Plane Electromagnetic Waves and Wave Propagation 1. Plane Monochromatic Waves in Nonconducting Media In the absence of free charge and current densities the Maxwell equations are π⋅π=0 π⋅π=0 ππ ππ‘ ππ π×π= − ππ‘ π×π= For uniform isotropic linear media we have π = ππ and π = ππ, where π and π are in general complex functions of frequency π. Then we obtain π 2π ππ‘ 2 π 2π π × π × π = −ππ 2 ππ‘ π × π × π = −ππ Since π × π × π = π(π ⋅ π) − π π π = −π π π and, similarly, π × π × π = −π π π, π 2π ππ‘ 2 π 2π π 2 π = ππ 2 ππ‘ π 2 π = ππ Monochromatic waves may be described as waves that are characterized by a single frequency. Assuming the fields with harmonic time dependence π −πππ‘ , so that π(π±, π‘) = π(π±)π −πππ‘ and π(π±, π‘) = π(π±)π −πππ‘ we get the Helmholtz wave equations π 2 π + πππ2 π = 0 π 2 π + πππ 2 π = 0 Electromagnetic plane wave of frequency π and wave vector π€ Suppose an electromagnetic plane wave with direction of propagation π§ to be constructed, π§ where is a unit vector. Then the variable π§ in the exponent must be replaced by π§ ⋅ π±, the projection of π± in the π§ direction. Thus an electromagnetic plane wave with direction of propagation π§ is described by π(π±, π‘) = ππ ππ€⋅π±−πππ‘ = ππ πππ§⋅π±−πππ‘ π(π±, π‘) = ππ ππ€⋅π±−πππ‘ = ππ πππ§⋅π±−πππ‘ where π and π are complex constant vector amplitudes of the plane wave. π and π satisfy the wave equations, therefore the dispersion relation is given as π 2 π π 2 = πππ2 = (π ) → π=π π π 1 Thus the Maxwell equations become π⋅π=0 π⋅π=0 ππ ππ‘ ππ π×π=− ππ‘ π×π = → π€ ⋅ π = 0 π€ × π = −ππππ π€⋅π=0 π€ × π = ππ where π€ = ππ§. The direction π§ and frequency π are completely arbitrary. The divergence equations demand that π§ ⋅ π = 0 and π§ ⋅ π = 0 This means that π and π are both perpendicular to the direction of propagation π§. The magnitude of π€ is determined by the refractive index of the material π = π(π) π π Then π is completely determined in magnitude and direction π = √ππ π§ × π = π π§×π π Note that in vacuum (π = 1), πΈ = ππ΅ in SI units. The phase velocity of the wave is π£ = π/π. Energy density and flux The time averaged energy density is π’= 1 1 1 (π ⋅ π∗ + π ⋅ π ∗ ) = (ππ ⋅ π∗ + π ⋅ π ∗ ) 4 4 π This gives π’= π 2 |π| 2 The time averaged energy flux is given by the real part of the complex Poyinting vector π= 1 (π × π ∗ ) 2 Thus the energy flow is 1 π 1 π = √ |π|2 π§ = π|π|2 ⋅ π£π§ = π’π― 2 π 2 where the speed of light in the uniform medium is π£= 1 π = √ππ π 2 2. Polarization There is more to be said about the complex vector amplitudes π and π. We introduce a righthanded set of orthogonal unit vectors (ππ , ππ , π§), as shown in the figure below, where we take π§ to be the propagation direction of the plane wave. In general, the electric field amplitude π can be written as π = ππ πΈ1 + ππ πΈ2 where the amplitudes πΈ1 and πΈ2 are arbitrary complex numbers. The two plane waves ππ = ππ πΈ1 π ππ€⋅π±−πππ‘ ππ = ππ πΈ2 π ππ€⋅π±−πππ‘ with π ππ = ππ π πΈ1 π ππ€⋅π±−πππ‘ π ππ = −ππ π πΈ2 π ππ€⋅π±−πππ‘ (if the index of refraction π is real, π and π have the same phase) are said to be linearly polarized with polarization vectors ππ and ππ . Thus the most general homogeneous plane wave propagating in the direction π€ = ππ§ is expressed as the superposition of two independent plane waves of linear polarization: π(π±, π‘) = ππ ππ€⋅π±−πππ‘ = (ππ πΈ1 + ππ πΈ2 )π ππ€⋅π±−πππ‘ 3 3. Reflection and Refraction at a Plane Interface between Dielectrics Reflection and refraction with polarization (a) perpendicular (s-polarization) and (b) parallel (ppolarization) to