SME
Engineering Electromagnetics · Problem type
Reflection at a boundary
Use this for “A plane wave in air hits a dielectric with εᵣ = 4 head-on. Find Γ, τ and the power passed.”
Diagram
Γ
=
η₂ − η₁
η₂ + η₁
=
188.3 − 376.7
565
Γ = −0.3333
τ = 1 + Γ = 0.6667: at the boundary 1 + Γ = τ, so E is the same on both sides (the dot)
Power:
Γ
2
=
11.11%
back,
1
−
Γ
2
=
88.89%
through;
beyond,
λ₂
=
λ₁
×
η₂
η₁
,
shorter.
This is an example. Type your own values to start a new problem.
η₁ (Ω)
=
η₀
÷
εᵣ₁
η₂ (Ω)
=
η₀
÷
εᵣ₂
Γ
=
η₂ (Ω)
−
η₁ (Ω)
η₂ (Ω)
+
η₁ (Ω)
τ
Transmission coefficient
calculated
R
Power reflected
calculated · %
T
Power transmitted
calculated · %
Formulas
η₁
=
376.7
÷
√
εᵣ₁
376.7
=
376.7
÷
√
1
η₂
=
376.7
÷
√
εᵣ₂
188.35
=
376.7
÷
√
4
Γ
=
η₂
−
η₁
η₂
+
η₁
−0.3333
=
188.35 − 376.7
188.35 + 376.7
τ
=
1
+
Γ
0.6667 = 1 + (−0.3333)
R
=
100
×
Γ
2
11.1111
=
100
×
(−0.3333)
2
T
=
100
×
(1
−
Γ
2
)
88.8889
=
100
×
(1
−
(−0.3333)
2
)
Assumptions
•
Normal incidence on a flat boundary between lossless, nonmagnetic media, so η = 376.7 ÷ √εᵣ Ω.
•
Γ and τ are ratios of the E amplitudes; Γ < 0 means the reflected E is flipped.
•
Medium 2 acts like a load η₂ on a line Z₀ = η₁, so medium 1 holds the same standing wave.
Step-by-step
WE KNOW
εᵣ₁ = 1, εᵣ₂ = 4
FIND
impedance 1 (η₁), impedance 2 (η₂), reflection coefficient (Γ), transmission coefficient (τ), power reflected (R), power transmitted (T)
Step 1 · Find impedance 1 (η₁)
Use η₁ = 376.7 ÷ √(εᵣ₁)
The vacuum’s η₀ = 376.7 Ω, smaller by √εᵣ in the first medium.
η₁
=
376.7
÷
√
1
η₁
=
376.7
Ω
Step 2 · Find impedance 2 (η₂)
Use η₂ = 376.7 ÷ √(εᵣ₂)
The same for the second medium.
η₂
=
376.7
÷
√
4
η₂
=
376.7
2
η₂
=
188.35
Ω
Step 3 · Find reflection coefficient (Γ)
Use Γ = (η₂ − η₁) ÷ (η₂ + η₁)
E and H must match across the boundary; solving the two conditions gives this share of E sent back.
Γ
=
188.35 − 376.7
188.35 + 376.7
Γ
=
−
188.35
565.05
Γ
=
−0.333333
Step 4 · Find transmission coefficient (τ)
Use τ = 1 + Γ
E just past the boundary equals the sum of the wave sent and the wave reflected.
τ
=
1
+
(−0.333333)
τ
=
0.6667
Step 5 · Find power reflected (R)
Use R = 100 × Γ²
Power
goes
as
E
2
,
so
Γ
2
of
it
comes
back,
as
a
percent.
R
=
100
×
(−0.333333)
2
R
=
100
×
0.1111
R
=
11.1111%
Step 6 · Find power transmitted (T)
Use T = 100 × (1 − Γ²)
No loss at the boundary: what is not reflected goes through.
T
=
100
×
(1
−
(−0.333333)
2
)
T
=
100
×
(1
−
0.1111)
T
=
100
×
0.8889
T
=
88.8889%
Check
Put the numbers back in. Both sides should match.
376.7
=
376.7
÷
√
1
✓
188.35
=
376.7
÷
√
4
✓
−0.333333
=
188.35 − 376.7
188.35 + 376.7
✓
0.6667 = 1 + (−0.333333) ✓
11.1111
=
100
×
(−0.333333)
2
✓
88.8889
=
100
×
(1
−
(−0.333333)
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.