Reflection and Refraction of Electromagnetic...
Transcript of Reflection and Refraction of Electromagnetic...
Reflection and Refraction of Electromagnetic Waves at a Plane
Interface Between Dielectrics Lu Wang
University of South Dakota
Nov 17,2014
Goals
Two properties of the electromagnetic wave at a plane interface
Incident Wave, Reflected Wave and Refracted Wave Boundary conditions Polarization perpendicular/parallel to the plane of
incidence
Two properties • Kinematic properties (a) Angle of reflection equals angle of incidence (b) Snell’s law: (sini)/(sinr)=n’/n, where i, r are the angles of incidence and refraction, where n, n’ are the corresponding indices of refraction. Do not depend on the detailed nature of waves or boundary conditions
• Dynamic properties: (a) Intensities of reflected and refracted radiation. (b) Phase changes and polarization. Depend on the specific nature of electromagnetic fields and boundary conditions
Electromagnetic Fields
Z>0
Z<0
Fig. 1
(7.30)
(7.31)
(7.32)
According to Eq.(7.18)
Incident 𝐄 = 𝐄𝟎𝑒−𝑖𝐤∙𝐱−𝑖𝜔𝑡
𝐁 = 𝜇𝜖𝐤×𝐄
𝑘
Refracted 𝐄′ = 𝑬𝟎’ 𝑒−𝑖𝐤′∙𝐱−𝑖𝜔𝑡
𝐁′ = 𝜇′𝜖′ 𝐤′×𝐄′
𝒌′
Reflected 𝐄" = 𝑬𝟎” 𝑒−𝑖𝐤"∙𝐱−𝑖𝜔𝑡
𝐁" = 𝜇𝜖 𝐤"×𝐄"
𝒌"
Kinematic Conditions
All fields vary in time as 𝑒−𝑖𝜔𝑡
The dependence of a plane wave on position is 𝑒𝑖𝐤∙𝐱
𝐤 ∙ 𝐱 𝑧=0 = 𝐤′ ∙ 𝐱 𝑧=0 = 𝐤" ∙ 𝐱 𝑧=0
Since the second kinematic properties, all waves must have wave vectors whose components lying in the plane of the interface are identical
k 𝑠𝑖𝑛 𝑖 = 𝑘′ 𝑠𝑖𝑛 𝑟 = 𝑘" 𝑠𝑖𝑛 𝑟′
Where 𝑖, 𝛾′𝑎𝑛𝑑 γ are called the angle of incidence, the angle of reflection, and the angle of refraction, correspondingly
(7.34)
(7.35)
Any reflected wave is a solution of the same wave equation as the incident wave.
𝑘 = 𝑘"
However, any transmitted wave
𝑘′ = 𝑘
According to Eq.(7.4), the wave numbers have the magnitudes 𝐤 = 𝐤" = 𝑘 = 𝜔 𝜇𝜖
𝐤′ = k′ = 𝜔 𝜇′𝜖′
With Eq. (7.35), we find 𝑖 = 𝑟’
The angle of incidence equals the angle of reflection.
