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). Ir i r t ni nt ûn x y . z b Ii It We will consider E-field polarizations which are (i) in the plane of incidence and (ii) perpendicular to the plane of incidence, as shown below.

E-field is perpendicular to the plane-of incidence E-field is parallel to the plane of incidence

Maxwell’s Equations for time-dependent fields in matter D – Displacement field H – Magnetic Intensity P – Polarization M – Magnetization  - Magnetic permeability  - Permittivity e - Dielectric Susceptibility m - Magnetic Susc. g – Conductivity j – Current density

Summary of the boundary conditions for fields at an interface Side 1 Maxwell’s equations in integral form allow for the derivation of the boundary conditions for the total fields on both sides of a boundary. Boundary Side 2 Normal component of D is discontinuous by the free surface charge density Tangential components of E are continuous Normal components of B are continuous Tangential components of H are discontinuous by the free surface current density

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 that i r ûn ni x y . z nt b t

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. Therefore Since then we have

Thus, at the interface plane  r cos  which is again Snell’s Law

Case 1: E  Plane of incidence Continuity of the tangential components of E and H give Cosines cancel Using H = -1 B, the tangential components are

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 as Note that t - r = 1

Case 2: E || Plane of Incidence Continuity of the tangential components of E: Continuity of the tangential components of -1 B:

If both media forming the interface are non-magnetic i  t  o then the amplitude coefficients become Using Snell’s law the Fresnel Equations for dielectric media become Note that t - r = 1 holds for all i , whereas t|| + r|| = 1 is only true for normal incidence, i.e., i = 0.

Consider limiting cases for nearly normal incidence: i  0. In which case, we have: since Also, using the following identity with Snell’s law Therefore, the amplitude reflection coefficient can be written as:

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: i ni = 1 We will consider a particular angle called the Brewster’s angle: p + t = 90 t nt = 1.5 External reflection nt > ni Internal reflection ni > nt At the polarization angle p, only the component of light polarized normal to the incident plane and therefore parallel to the surface will be reflected.

External Reflection (nt > ni) Internal Reflection (nt < ni)

Concept of Phase Shifts () in E-M waves: Since when nt > ni and t < i as in the Air  Glass interface, we expect a reversal of sign in the electric field for the Ecase when We 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.

Examples of Phase shifts for two particular cases: (b)

Analogy between a wave on a string and an E-M wave traversing the air-glass interface. Glass (n = 1.5) Air (n = 1)  = 0  = 0 Air (n = 1) Glass (n = 1.5)  =   = 0 Compare with the case of

Examples of phase-shifts using our air-glass interface: In order to understand these phase shifts, it’s important to understand the definition of .

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.

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.

Reflectance and Transmittance Remember that the power/area crossing a surface in vacuum (whose normal is parallel to the Poynting vector) is given by The radiant flux density or irradiance (W/m2) is Phase velocity From 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:

Define Reflectance and Transmittance: Note that Conservation of Energy at the interface yields:

Therefore, We can write this expression in the form of componets  and ||: We must use the previously calculated values for

It’s possible to verify for the special case of normal incidence: Consider the case of Total Internal Reflection (TIR): t nt = 1 ni = 1.5 i