Lecture 4 - Fall 2024 - EEE 314 - Oblique Incidence I PDF

Lecture 4 Oblique Incidence I EEE 314 WAVES PROPAGATION 3rd year- Fall 2024 Electronics and Electrical Communications Assoc. Prof. Ahmed Farghal Dept. of Electrical Engineering,...

Lecture 4 Oblique Incidence I EEE 314 WAVES PROPAGATION 3rd year- Fall 2024 Electronics and Electrical Communications Assoc. Prof. Ahmed Farghal Dept. of Electrical Engineering, Faculty of Engineering, Sohag University 22/10/2024  Introduction Lecture 4 A plane wave, obliquely incident on an interface between two materials, undergoes changes similar to those for normal incidence. Part of the wave is transmitted and part of it is reflected. In some cases, there is either only transmission or only Dr. Ahmed Farghal reflection. The behavior of the wave @ the interface depends on the polarization of the wave. Oblique incidence Wave arrives at an angle ◼ Snell’s Law and Critical angle Fall 2024 ◼ Two types: Parallel or Perpendicular ◼ Brewster angle 2  Introduction Uniform plane wave in general form Lecture 4 𝐄 𝐫, 𝑡 = 𝐸𝑜 cos(𝜔𝑡 − 𝐤 ⋅ 𝐫) 𝐚𝐸 General phasor form 𝐄 𝐫 = 𝐚𝐸 𝐸𝑜 𝑒 −𝑗𝐤⋅𝐫 position vector Dr. Ahmed Farghal 𝐫 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 wave number or propagation vector 𝐤 = 𝑘𝑥 𝐚𝑥 + 𝑘𝑦 𝐚𝑦 + 𝑘𝑧 𝐚𝑧 where 𝑘= 𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 = ω 𝜇𝜖 Fall 2024 For lossless unbounded media, k =  𝛃 = 𝛽𝑥 𝐚𝑥 + 𝛽𝑦 𝐚𝑦 + 𝛽𝑧 𝐚𝑧 3  Introduction 𝐚𝑛 Lecture 4 Plane of incidence: the plane described by the direction of i y propagation of the incident wave 𝐤 𝑖 Ei (i.e., the Poynting vector) and the ki z=0 normal to the surface @ the interface Dr. Ahmed Farghal 𝐻𝑖 𝐚𝑛. Parallel polarization (||): E is parallel to the plane of incidence. Perpendicular polarization (⊥): E is ⊥ to the plane of incidence. Arbitrary polarization can be treated by separating the E into its normal and parallel components as a combination of || & ⊥ polarizations. 𝐚𝐸 Fall 2024 𝐚𝑘 𝐚𝐻 4  Introduction Lecture 4 Medium 1 : 𝜀1 , 𝜇1 , 𝜎1 Medium 2: 𝜀2 , 𝜇2 , 𝜎2 𝐤𝑟 𝛽1 sin 𝜃𝑟 𝛽2 sin 𝜃𝑡 kt Dr. Ahmed Farghal 𝛽1 cos 𝜃𝑟 r 𝜃𝑡 𝛽2 cos 𝜃𝑡 i 𝐤𝑖 z 𝛽1 sin 𝜃𝑖 x 𝛽1 cos 𝜃𝑖 y z Fall 2024 𝛽1 = 𝜔 𝜖1 𝜇1 𝛽2 = 𝜔 𝜖2 𝜇2 z=0 5  Introduction Incident wave can be viewed as two components: Lecture 4 one propagating in the +ve 𝑥 direction with phase constant 𝑘𝑖𝑥 one propagating in the +ve 𝑧 direction with phase constant 𝑘𝑖𝑧 where 𝐤 𝑖 = 𝑘𝑖𝑥 𝐚𝑥 + 𝑘𝑖𝑧 𝐚𝑧 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 + 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝑘𝑖𝑥 = 𝛽1 sin 𝜃𝑖 , 𝑘𝑖𝑧 = 𝛽1 cos 𝜃𝑖 [radΤm] Dr. Ahmed Farghal The unit vector for the incident wave 𝐤𝑖 𝛽1 sin 𝜃𝑖 𝐚𝑥 + β1 cos 𝜃𝑖 𝐚𝑧 𝒂𝐤 𝑖 = = 𝐤𝑖 𝛽 2 sin2 𝜃𝑖 + cos 2 𝜃𝑖 1 ∴ 𝒂𝐤 𝑖 = 𝐚𝑥 sin 𝜃𝑖 + 𝐚𝑧 cos 𝜃𝑖 𝑘𝑖 = 𝛽1 Fall 2024 𝑘𝑖𝑥 = 𝛽1 sin 𝜃𝑖 i 𝑘𝑖𝑧 = 𝛽1 cos 𝜃𝑖 6  Introduction Transmitted wave can be viewed as two components: Lecture 4 one propagating in the +ve 𝑥 direction with phase constant 𝑘𝑡𝑥 one propagating in the +ve 𝑧 direction with phase constant 𝑘𝑡𝑧 where 𝐤 𝑡 = 𝑘𝑡𝑥 𝐚𝑥 + 𝑘𝑡𝑧 𝐚𝑧 = 𝛽2 sin 𝜃𝑡 𝐚𝑥 + 𝛽2 cos 𝜃𝑡 𝐚𝑧 𝑘𝑡𝑥 = 𝛽2 sin 𝜃𝑡 , 𝑘𝑡𝑧 = 𝛽2 cos 𝜃𝑡 [radΤm] Dr. Ahmed Farghal The unit vector for the incident wave 𝐤𝑡 𝛽2 sin 𝜃𝑡 𝐚𝑥 + 𝛽2 cos 𝜃𝑡 𝐚𝑧 𝒂𝐤 𝑡 = = 𝐤𝑡 𝛽 2 sin2 𝜃𝑡 + cos 2 𝜃𝑡 2 ∴ 𝒂𝐤 𝑡 = 𝐚𝑥 sin 𝜃𝑡 + 𝐚𝑧 cos 𝜃𝑡 𝑘𝑡 = 𝛽2 Fall 2024 𝛽2 sin 𝜃𝑡 t 𝑘𝑡𝑧 = 𝛽2 cos 𝜃𝑡 7  Introduction Similarly, reflected wave can be viewed as two components: Lecture 4 one propagating in the +ve 𝑥 direction with phase constant 𝛽1𝑥 one propagating in the −ve 𝑧 direction with phase constant 𝛽1𝑧 where 𝐤 = 𝑘 𝐚 + 𝑘 𝐚 = 𝛽 sin 𝜃 𝐚 − 𝛽 cos 𝜃 𝐚 𝑟 𝑟𝑥 𝑥 𝑟𝑧 𝑧 1 𝑟 𝑥 1 𝑟 𝑧 𝑘𝑟𝑥 = 𝛽1 sin 𝜃𝑟 , 𝑘𝑟𝑧 = 𝛽1 cos 𝜃𝑟 [radΤm] Dr. Ahmed Farghal The unit vector for the reflected wave 𝐤𝑟 𝛽1 sin 𝜃𝑟 𝐚𝑥 − 𝛽1 cos 𝜃𝑟 𝐚𝑧 𝒂𝐤 𝑟 = = 𝐤𝑟 𝛽12 sin2 𝜃𝑟 + cos 2 𝜃𝑟 ∴ 𝒂𝐤 𝑟 = 𝐚𝑥 sin 𝜃𝑟 − 𝐚𝑧 cos 𝜃𝑟 𝑘𝑟𝑧 = 𝛽1 cos 𝜃𝑟 Fall 2024 r 𝑘𝑟𝑥 = 𝛽1 sin 𝜃𝑟 𝑘𝑟 = 𝛽1 8  Introduction The trick in deriving the components is to first get the propagation Lecture 4 vector k for incident, reflected, and transmitted waves. 𝑘𝑖 = 𝛽1 𝑘𝑟𝑧 = 𝛽1 cos 𝜃𝑟 𝑘𝑡 = 𝛽2 𝜃𝑟 𝑘𝑖𝑥 = 𝛽1 sin 𝜃𝑖 𝛽2 sin 𝜃𝑡 i 𝑘𝑟𝑥 = 𝛽1 sin 𝜃𝑟 t 𝑘𝑟 = 𝛽1 Dr. Ahmed Farghal 𝑘𝑖𝑧 = 𝛽1 cos 𝜃𝑖 𝑘𝑡𝑧 = 𝛽2 cos 𝜃𝑡 𝐤 𝑖 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 + 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝒂𝐤 𝑖 = 𝐚𝑥 sin 𝜃𝑖 + 𝐚𝑧 cos 𝜃𝑖 x 𝐤 𝑟 = 𝛽1 sin 𝜃𝑟 𝐚𝑥 − 𝛽1 cos 𝜃𝑟 𝐚𝑧 y z 𝒂𝐤 𝑟 = 𝐚𝑥 sin 𝜃𝑟 − 𝐚𝑧 cos 𝜃𝑟 Fall 2024 𝐤 𝑡 = 𝛽2 sin 𝜃𝑡 𝐚𝑥 + 𝛽2 cos 𝜃𝑡 𝐚𝑧 𝒂𝐤 𝑡 = 𝐚𝑥 sin 𝜃𝑡 + 𝐚𝑧 cos 𝜃𝑡 9  Introduction Once k is known, we define 𝐄 such that 𝐚𝑘 ∙ 𝐄 = 0 Lecture 4 General phasor form 𝐄 𝐫 = 𝐚𝐸 𝐸𝑜 𝑒 −𝑗𝐤⋅𝐫 position vector 𝐫 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 Dr. Ahmed Farghal wave number or propagation vector 𝐤 = 𝑘𝑥 𝐚𝑥 + 𝑘𝑦 𝐚𝑦 + 𝑘𝑧 𝐚𝑧 𝑘= 𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 = ω 𝜇𝜖 = 𝛽 𝐤 𝑖 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑖 + 𝛽1 𝑧 cos 𝜃𝑖 𝐤 𝑟 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑟 − 𝛽1 𝑧 cos 𝜃𝑟 𝐤 𝑡 ∙ 𝐫 = 𝛽2 𝑥 sin 𝜃𝑡 + 𝛽2 𝑧 cos 𝜃𝑡 Then 𝐇 is obtained from Fall 2024 1 𝐇 𝑥, 𝑧 = 𝐚𝑘 × 𝐄(𝑥, 𝑧) 𝜂 10 Lecture 4 OBLIQUE INCIDENCE ON A CONDUCTING INTERFACE x Dr. Ahmed Farghal Reflected Incidence is from a dielectric with wave Perfect conductor 𝐤 Hr1 permittivity 𝜖1 on a perfect 𝑟 E conductor surface. r1 y. z Incident 𝐤𝑖 wave E Fall 2024 i1 Hi1 Medium 1 Medium 2 ( 1 = 0) ( 2 = ) z=0 11  Perpendicular Polarization E is ⊥ to the plane of incidence. 