30π x)
(V/m),
377
η0
η0
η0
=
= 251.33 Ω,
=
η1 = √ = √
εr1
1.5
2.25 1.5
η0
η0
377
= 188.5 Ω,
η2 = √ = √ =
εr2
2
4
Γ=
η2 − η1 1/2 − 1/1.5
=
= −0.143,
η2 + η1 1/2 + 1/1.5
τ = 1 + Γ = 1 − 0.143 = 0.857,
Er = ΓEi = −1.14 ŷ cos(6π × 109 t + 30π x)
(V/m).
Note that the coefficient of x is positive, denoting the fact that Er belongs to a wave
traveling in −x-direction.
E1 = Ei + Er = ŷ [8 cos(6π × 109 t − 30π x) − 1.14 cos(6π × 109 t + 30π x)]
8
cos(6π × 109 t − 30π x) = ẑ 31.83 cos(6π × 109t − 30π x)
η1
1.14
cos(6π × 109t + 30π x) = ẑ 4.54 cos(6π × 109 t + 30π x)
Hr = ẑ
η1
Hi = ẑ
(A/m),
(mA/m),
(mA/m),
H1 = Hi + Hr
= ẑ [31.83 cos(6π × 109t − 30π x) + 4.54 cos(6π × 109t + 30π x)]
√
√
Since k1 = ω µε1 and k2 = ω µε2 ,
r
r
ε2
4
k1 =
30π = 40π
(rad/m),
k2 =
ε1
2.25
E2 = Et = ŷ 8τ cos(6π × 109t − 40π x) = ŷ 6.86 cos(6π × 109t − 40π x)
H2 = Ht = ẑ
8τ
cos(6π × 109t − 40π x) = ẑ 36.38 cos(6π × 109t − 40π x)
η2
(mA/m).
(V/m),
(mA/m).
(b)
Siav = x̂
82
64
=
= x̂ 127.3
2η1 2 × 251.33
(mW/m2 ),
Srav = −|Γ|2 Siav = −x̂ (0.143)2 × 0.127 = −x̂ 2.6
Stav =
(mW/m2 ),
|E0t |2
2η2
= x̂ τ 2
(8)2
(0.86)2 64
= x̂ 124.7
= x̂
2η2
2 × 188.5
Within calculation error, Siav + Srav = Stav .
(mW/m2 ).