Snell’s law: sin 𝑖
sin 𝑟=𝑘′
𝑘=
𝜇′𝜖′
𝜇𝜖=𝑛′
𝑛
(7.33)
(7.36)
Conditions from Maxwell’s Equations
𝛁 ∙ 𝐃 = 𝜌𝑓
𝐃 ∙ 𝑑𝒂 = 4𝜋𝜎
𝛁 ∙ 𝐁 = 0 𝐁 ∙ 𝑑𝒂 = 0 𝐁𝟐 − 𝐁𝟏 ∙ 𝐧 = 𝟎
𝛁 × 𝐇 = 𝐉 +𝜕𝐃
𝜕𝑡 𝐇 ∙ 𝑑𝐥 = 𝐉 +
𝜕𝐃
𝜕𝑡∙ 𝐧′𝑑𝑎
𝛁 × 𝐄 = −𝜕𝐁
𝜕𝑡 𝐄 ∙ 𝑑𝐥 = −
𝜕𝐁
𝜕𝑡∙ 𝐧′𝑑𝑎 𝐄𝟐 − 𝐄𝟏 × 𝐧 = 𝟎
𝐧 × 𝐇𝟐 − 𝐇𝟏 = 𝐊
(𝐃𝟐−𝐃𝟏) ∙ 𝐧 = 4𝜋𝜎
For uncharged insulators, , the idealized surface charge 𝜎 = 0 and microscopic current density 𝐾 = 0, so we have four continuous conditions: (𝐃𝟐−𝐃𝟏) ∙ 𝐧 = 0
𝐁𝟐 − 𝐁𝟏 ∙ 𝐧 = 𝟎
𝐧 × 𝐇𝟐 − 𝐇𝟏 = 𝟎
𝐄𝟐 − 𝐄𝟏 × 𝐧 = 𝟎
𝛜 𝐄𝟎 + 𝐄𝟎" − 𝛜′𝐄𝟎′ ∙ 𝐧 = 𝟎
𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" − 𝐤′ × 𝐄𝟎′ ∙ 𝐧 = 𝟎
𝐄𝟎 + 𝐄𝟎" − 𝐄𝟎′ × 𝐧 = 𝟎
𝟏
μ𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" −
𝟏
μ′𝐤′ × 𝐄𝟎′ × 𝐧 = 𝟎
1.𝐷𝑛 continuous
2. 𝐵𝑛 continuous
3. 𝐸𝑡 continuous
4.𝐻𝑡 continuous
(7.37)
Polarization of E perpendicular to the plane of incidence
Incident plane defined by k and n
𝛜 𝐄𝟎 + 𝐄𝟎" − 𝛜′𝐄𝟎′ ∙ 𝐧 = 𝟎 yields nothing
𝛜 𝐄𝟎 + 𝐄𝟎" − 𝛜′𝐄𝟎′ ∙ 𝐧 = 𝟎
𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" − 𝐤′ × 𝐄𝟎′ ∙ 𝐧 = 𝟎
𝐄𝟎 + 𝐄𝟎" − 𝐄𝟎′ × 𝐧 = 𝟎
𝟏
μ𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" −
𝟏
μ′𝐤′ × 𝐄𝟎′ × 𝐧 = 𝟎
𝛜 ∙ 𝐧 = 𝟎 and 𝛜′ ∙ 𝐧 = 𝟎
Boundary conditions:
𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" − 𝐤′ × 𝐄𝟎′ ∙ 𝐧 = 𝟎 (2)
Since Eq.(7.33) 𝐤 = 𝐤" = 𝑘 = 𝜔 𝜇𝜖
𝐤′ = k′ = 𝜔 𝜇′𝜖′
𝐸0𝑘 cos𝜋
2− 𝑖 + 𝐸0"𝑘 cos
𝜋
2− 𝑖 − 𝐸0
′𝑘′ cos𝜋
2− 𝑟 = 0
Due to Snell’s law : 𝑘 sin 𝑖 = 𝑘′ sin 𝑟 𝐸0+𝐸0”−𝐸0’= 0
𝐸0𝑘 sin 𝑖 +𝐸0"𝑘 sin 𝑖 −𝐸0′𝑘′ sin 𝑟 = 0
Due to 𝛜 × 𝐧 = 𝟏 and 𝛜′ × 𝐧 = 𝟏,
𝐄𝟎 + 𝐄𝟎" − 𝐄𝟎′ × 𝐧 = 𝟎 (3) 𝐸0 + 𝐸0”− 𝐸0’ = 0