𝐚𝑘 ∙ 𝐄 = 0 Lecture 4 Incident E is in the +ve 𝑦 direction 𝐤𝑟 𝐄𝑖 𝐫 = 𝐚𝐸𝑖 𝐸𝑜 𝑒 −𝑗𝐤 𝑖 ⋅𝐫 𝐤 𝑖 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 + 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝐫 = 𝑥 𝐚𝑥 + 𝑦 𝐚𝑦 + 𝑧 𝐚𝑧 𝐤𝑖 Dr. Ahmed Farghal ∴ The phase of the incident wave is 𝐤 𝑖 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑖 + 𝛽1 𝑧 cos 𝜃𝑖 𝐄𝑖 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑖𝑜 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖+𝑧 cos 𝜃𝑖 ) VΤm Incident H has both -ve 𝑥 and +ve 𝑧 component. 1 𝐇𝑖 𝑥, 𝑧 = 𝒂𝐤 𝑖 × 𝐄𝑖 (𝑥, 𝑧) Fall 2024 𝒂𝐤𝑖 = 𝐚𝑥 sin 𝜃𝑖 + 𝐚𝑧 cos 𝜃𝑖 𝜂1 𝐸𝑖𝑜 𝐇𝑖 𝑥, 𝑧 = (−𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖 ) AΤm 𝜂1 12  Perpendicular Polarization Reflected E is in the +ve 𝑦 direction 𝐚𝑘 ∙ 𝐄 = 0 Lecture 4 𝐄𝑟 𝐫 = 𝐚𝐸𝑟 𝐸𝑜 𝑒 −𝑗𝐤𝑟⋅𝐫 𝐤𝑟 𝐤 𝑟 = 𝛽1 sin 𝜃𝑟 𝐚𝑥 − 𝛽1 cos 𝜃𝑟 𝐚𝑧 𝐫 = 𝑥 𝐚𝑥 + 𝑦 𝐚𝑦 + 𝑧 𝐚𝑧 𝐤𝑖 Dr. Ahmed Farghal ∴ The phase of the reflected wave is 𝐤 𝑟 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑟 − 𝛽1 𝑧 cos 𝜃𝑟 𝐄𝑟 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑟𝑜 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑟−𝑧 cos 𝜃𝑟 ) VΤm Reflected H has both +ve 𝑥 and +ve 𝑧 component. 1 𝐇𝑟 𝑥, 𝑧 = 𝐚𝑘𝐫 × 𝐄𝑟 (𝑥, 𝑧) 𝒂𝐤𝑟 = 𝐚𝑥 sin 𝜃𝑟 − 𝐚𝑧 cos 𝜃𝑟 𝜂1 Fall 2024 𝐸𝑟𝑜 𝐇𝑟 𝑥, 𝑧 = (𝐚𝑥 cos 𝜃𝑟 + 𝐚𝑧 sin 𝜃𝑟 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑟 −𝑧 cos 𝜃𝑟) AΤm 𝜂1 13  Perpendicular Polarization 𝐄𝑖 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑖0 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖 ) Lecture 4 𝐄𝑟 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑟0 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑟 −𝑧 cos 𝜃𝑟) ❖ In a perfect conductor, there is no transmitted wave; @ the interface (𝑧 = 0), the total tangential E must be zero: Dr. Ahmed Farghal 𝐄𝑖 𝑥, 0 + 𝐄𝑟 𝑥, 0 = 𝐚𝑦 𝐸𝑖𝑜 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 + 𝐸𝑟𝑜 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑟 = 0 For this to be satisfied, the following must hold: 𝐸𝑟𝑜 = −𝐸𝑖𝑜 and 𝜃𝑟 = 𝜃𝑖 𝐸 The reflection coefficient in this case is equal to Γ ≡ 𝑟𝑜 = −1. 𝐸𝑖𝑜 The relation, 𝜃𝑟 = 𝜃𝑖 is called Snell’s law of reflection. Fall 2024 Snell’s law of reflection » the angle of reflection equals the angle of incidence. 