B E  Chapter 4 The Propagation of Light September 19,22 Scattering and interference 4.1 Introduction Scattering: The absorption and prompt re-emission of electromagnetic radiation by electrons in atoms and molecules. Scattering plays an essential rule in light propagation. Transmission, reflection and refraction are all macroscopic manifestations of scattering by atoms and molecules. 4.2 Rayleigh scattering Rayleigh scattering: Elastic scattering of light by molecules which are much smaller than the wavelength of the light. Re-emission of light from an induced dipole: Rayleigh’s scattering law: Why is the sky blue? Why is the sunset red?

Scattering and interference Scattering in tenuous media (d > ): Lateral scattering: random phase  m Summation is a random walk (no interference). Forward scattering: in phase Summation is a constructive interference. P P Primary light Lateral scattering Forward scattering

The transmission of light through dense media Scattering in dense media (d<< ): Lateral scattering: destructive interference Wherever there is an atom A, there will be an atom A' /2 away laterally. There is little or no light scattered laterally or backwards in a dense homogenous media. Forward scattering: in phase, constructive interference. Tyndall and Mie scattering: When the size of the molecules increases, scattering of the longer wavelengths increases proportionally. (Destructive interference starts for the scatter of short wavelength). Why is the cloud (or milk) white? P P Primary light Lateral scattering Forward scattering

4 1)Phase lag of the oscillation of the damped oscillator: (It is easier to use the phase factor e i  t when discussing phase lag or phase lead.) Transmission and the index of refraction Why does n >1 lead to v < c? Both the primary and the secondary waves propagate with speed c in the space between atoms. They overlap and generate the transmitted wave. + - Primary Primary + secondary

5 Note that the “+  ” comes from our initial definition of If we instead define then 2) Additional phase lag of  /2 for emitting the secondary wave (Problem4.5): P x xx

6 Total phase lag  =  +  /2,  < 3 . Note that a phase lag of  >  is equivalent to a phase lead of 2  Transmitted = primary + secondary Successive phase shift at each atom is tantamount to a difference in phase velocity. Re(n)   Primary Secondary Transmitted

7 Read: Ch4: 1-2 Homework: Ch4: 2,4,5 Due: September 26

8 September 24 Reflection and refraction 4.3 Reflection Reflection: Backward scattering at an interface. Mechanism: Consider the sum of all the backward scattered wavelets at an observation point close to the interface. The backward wavelets from different dipoles inside a uniform dense medium are paired and cancel each other because of their phases. However, at the interface, this cancellation is incomplete because of their amplitudes, which results in a net reflection from a thin layer of about /2 deep on the surface. External reflection: n incident < n transmitting Internal reflection: n incident > n transmitting P

The law of reflection Ray: A line drawn on the direction of the flow of the radiant energy. Perpendicular to the wavefronts in an isotropic medium. 1) The angle of incidence equals the angle of reflection. 2) The incident ray, the normal of the surface, and the reflected ray are all in one plane (called the plane of incidence). Normal incidence: Glancing incidence: The law of reflection: Specular reflection Diffuse reflection A B C D ii rr

Refraction Refraction: Forward scattering on an interface + primary beam The law of refraction Snell’s law: 1) 2) The incident, reflected, and refracted rays all lie in the plane of incidence. A B C D ii tt vitvit vttvtt

11  '' ""  '' vt  '' Huygens’s principle Huygens’s principle (1690): How can we determine a new wavefront  ' from a known wavefront  ? 1) Every point on a propagating wavefront acts as the source of spherical secondary wavelets, the wavefront at a later time is the envelope of these wavelets. 2) The secondary wavelets have the same frequency and speed as the propagating wave. Proposed long before Maxwell First step toward scattering theory Interference not included Primary wave completely scattered Still useful in inhomogeneous media

12 Read: Ch4: 3-4 Homework: Ch4: 8,12,22,23,25 Due: October 3

Fermat’s principle Fermat’s principle (1657): The actual path between two points taken by a beam of light is the one that is traversed in the least time. Principle of least time. Example: refraction S O P h b a x ii tt nini ntnt Optical path length (OPL): Light traverses the route that have the smallest optical path length. nini September 26 Fermat’s principle

14 Optical mirages in the atmosphere: Inferior mirage (road, desert) Superior mirage (sea, airplane)

15 Modern formulation of Fermat’s principle: The actual light ray going from point S to point P must traverse an optical path length that is stationary with respect to variations of that path. Stationary path  stationary phase  constructive interference Non-stationary path  varying phase  destructive interference Stationary paths: An actual OPL for a ray can be stationary, minimum, or maximum. S P S P OPL×  /c= 

16 Read: Ch4: 5 Homework: Ch4: 32,34 Due: October 3

17 September 29,October 1 Electromagnetic approach 4.6 The electromagnetic approach Electromagnetic theory provides a complete description on light propagation in media, regarding its amplitude, polarization, phase, wavelength, direction, and frequency.  Waves at an interface Boundary conditions: 1) The tangent component of the electric field E is continuous on the interface. 2) Similarly, the tangent component of H (= B/  ) is continuous on the interface. y x E 1tx E 2tx

