Presentation on theme: "Electromagnetic (E-M) theory of waves at a dielectric interface"— Presentation transcript:
1 Electromagnetic (E-M) theory of waves at a dielectric interface While it is possible to understand reflection and refraction from Fermat’s principle, we need to use E-M theory in order to understand quantitatively the relationship between the incident, reflected, and transmitted radiant flux densities:We can accomplish this treatment by assuming incident monochromatic light waves which form plane waves with well defined k-vectors as shown in the diagram. The interface is shown with an origin and coordinates (x,y,z).Irirtnintûnxy.zbIiItWe will consider E-field polarizations which are (i) in the plane of incidence and (ii) perpendicular to the plane of incidence, as shown below.
2 E-field is perpendicular to the plane-of incidence E-field is parallel to the plane of incidence
3 Maxwell’s Equations for time-dependent fields in matter D – Displacement fieldH – Magnetic IntensityP – PolarizationM – Magnetization - Magnetic permeability - Permittivitye - Dielectric Susceptibilitym - Magnetic Susc.g – Conductivityj – Current density
4 Summary of the boundary conditions for fields at an interface Side 1Maxwell’s equations in integral form allow for the derivation of the boundary conditions for the total fields on both sides of a boundary.BoundarySide 2Normal component of D is discontinuous by the free surface charge densityTangential components of E are continuousNormal components of B are continuousTangential components of H are discontinuous by the free surface current density
5 For dielectrics, j = 0. Therefore, the components of E and H that are tangent to the interface must be continuous across it. Since we have Ei, Er, and Et the continuity of E components yield:Note thatirûnnixy.zntbt
6 Consider the expression on the interface (y = b) for all x, z and t Consider the expression on the interface (y = b) for all x, z and t. The above relationship must hold at all points and at any instant in time on the interface. ThereforeSince then we have
7 Thus, at the interface plane r cos which is again Snell’s Law
8 Case 1: E Plane of incidence Continuity of the tangential components of E and H giveCosines cancelUsing H = -1 B, the tangential components are
9 The last two equations give The symbol means E Plane of incidence. These are called the Fresnel equations; most often i t o.Let r = amplitude reflection coefficient and t = amplitude transmission coefficient. Then, the Fresnel equations appear asNote that t - r = 1
10 Case 2: E || Plane of Incidence Continuity of the tangential components of E:Continuity of the tangential components of -1 B:
11 If both media forming the interface are non-magnetic i t o then the amplitude coefficients becomeUsing Snell’s lawthe Fresnel Equations for dielectric media becomeNote that t - r = 1 holds for all i , whereas t|| + r|| = 1 is only true for normal incidence, i.e., i = 0.
12 Consider limiting cases for nearly normal incidence: i 0. In which case, we have:sinceAlso, using the following identity with Snell’s lawTherefore, the amplitude reflection coefficient can be written as:
13 In the limiting case for normal incidence i=t = 0, we have : Note that these equalities occur for near normal incidence as a consequence of the fact that the plane of incidence is no longer specified when i t 0.Consider a specific example of an air-glass interface:ini = 1We will consider a particular angle called the Brewster’s angle: p + t = 90tnt = 1.5External reflection nt > ni Internal reflection ni > ntAt the polarization angle p, only the component of light polarized normal to the incident plane and therefore parallel to the surface will be reflected.
14 External Reflection (nt > ni) Internal Reflection (nt < ni)
15 Concept of Phase Shifts () in E-M waves: Sincewhen nt > ni and t < ias in the Air Glass interface,we expect a reversal of sign in the electric field for the Ecase whenWe need to define phase shift for two cases:When two fields E or B are to the plane of incidence, they are said to be (i) in-phase (=0) if the two E or B fields are parallel and (ii) out-of-phase ( = ) if the fields are anti-parallel.When two fields E or B are parallel to the plane of incidence, the fields are (i) in-phase if the y-components of the field are parallel and (ii) out-of-phase if the y-components of the field are anti-parallel.
16 Examples of Phase shifts for two particular cases: (b)
17 Analogy between a wave on a string and an E-M wave traversing the air-glass interface. Glass (n = 1.5) Air (n = 1) = = 0Air (n = 1) Glass (n = 1.5) = = 0Compare with the case of
18 Examples of phase-shifts using our air-glass interface: In order to understand these phase shifts, it’s important to understand the definition of .
19 Reflected E-field orientations at various angles for the case of External Reflection (ni < nt). It is worth checking and comparing with the various plots for the phase shift on the previous slides.
20 Reflected E-field orientations at various angles for the case of Internal Reflection (ni > nt). It is worth checking and comparing with the various plots for the phase shift on the previous slides.
21 Reflectance and Transmittance Remember that the power/area crossing a surface in vacuum (whose normal is parallel to the Poynting vector) is given byThe radiant flux density or irradiance (W/m2) isPhase velocityFrom the geometry and total area A of the beam at the interface, the power (P) for the (i) incident, (ii) reflected and (ii) transmitted beams are:
22 Define Reflectance and Transmittance: Note thatConservation of Energy at the interface yields:
23 Therefore,We can write this expression in the form of componets and ||:We must use the previously calculated values for
24 It’s possible to verify for the special case of normal incidence: Consider the case of Total Internal Reflection (TIR):tnt = 1ni = 1.5i