14  Perpendicular Polarization Lecture 4 𝐄𝑟 and 𝐇𝑟 can be written in terms of the incident field using Γ = −1 (𝐸𝑟𝑜 = −𝐸𝑖𝑜 ) and 𝜃𝑟 = 𝜃𝑖 as 𝐄𝑟 𝑥, 𝑧 = −𝐚𝑦 𝐸𝑖𝑜 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖) 𝐸𝑖𝑜 𝐇𝑟 𝑥, 𝑧 = − (𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) 𝜂1 Dr. Ahmed Farghal The total E is 𝐄1 𝑥, 𝑧 = 𝐄𝑖 𝑥, 𝑧 + 𝐄𝑟 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑖𝑜 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 − 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 = −𝐚𝑦 𝑗2𝐸𝑖𝑜 sin(𝛽1 𝑧 cos 𝜃𝑖 ) 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 Fall 2024 1 𝑗𝜃 sin 𝜃 = 𝑒 − 𝑒 −𝑗𝜃 2𝑗 15  Perpendicular Polarization 𝐸𝑖𝑜 (−𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖 ) Lecture 4 𝐇𝑖 𝑥, 𝑧 = 𝜂1 𝐸𝑖𝑜 𝐇𝑟 𝑥, 𝑧 = − (𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) 𝜂1 The total H is Dr. Ahmed Farghal 𝐇1 𝑥, 𝑧 = 𝐇𝑖 𝑥, 𝑧 + 𝐇𝑟 𝑥, 𝑧 𝐸𝑖𝑜 = ቂ−𝐚𝑥 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 + 𝑒 +𝑗𝛽1 𝑧 cos 𝜃𝑖 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝜂1 +𝐚𝑧 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 − 𝑒 +𝑗𝛽1 𝑧 cos 𝜃𝑖 sin 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 ൧ 𝐸𝑖𝑜 = −2 𝐚𝑥 cos 𝜃𝑖 cos(𝛽1 𝑧 cos 𝜃𝑖 ) + 𝐚𝑧 𝑗 sin 𝜃𝑖 sin(𝛽1 𝑧 cos 𝜃𝑖 ) 𝑒 −𝑗𝛽1𝑥 sin 𝜃𝑖 𝜂1 𝐸𝑖𝑜 Fall 2024 = −2 𝐚𝑥 𝐻1𝑥 + 𝐚𝑧 𝑗𝐻1𝑧 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝜂1 1 𝑗𝜃 1 𝑗𝜃 cos 𝜃 = 𝑒 + 𝑒 −𝑗𝜃 & sin 𝜃 = 𝑒 − 𝑒 −𝑗𝜃 2 2𝑗 16  Perpendicular Polarization The time-averaged Poynting vector: Lecture 4 1 ∗ 1 𝐒 = Re 𝐄1 × 𝐇1 = Re൛ −𝐚𝑦 2𝑗𝐸𝑖𝑜 sin(𝛽1 𝑧 cos 𝜃𝑖 ) 𝑒 −𝑗𝛽1𝑥 sin 𝜃𝑖 2 2 𝐸𝑖𝑜 × −2 𝐚𝑥 cos 𝜃𝑖 cos(𝛽1 𝑧 cos 𝜃𝑖 ) − 𝐚𝑧 𝑗 sin 𝜃𝑖 sin(𝛽1 𝑧 cos 𝜃𝑖 ) 𝑒 +𝑗𝛽1𝑥 sin 𝜃𝑖 ቋ 𝜂1 2 𝐸𝑖𝑜 = Re ൝(𝐚𝑦 × 𝐚𝑥 )2𝑗 sin(𝛽1 𝑧 cos 𝜃𝑖 ) cos(𝛽1 𝑧 cos 𝜃𝑖 ) cos 𝜃𝑖 𝜂1 Dr. Ahmed Farghal 2 𝐸 𝑖𝑜 + −𝐚𝑦 × 𝐚𝑧 2𝑗 2 sin(𝛽1 𝑧 cos 𝜃𝑖 ) sin(𝛽1 𝑧 cos 𝜃𝑖 ) sin 𝜃𝑖 ቋ 𝜂1 𝐚𝑦 × 𝐚𝑥 = −𝐚𝑧 , 𝐚𝑦 × 𝐚𝑧 = 𝐚𝑥 , 𝑗 2 = −1 1 1 sin(𝑥) cos(𝑦) = sin(𝑥 + 𝑦) + sin(𝑥 − 𝑦) 2 2 ∴ sin 𝛽1 𝑧 cos 𝜃𝑖 cos 𝛽1 𝑧 cos 𝜃𝑖 = 1Τ2 sin 2𝛽1 𝑧 cos 𝜃𝑖 2 2 𝐸𝑖𝑜 2𝐸𝑖𝑜 Fall 2024 𝐒 = Re −𝐚𝑧 𝑗 sin 2𝛽1 𝑧 cos 𝜃𝑖 cos 𝜃𝑖 + 𝐚𝑥 sin2 𝛽1 𝑧 cos 𝜃𝑖 sin 𝜃𝑖 𝜂1 𝜂1 standing wave power density is purely imaginary propagating wave 17  Perpendicular Polarization 2 2𝐸𝑖1 Lecture 4 𝐒 = 𝐚𝑥 sin2 𝛽1 𝑧 cos 𝜃𝑖 sin 𝜃𝑖 𝜂1 Note: (1) For 0 < 𝜃𝑖 < 𝜋/2, there is a propagating term & standing wave term. standing wave becomes smaller as 𝜃𝑖 ↓, whereas the propagating term becomes smaller as 𝜃𝑖 ↑. Dr. Ahmed Farghal (2) For normal incidence (𝜃𝑖 = 0), the propagating term is zero and the wave is a purely standing wave. Time-averaged power density is zero. (3) The wave