--- title: "Normal incidence, reflection and transmission" subject: "Electromagnetism" chapter: 4 key: "semester_3/electromagnetism:normal-incidence-reflection-transmission" source: https://www.bctengineer.com/q/semester_3/electromagnetism/normal-incidence-reflection-transmission site: BCT Engineering --- # Normal incidence, reflection and transmission Electromagnetism · Chapter 4: Wave equation and Wave propagation - Asked in: 2073 Shrawan, 8 marks - Video: [Reflection of Plane Wave At Normal Incidence](https://www.youtube.com/watch?v=_GuLwk-k4KQ) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/normal-incidence-reflection-transmission - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4 ## Question Explain the phenomena when a plane wave is incident normally on the interface between two different media. Derive the expression for reflection and transmission coefficient. ## Short answer At normal incidence on a dielectric interface, part of the wave reflects and part transmits. The reflection coefficient is Γ = (η₂ − η₁)/(η₂ + η₁), and the transmission coefficient is τ = 2η₂/(η₂ + η₁), where η is the intrinsic impedance of each medium. ## Answer #### Phenomenon When a uniform plane wave meets the boundary between two different media, part of its energy is **reflected** back into the first medium and the rest is **transmitted** into the second. For normal (perpendicular) incidence, the incident, reflected and transmitted waves all travel along the normal to the interface, with E and H tangential to the boundary. The reflection produces a standing wave in medium 1 (a mixture of the incident and reflected waves), while medium 2 carries only a travelling wave. ```text Medium 1 (ε1, μ1, η1) | Medium 2 (ε2, μ2, η2) | Ei ───────────► | Et ───────────► Er ◄─────────── | | z = 0 ``` #### Derivation Take the boundary at z = 0, media 1 (z < 0) and 2 (z > 0), perfect dielectrics, and E along a_x (H along a_y). ```text Incident: E_i = E_i0 e^{−jβ₁z} a_x, H_i = (E_i0/η₁) e^{−jβ₁z} a_y Reflected: E_r = E_r0 e^{+jβ₁z} a_x, H_r = −(E_r0/η₁) e^{+jβ₁z} a_y Transmitted: E_t = E_t0 e^{−jβ₂z} a_x, H_t = (E_t0/η₂) e^{−jβ₂z} a_y ``` The reflected H has a negative sign because the wave travels in −z (E × H must point in −z). **Boundary conditions at z = 0** (tangential E and H continuous, no surface current): ```text E: E_i0 + E_r0 = E_t0 ... (1) H: (E_i0 − E_r0)/η₁ = E_t0/η₂ ... (2) ``` Substitute (1) into (2): (E_i0 − E_r0) η₂ = (E_i0 + E_r0) η₁, so ```text E_r0 (η₁ + η₂) = E_i0 (η₂ − η₁) ``` **Reflection coefficient** **Γ = E_r0/E_i0 = (η₂ − η₁)/(η₂ + η₁)** **Transmission coefficient** (from (1): τ = 1 + Γ) **τ = E_t0/E_i0 = 2η₂/(η₂ + η₁)** #### Remarks - Relation: 1 + Γ = τ. - For non-magnetic dielectrics η = η₀/√ε_r, so Γ = (√ε_r1 − √ε_r2)/(√ε_r1 + √ε_r2). - If η₂ = η₁, Γ = 0 (matched, no reflection). Perfect conductor (η₂ = 0): Γ = −1, τ = 0, total reflection. - Power: reflected fraction = |Γ|², transmitted fraction = 1 − |Γ|² = (η₁/η₂)|τ|². - SWR in medium 1: s = (1 + |Γ|)/(1 − |Γ|). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When a plane wave hits a boundary at normal incidence: A. It is fully absorbed B. Part is reflected into medium 1 and part transmitted into medium 2 C. It turns 90 degrees D. Only reflection occurs for any media Answer: B. The reflected wave combines with the incident wave to give a standing wave in medium 1, while medium 2 has only the transmitted travelling wave. (Key point: Phenomenon) 2. Which boundary conditions at z = 0 are used? A. Normal E and tangential H continuous B. Tangential E and tangential H continuous (no surface current) C. Only E continuous D. Normal D and tangential B continuous Answer: B. E_i0 + E_r0 = E_t0 and (E_i0 - E_r0)/eta1 = E_t0/eta2. (Key point: Boundary conditions) 3. The reflection coefficient is: A. (eta1 - eta2)/(eta1 + eta2) B. 2 eta2/(eta1 + eta2) C. (eta2 + eta1)/(eta2 - eta1) D. (eta2 - eta1)/(eta2 + eta1) Answer: D. Solving the two boundary equations gives Gamma = (eta2 - eta1)/(eta2 + eta1). (Key point: Reflection coefficient) 4. The transmission coefficient and its relation to Gamma are: A. tau = 1 + Gamma = 2 eta2/(eta2 + eta1) B. tau = 1 - Gamma = 2 eta1/(eta2 + eta1) C. tau = Gamma^2 D. tau = Gamma/(1 + Gamma) Answer: A. From E_i0 + E_r0 = E_t0, tau = 1 + Gamma = 2 eta2/(eta2 + eta1). 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