the plane of incidence Phase matching on the boundary indicates that (i) All three vectors, π€, π€ ′ and π€ ′′ , lie in a plane, i.e., π€ ′ and π€ ′′ lie in the plane of incidence; (ii) Law of reflection: |π€ × π§| = |π€ ′′ × π§| → π sin ππ = π sin ππ , thus ππ = ππ ′ (iii) Snell’s Law: |π€ × π§| = |π€ × π§| → π sin ππ = π ′ sin ππ‘ , thus π sin ππ = π′ sin ππ‘ Reflection coefficients for s- and p-polarization: ππ = πΈ0′′ π cos ππ − π′ cos ππ‘ π cos ππ − √π′2 − π2 sin2 ππ = = πΈ0 π cos ππ + π′ cos ππ‘ π cos ππ + √π′2 − π2 sin2 ππ 2 πΈ0′′ π′ cos ππ − π cos ππ‘ π′ cos ππ − π√π′2 − π2 sin2 ππ ππ = = = πΈ0 π′ cos ππ + π cos ππ‘ π′ 2 cos ππ + π√π′2 − π2 sin2 ππ 4 Brewster’s angle and total internal reflection We consider the dependence of π and π on the angle of incidence, using the Fresnel coefficients. Reflectance for s- and p-polarization at an air-glass interface. Brewster’s angle is π½π© = ππβ π=π π′ = π. π Brewster angle Brewster’s angle ππ΅ = ππ at which the p-polarized reflected wave is zero: π π sin ππ΅ = π′ sin ( − ππ΅ ) = π′ cos ππ΅ 2 or π′ tan ππ΅ = π Polarization at the Brewster angle is a practical means of producing polarized radiation. If a plane wave of mixed polarization is incident on a plane interface at the Brewster angle, the reflected radiation is completely s-polarized. The generally lower reflectance for p-polarized lights accounts for the usefulness of polarized sunglasses. Since most outdoor reflecting surfaces are horizontal, the plane of incidence for most reflected glare reaching the eyes is vertical. The polarized lenses are oriented to eliminate the strongly reflected s-component. The figure above shows π π and π π as a function of ππ with π = 1 and π′ = 1.5, as for an air-glass interface. The Brewster angle is ππ΅ = 56β for this case. 5 Total internal reflection There is another case in which π π = π π = 1. Perfect reflection occurs for ππ‘ = π/2. The incident angle for which ππ‘ = π/2 is called the critical angle, ππ = ππ . From Snell’s law π′ sin ππ = π ππ can exist only if π > π′, i.e., the incident and reflected waves are in a medium of larger index of refraction than the refracted wave. Reflectance for s- and p-polarzation at an air-glass interface. Brewster’s angle is π½π© = ππβ and the critical angle is π½π = ππβ π = π. π π′ = π For waves incident at ππ , the refracted wave is propagated parallel to the surface. There can be no energy flow across the surface. Hence at that angle of incidence there must be total reflection. For incident angles greater than the critical angle ππ > ππ , Snell’s law gives sin ππ‘ = π π sin π > sin ππ = 1 π π′ π′ This means that ππ‘ is a complex angle with a purely imaginary cosine. sin ππ 2 √ cos ππ‘ = π ( ) −1 sin ππ Then the reflection coefficients ππ and ππ both take the form π= π − ππ π + ππ 6 where π and π are real, therefore, π = |π|2 π − ππ 2 =| | =1 π + ππ The result is that π π = π π = 1 for all ππ > ππ . This perfect reflection is called total internal reflection. The meaning of this total internal reflection becomes clear when we consider the propagation factor for the refracted wave: π ππ€ ′ ⋅π± = π ππ ′ (π₯ sin π +π§ cos π ) π‘ π‘ π§ ππ ′ (sin ππ )π₯ sin ππ = π− πΏπ where 1 2ππ √sin2 ππ − sin2 ππ = −ππ ′ cos ππ‘ = π√sin2 ππ − sin2 ππ = πΏ π with the wavelength of the radiation π in vacuum. This shows that, for ππ > ππ , the refracted wave is propagating only parallel to the surface and is attenuated exponentially beyond the interface. The attenuation occurs within a few wavelengths of the boundary except for ππ ≈ ππ . 7