propagates ∥the surface of the conductor (in this case, in +ve 𝒙 direction) for any angle 0 < 𝜃𝑖 < 𝜋/2. The conducting surface has the net effect of guiding the waves ∥ its surface. (4) The amplitude of the propagating wave depends on the 𝒛-coordinate. amplitude is not constant on the plane ⊥ direction of propagation and Fall 2024 hence this is not a uniform plane wave. Note in particular that whereas the incident and reflected waves are uniform plane waves, their sum is not. 18  Parallel Polarization The 𝐄 now lies in the incidence plane. Lecture 4 Incident 𝐄𝑖 now has two components; 𝐚𝑘 ∙ 𝐄 = 0 one is in +ve 𝑥 direction, second in −ve 𝑧 direction, Incident 𝐇 is ⊥ to the plane of incidence (y direction). 𝐤 𝑖 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 + 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝐫 = 𝑥 𝐚𝑥 + 𝑦 𝐚𝑦 + 𝑧 𝐚𝑧 Dr. Ahmed Farghal x Reflected 𝐤 𝑖 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑖 + 𝛽1 𝑧 cos 𝜃𝑖 wave 𝐤𝑟 Er Perfect conductor 𝐄𝑖 𝑥, 𝑧 Hr = 𝐸𝑖𝑜 (𝐚𝑥 cos 𝜃𝑖 − 𝒂𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖 ) r y. z 𝐇𝑖 𝑥, 𝑧 = 1 𝒂𝐤𝑖 × 𝐄𝑖 (𝑥, 𝑧) Ei i 𝜂1 𝐤𝑖 𝒂𝐤 𝑖 = 𝐚𝑥 sin 𝜃𝑖 + 𝐚𝑧 cos 𝜃𝑖 Fall 2024 Incident wave Hi cos 2 𝜃 + sin2 𝜃 = 1 Medium 1 Medium 2 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) ( 1 = 0) ( 2 = ) 𝐇𝑖 𝑥, 𝑧 = 𝒂𝑦 𝑒 1 𝑖 𝑖 z=0 𝜂1 19  Parallel Polarization The reflected fields (inside medium 1) Lecture 4 𝐤 𝑟 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 − 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝐫 = 𝑥 𝐚𝑥 + 𝑦 𝐚𝑦 + 𝑧 𝐚𝑧 𝐤 𝑟 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑟 − 𝛽1 𝑧 cos 𝜃𝑟 𝐄𝑟 𝑥, 𝑧 = 𝐸𝑟𝑜 (𝐚𝑥 cos 𝜃𝑟 + 𝐚𝑧 sin 𝜃𝑟 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑟−𝑧 cos 𝜃𝑟 ) Dr. Ahmed Farghal 1 𝐇𝑟 𝑥, 𝑧 = 𝒂𝐤 𝑟 × 𝐄𝑟 (𝑥, 𝑧) 𝒂𝐤 𝑟 = 𝐚𝑥 sin 𝜃𝑟 − 𝐚𝑧 cos 𝜃𝑟 𝜂1 𝐸𝑟𝑜 −𝑗𝛽 (𝑥 sin 𝜃 −𝑧 cos 𝜃 ) 𝐇𝑟 𝑥, 𝑧 = −𝐚𝑦 𝑒 1 𝑟 𝑟 𝜃𝑟 =? 𝐸𝑟𝑜 =? 𝜂1 B. C. (@ 𝑧 = 0), the tangential components (i.e., 𝑥 components) of the electric field intensity 𝐄 𝑥, 0 = 𝐄𝑖 (𝑥, 0) + 𝐄𝑟 (𝑥, 0) must be zero. Fall 2024 𝐸𝑖𝑜 𝑎𝑥 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 + 𝐸𝑟𝑜 𝑎𝑥 cos 𝜃𝑟 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑟 =0 ⟶ 𝐸𝑟1 = −𝐸𝑖1 , 𝜃𝑟 = 𝜃𝑖 20  Parallel Polarization The total fields in material (1): 𝐸𝑟1 = −𝐸𝑖1 and 𝜃𝑟 = 𝜃𝑖 Lecture 4 𝐄1 𝑥, 𝑧 = 𝐄𝑖 𝑥, 𝑧 + 𝐄𝑟 𝑥, 𝑧 𝐄𝑖 𝑥, 𝑧 = 𝐸𝑖𝑜 (𝐚𝑥 cos 𝜃𝑖 − 𝒂𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖 ) 𝐄𝑟 𝑥, 𝑧 = −𝐸𝑖𝑜 (𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) Dr. Ahmed Farghal 𝐄1 𝑥, 𝑧 = −2𝐸𝑖𝑜 ൣ𝐚𝑥 𝑗 cos 𝜃𝑖 sin 𝛽1 𝑧 cos 𝜃𝑖 +𝐚𝑧 