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?