Also since Eq.(7.33) 𝐤 = 𝐤" = 𝑘 = 𝜔 𝜇𝜖
𝐤′ = k′ = 𝜔 𝜇′𝜖′ 𝟏
𝛍𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" −
𝟏
𝛍′𝐤′ × 𝐄𝟎′ × 𝐧 = 𝟎 (𝟒)
1
𝜇𝜔 𝜇𝜖𝐸0 𝑠𝑖𝑛
𝜋
2− 𝑖 −𝜔 𝜇𝜖𝐸0" 𝑠𝑖𝑛
𝜋
2− 𝑖 −
1
𝜇′𝜔 𝜇′𝜖′𝐸0’𝑠𝑖𝑛
𝜋
2− 𝑟 = 0
𝜖
𝜇𝐸0 − 𝐸0" cos 𝑖 −
𝜖′
𝜇′𝐸0’cos 𝑟 = 0
𝐸0 + 𝐸0”− 𝐸0’ = 0
𝜖
𝜇𝐸0 − 𝐸0" cos 𝑖 −
𝜖′
𝜇′𝐸0’cos 𝑟 = 0
(7.38)
E perpendicular to plane of incidence:
(7.39)
Polarization of E parallel to the plane of incidence
𝛜 𝐄𝟎 + 𝐄𝟎" − 𝛜′𝐄𝟎′ ∙ 𝐧 = 𝟎 (1) D
𝐄𝟎 + 𝐄𝟎" − 𝐄𝟎′ × 𝐧 = 𝟎 (3) E
𝟏
𝛍𝐤 × 𝐄𝟎 + 𝐤" × 𝐄𝟎" −
𝟏
𝝁′𝐤′ × 𝐄𝟎′ × 𝐧 = 𝟎 (4) H
Boundary conditions:
𝛜 ∙ 𝐧 = cos𝜋
2− 𝑖 and 𝛜′ ∙ 𝐧 = cos
𝜋
2− 𝛾
(1) 𝐸0 + 𝐸0" 𝜖 cos𝜋
2− 𝑖 − 𝐸0’𝜖’ cos
𝜋
2− 𝛾 = 0
𝐸0 + 𝐸0" 𝜖 sin 𝑖 − 𝐸0’𝜖’ sin 𝛾 = 0
Snell’s law: sin 𝑖
sin 𝑟=
𝜇′𝜖′
𝜇𝜖
Divided by sin 𝜸 on both sides: 𝐸0 + 𝐸0" 𝜖𝜇′𝜖′
𝜇𝜖− 𝐸0’𝜖’ = 0
Divided by 𝜇′𝜖′ on both sides: 𝜖
𝜇𝐸0 + 𝐸0" −
𝜖′
𝜇′𝐸0’= 0
Also since Eq.(7.33) 𝐤 = 𝐤" = 𝑘 = 𝜔 𝜇𝜖
𝐤′ = k′ = 𝜔 𝜇′𝜖′
(4) 1
𝜇−𝜔 𝜇𝜖𝐸0 sin
𝜋
2− 𝜔 𝜇𝜖𝐸0" sin
𝜋
2+1
𝜇′𝜔 𝜇′𝜖′𝐸0’𝑠𝑖𝑛
𝜋
2= 0
𝜖
𝜇𝐸0 + 𝐸0" −
𝜖′
𝜇′𝐸0’= 0
Due to 𝛜 × 𝐧 = sin𝜋
2− 𝑖 and 𝛜′ × 𝐧 = sin
𝜋
2− 𝛾
(3) cos 𝑖 𝐸0 − 𝐸0" − cos 𝛾𝐸0’= 0
cos 𝑖 𝐸0 − 𝐸0" − cos 𝛾𝐸0’= 0
𝜖
𝜇𝐸0 + 𝐸0" −
𝜖′
𝜇′𝐸0’= 0
(7.40)
E parallel to plane of incidence
(7.41)
Equation (7.39) and Eq.(7.41) are known as Fresnel’s equations
Special case
𝑖 = 0, then γ′ = 0
Fresnel’s equations reduce to
Normal incidence
(7.42)
If 𝑛′ > 𝑛, 𝐸0” is opposite in sign to the incident one 𝐸0
which means the elctric field of the reflected wave is phase shifted by π radius relative to that of the incident one.
𝜇′ = 𝜇
Thank you
Questions?