sin 𝜃𝑖 cos(𝛽1 𝑧 cos 𝜃𝑖 )]𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝐇1 𝑥, 𝑧 = 𝐇𝑖 𝑥, 𝑧 + 𝐇𝑟 𝑥, 𝑧 1 𝑗𝜃 cos 𝜃 = 𝑒 + 𝑒 −𝑗𝜃 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 2 𝐇𝑖 𝑥, 𝑧 = 𝒂𝑦 𝑒 1 𝑖 𝑖 1 𝑗𝜃 𝜂1 sin 𝜃 = 𝑒 − 𝑒 −𝑗𝜃 2𝑗 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 −𝑧 cos 𝜃 ) Fall 2024 𝐇𝑟 𝑥, 𝑧 = +𝐚𝑦 𝑒 1 𝑖 𝑖 𝜂1 𝐸𝑖𝑜 𝐇1 𝑥, 𝑧 = 𝐚𝑦 2 cos(𝛽1 𝑧 cos 𝜃𝑖 ) 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝜂1 21 Lecture 4 Dr. Ahmed Farghal OBLIQUE INCIDENCE ON DIELECTRIC INTERFACES Fall 2024 22  Oblique Incidence on Dielectric Interfaces An incident wave gives rise to a reflected wave, both propagating Lecture 4 in the same material. Unlike the perfect conductors, there is also a wave propagating in material (2). Since the incident wave is @ an angle 𝜃𝑖 to the normal, we expect Dr. Ahmed Farghal the wave in material (2) to also propagate at an angle to the normal (i.e., refraction angle 𝜽𝒕 ). Reflection angle 𝜃𝑟 depends on incident angle 𝜃𝑖. 𝜃𝑟 = 𝜃𝑖 (Snell’s law of reflection) To be able to describe all wave properties in terms of the incident wave alone we must also define a relation between refraction angle 𝜽𝒕 Fall 2024 and incidence angle 𝜽𝒊. Also, we must expect that Γ and 𝜏 should be different than those obtained for normal incidence. 23  Perpendicular Polarization The E is ⊥ to the plane of incidence in 𝑦 direction. Lecture 4 𝐚𝑘 ∙ 𝐄 = 0 Dr. Ahmed Farghal Fall 2024 24  Perpendicular Polarization 𝐄𝑖 and 𝐇𝑖 : Lecture 4 𝐄𝑖 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑖𝑜 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖+𝑧 cos 𝜃𝑖 ) 𝐸𝑖𝑜 𝐇𝑖 𝑥, 𝑧 = (−𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖 ) 𝜂1 Dr. Ahmed Farghal 𝐄𝑟 and 𝐇𝑟 : 𝐄𝑟 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑟𝑜 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) 𝐸𝑟𝑜 𝐇𝑟 𝑥, 𝑧 = (𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) 𝜂1 𝐄𝑡 and 𝐇𝑡 have the same form as the incident wave but with different amplitudes and propagate @ a different angle: 𝐄𝑡 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑡𝑜 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) Fall 2024 𝐸𝑡𝑜 𝐇𝑡 𝑥, 𝑧 = (−𝐚𝑥 cos 𝜃𝑡 + 𝐚𝑧 sin 𝜃𝑡 ) 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝜂2 25  Perpendicular Polarization To determine Γ and 𝜏, the tangential components of E and H on Lecture 4 both sides of the interface (i.e., @ 𝑧 = 0) are equated. Which components are tangent to the interface between two surfaces? 