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Electromagnetic (E-M) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b While it is possible to understand reflection and.

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Presentation on theme: "Electromagnetic (E-M) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b While it is possible to understand reflection and."— Presentation transcript:

1 Electromagnetic (E-M) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b 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). 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. IiIi IrIr ItIt

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 field H – Magnetic Intensity P – Polarization M – Magnetization  - Magnetic permeability  - Permittivity  e - Dielectric Susceptibility  m - Magnetic Susc. g – Conductivity j – Current density

4 Summary of the boundary conditions for fields at an interface Boundary Side 1 Side 2 Maxwell’s equations in integral form allow for the derivation of the boundary conditions for the total fields on both sides of a boundary. 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

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 E i, E r, and E t the continuity of E components yield: ii rr tt nini ntnt ûnûn x y. z b   Note that

6 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

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 give Cosines cancel Using 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 as Note 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 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.

12 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:

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: n i = 1 n t = 1.5 We will consider a particular angle called the Brewster’s angle:  p +  t = 90  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. ii tt External reflection n t > n i Internal reflection n i > n t

14 External Reflection (n t > n i ) Internal Reflection (n t < n i )

15 Concept of Phase Shifts (  ) in E-M waves: Since when n t > n i 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: A.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. B.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: (a) (b)

17 Glass (n = 1.5) Air (n = 1) Air (n = 1) Glass (n = 1.5) Analogy between a wave on a string and an E-M wave traversing the air-glass interface.  = 0  =   = 0 Compare 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 (n i < n t ). 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 (n i > n t ). 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 by The radiant flux density or irradiance (W/m 2 ) 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:

22 Define Reflectance and Transmittance: Note that Conservation 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): tt ii n t = 1 n i = 1.5


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