𝑦 and 𝑥 components. Dr. Ahmed Farghal Taking only the tangential components (y component for E and 𝑥 component for H) @ 𝑧 = 0, we have 𝐸𝑖𝑜 + 𝐸𝑟𝑜 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 = 𝐸𝑡𝑜 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 𝐸𝑖𝑜 𝐸𝑟𝑜 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝐸𝑡𝑜 − + cos 𝜃𝑖 𝑒 =− cos 𝜃𝑡 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 𝜂1 𝜂1 𝜂2 There are three relations that must be satisfied. 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 = 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 or 𝛽1 sin 𝜃𝑖 = 𝛽2 sin 𝜃𝑡 01 Fall 2024 𝐸𝑖𝑜 𝐸𝑟𝑜 𝐸𝑡𝑜 𝐸𝑖𝑜 + 𝐸𝑟𝑜 = 𝐸𝑡𝑜 02 − + cos 𝜃𝑖 = − cos 𝜃𝑡 03 𝜂1 𝜂1 𝜂2 26  Perpendicular Polarization 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 = 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 or 𝛽1 sin 𝜃𝑖 = 𝛽2 sin 𝜃𝑡 01 Lecture 4 ∵ 𝛽1 = 𝜔 𝜖1 𝜇1 and 𝛽2 = 𝜔 𝜖2 𝜇2 , we get 𝜖1 𝜇1 𝜔 𝜖1 𝜇1 sin 𝜃𝑖 = 𝜔 𝜖2 𝜇2 sin 𝜃𝑡 ⟹ sin 𝜃𝑡 = sin 𝜃𝑖 𝜖2 𝜇 2 Dr. Ahmed Farghal This relation between 𝜃𝑖 and 𝜃𝑡 is Snell’s law of refraction. ∵ 𝜖1 = 𝜖0 𝜖𝑟1 , 𝜖2 = 𝜖0 𝜖𝑟2 , 𝜇1 = 𝜇0 𝜇𝑟1 , 𝜇2 = 𝜇0 𝜇𝑟2 , 𝑣𝑝1 = 1Τ 𝜖1 𝜇1 and 𝑣𝑝2 = 1Τ 𝜖2 𝜇2 , we can also write Snell’s law of refraction as sin 𝜃𝑡 𝜖1 𝜇1 𝑛1 𝑣𝑝2 = = = sin 𝜃𝑖 𝜖2 𝜇2 𝑛2 𝑣𝑝1 Fall 2024 𝑐 𝑛1 = = 𝜖𝑟1 𝜇𝑟1 is the (optical) index of refraction in medium (1) 𝑣𝑝1 𝑐 𝑛2 = = 𝜖𝑟2 𝜇𝑟2 is the (optical) index of refraction in medium (2). 𝑣𝑝2 27  Perpendicular Polarization Lecture 4 𝐸𝑖𝑜 𝐸𝑟𝑜 𝐸𝑡𝑜 𝐸𝑖𝑜 + 𝐸𝑟𝑜 = 𝐸𝑡2 02 − + cos 𝜃𝑖 = − cos 𝜃𝑡 03 𝜂1 𝜂1 𝜂2 The solution of Eqs. (2) and (3) for 𝐸𝑟𝑜 and 𝐸𝑡𝑜 gives 𝜂2 cos 𝜃𝑖 − 𝜂1 cos 𝜃𝑡 2𝜂2 cos 𝜃𝑖 Dr. Ahmed Farghal 𝐸𝑟𝑜 = 𝐸𝑖𝑜 , 𝐸𝑡𝑜 = 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 Γ and 𝜏 for perpendicular polarization:- 𝐸𝑟𝑜 𝜂2 cos 𝜃𝑖 − 𝜂1 cos 𝜃𝑡 Γ⊥ = = 𝐸𝑟𝑜 = Γ⊥ 𝐸𝑖𝑜 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 𝐸𝑡𝑜 2𝜂2 cos 𝜃𝑖 𝜏⊥ = = 𝐸𝑡𝑜 = 𝜏⊥ 𝐸𝑖𝑜 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 Fall 2024 The notation ⊥ indicates these are the reflection and transmission coefficients for perpendicular polarization, because, as we shall see, the coefficients for parallel polarization differ. 28  Perpendicular Polarization In medium (1), the total incident and reflected waves: Lecture 4 𝐸𝑟𝑜 = Γ⊥ 𝐸𝑖𝑜 𝐄1 𝑥, 𝑧 = 𝐚𝑦 𝐸𝑖𝑜 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 + Γ⊥ 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝐸𝑖𝑜 cos 𝜃𝑖 𝐇1 𝑥, 𝑧 = 𝐚𝑥 Γ⊥ 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 − 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 Dr. Ahmed Farghal 𝜂1 𝐸𝑖𝑜 sin 𝜃𝑖 −𝑗𝛽 𝑧 cos 𝜃 +𝐚𝑧 𝑒 1 𝑖 + Γ 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 ⊥ 𝜂1 In medium (2), where the only wave is the transmitted wave. Using the transmission coefficient, we can write 𝐸𝑡𝑜 = 𝜏⊥ 𝐸𝑖𝑜 𝐄𝑡 𝑥, 𝑧 = 𝐚𝑦 𝜏⊥ 𝐸𝑖𝑜 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) Fall 2024 𝜏⊥ 𝐸𝑖𝑜 𝐇𝑡 𝑥, 𝑧 = (−𝐚𝑥 cos 𝜃𝑡 + 𝒂𝑧 sin 𝜃𝑡 ) 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝜂2 29  Perpendicular Polarization Lecture 4 𝜂2 cos 𝜃𝑖 − 𝜂1 cos 𝜃𝑡 2𝜂2 cos 𝜃𝑖 Γ⊥ = 𝜏⊥ = 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 It is easily shown that 1 + Γ⊥ = 𝜏⊥ Dr. Ahmed Farghal When 𝜃𝑖 = 𝜃𝑡 = 0, the equations reduce to 𝜂2 − 𝜂1 2𝜂2 Γ⊥ = Γ = & 𝜏⊥ = 𝜏 = as expected, 𝜂2 + 𝜂1 𝜂2 + 𝜂1 Since 𝜃𝑖 & 𝜃𝑡 are related according to Snell’s law sin 𝜃𝑡 𝑛1 𝑣𝑝2 = = sin 𝜃𝑖 𝑛2 𝑣𝑝1 Γ⊥ and 𝜏⊥ can be written in terms of 𝜃𝑖 by substituting Fall 2024 2 cos 𝜃𝑡 = 1 − sin2 𝜃𝑡 = 1 − 𝑣𝑝2 Τ𝑣𝑝1 sin2 𝜃𝑖 30  Example A perpendicularly polarized plane wave impinges on a very thick Lecture 4 sheet of plastic @ 𝜃𝑖 = 30° from free space. The relative permittivity of the plastic is 𝜀𝑟 = 4 and its relative permeability is 𝜇𝑟 = 1. If the amplitude of the incident electric field intensity is 𝐸𝑖𝑜 = 100 V/m, calculate the time-averaged power density transmitted into the plastic. Dr. Ahmed Farghal Solution: 𝜇1 = 𝜇2 = 𝜇0 𝜀1 = 𝜀𝑜 , 𝜀2 = 𝜀𝑟2 𝜀𝑜 𝜃𝑟 and 𝜃𝑡 are evaluated from the following relations: 𝜀1 sin 𝜃𝑖 0.5 𝜃𝑟 = 𝜃𝑖 , sin 𝜃𝑡 = sin 𝜃𝑖 = = = 0.25 Fall 2024 𝜀2 𝜀𝑟2 4 𝜇 𝜂0 𝜂0 𝜂1 = 𝜂𝑜 𝜂2 = = = 𝜖 𝜀𝑟2 2 31  Example 𝜂1 = 𝜂𝑜 = 2𝜂2 2𝜂2 cos 𝜃𝑖 2 cos 𝜃𝑖 Lecture 4 𝜏⊥ = = 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 cos 𝜃𝑖 + 2 cos 𝜃𝑡 sin 𝜃𝑖 sin 𝜃𝑡 = ∵ cos 𝜃𝑡 = 1 − sin2 𝜃𝑡 = 1 − sin2 𝜃𝑖 Τ𝜀𝑟2 , we get 𝜀𝑟2 2 cos 𝜃𝑖 1.732 𝜏⊥ = = = 0.618 cos 𝜃𝑖 + 2 1 − sin2 𝜃𝑖 Τ𝜀𝑟2 0.866 + 4 − 0.25 Dr. Ahmed Farghal ∴ 𝐄𝑡 and 𝐇𝑡 are 𝐄𝑡 𝑥, 𝑧 = 𝒂𝑦 𝜏⊥ 𝐸𝑖𝑜 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝜏⊥ 𝐸𝑖𝑜 𝐇𝑡 𝑥, 𝑧 = (−𝐚𝑥 cos 𝜃𝑡 + 𝐚𝑧 sin 𝜃𝑡 ) 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝜂2 The time-averaged Poynting vector: 1 ∗ 𝜏⊥ 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝐒𝑡 = Re 𝐄𝑡 𝑥, 𝑧 × 𝐇𝑡 𝑥, 𝑧 = 𝐚𝑦 𝑒 2 𝑡 𝑡 2 2 Fall 2024 𝜏⊥ 𝐸𝑖𝑜 × (−𝐚𝑥 cos 𝜃𝑡 + 𝐚𝑧 sin 𝜃𝑡 ) 𝑒 +𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝜂2 32  Example Lecture 4 2 𝜏⊥2 𝐸𝑖𝑜 𝐒𝑡 = 𝐚𝑦 × −𝐚𝑥 cos 𝜃𝑡 + 𝐚𝑧 sin 𝜃𝑡 2𝜂2 2 𝜏⊥2 𝐸𝑖𝑜 = 𝐚𝑧 cos 𝜃𝑡 + 𝐚𝑥 sin 𝜃𝑡 2𝜂2 Dr. Ahmed Farghal ❑ With sin 𝜃𝑡 = 0.25, cos 𝜃𝑡 = 1 − 0.25 2 = 0.968 , and 𝜂2 = 𝜂𝑜 Τ2 = 377Τ2 Ω, the time-averaged power density is 0.6182 × 1002 𝐒𝑡 = 𝐚𝑧 0.968 + 𝐚𝑥 0.25 377 W Fall 2024 = 𝐚𝑥 2.533 − 𝐚𝑧 9.8 2 m 33  Have a nice day. Dr. Ahmed Farghal 34