18 Deriving the laws of reflection and refraction using the boundary conditions: Let y =0 be the interface plane, and be the unit vector of the normal of the interface. Let z=0 be the plane-of-incidence, so that and k iz =0. The three participating waves are: 1) Incident wave: 2) Reflected wave: 3) Refracted wave: Here E 0i, E 0r, E 0t are complex amplitudes. The boundary condition (at the interface y = 0) is now ii rr tt krkr kiki ktkt y=0 x y

19 At the (x=0, z=0) point on the interface, this indicates a phasor sum of a(t)+ b(t) = c(t) at all time with fixed lengths a, b, and c, which is only possible when the three phasor vectors have an identical angular velocity (but may have different initial phases). Therefore  r =  t =  i. Similarly, we have k rx =k tx =k ix by fixing (z=0,t=0) and varying x on the interface. Similarly, we have k rz =k tz (=k iz )=0 by fixing (x=0,t=0) and varying z on the interface. The laws of reflection and refraction: 1) k rx =k ix, k rz =k iz =0  a) The incident beam, the reflected beam and the normal are in one plane; b) k i sin  i = k r sin  r. Since k i = k r, we have  i =  r. 2) k tx =k ix, k tz =k iz =0  a) The incident beam, the refracted beam and the normal are in one plane; b) k i sin  i = k t sin  t. Since k i /n i = k t /n t, we have n i sin  i = n t sin  t.

The Fresnel equations How much are E 0r and E 0t for a given E 0i ? Answer: Solving the two boundary conditions. s polarization (TE mode) and p polarization (TM mode): EiEi BiBi kiki ii x y s polarization BiBi EiEi kiki ii x y p polarization E is B is kiki ii rr tt nini E rs B rs krkr E ts B ts ktkt ntnt x y B rp E rp B tp E tp Question: will an s-polarized incident light produce both p and s polarized reflected and refracted light? Choosing the positive directions of the fields so that at large angles of incidence they appear to be continuous. Boundary condition 1, continuity of E tan : B ip E ip Boundary condition 2, continuity of B tan /  : Solution for the p-polarization: Therefore, s-polarized light won’t produce any p-polarized reflection or refraction, and vise versa.

21 I) E perpendicular to the plane-of-incidence (s polarization): Boundary condition 1, continuity of E tan : Boundary condition 2, continuity of B tan /  : EiEi BiBi kiki ii rr tt nini ErEr BrBr krkr EtEt BtBt ktkt ntnt x y Solving these two equations for E 0r and E 0t, using Cramer’s rule, we have For dielectrics with  i ≈  t ≈  0, we have the Amplitude reflection (and transmission) coefficients:

22 II) E parallel to the plane-of-incidence ( p polarization): Boundary condition 1, continuity of E tan : Boundary condition 2, continuity of B tan /  : Solving these two equations for E 0r and E 0t, using Cramer’s rule, we have For nonmagnetic dielectrics, the amplitude reflection (and transmission) coefficients: BiBi EiEi kiki ii rr tt nini BrBr ErEr krkr BtBt EtEt ktkt ntnt x y

23 Finally using Snell’s law we have the Fresnel equations: Note that the negative sign is from our definition of the positive E field direction for each beam.

24 Read: Ch4: 6 Homework: Ch4: 37,39,40,41 Due: October 10

25 Sum and difference formulas: Double-angle formulas: Half-angle formulas: Product to sum formulas: Sum to product formulas: Trigonometric Formulas

26 October 3 Reflectance and transmittance Interpretation of the Fresnel equations I. Amplitude coefficients External reflection Internal reflection

27 1) At normal incidence, 2) When is the polarization angle (Brewster angle), where r // = 0. 3) For external reflection, at glancing incidence, 4) For internal reflection, when is the critical angle, where E r // = 0 Brewster angle ii External reflection Internal reflection

28 II. Phase shift: External reflection

29 III. Reflectance and transmittance: A Acos  i External reflection Internal reflection

30 Read: Ch4: 6 Homework: Ch4: 45,49(Optional),56,58 Due: October 10

31 October 6 Total internal reflection 4.7 Total internal reflection Total internal reflection: For internal reflection, when  i ≥  c = arcsin(n t /n i ), all the incoming energy is reflected back into the incident medium. I) Reflection 1) Reflectance: total reflectance. 2) Phase:

32 Relative phase difference between r ┴ and r // : Maximum relative phase difference  m :

33 * (Reading) Producing a circular polarization: 1) Let  m =  /2, we have n ti = 0.414, or n i = if n t = 1 (air). This means n i ≥ is required. These materials are not easily obtainable. 2) Let us try two reflections. Let  m =  /4, we have n i = For a glass of n =1.51,  i = 48º37' and 54º37' will make  =  /4. Fresnel’s rhomb: 54º37' Linear polarization Circular polarization

34 II) Transmission: the evanescent wave tt ii ktkt k t cos  t k t sin  t x y 1)The wave propagates along the x axis. 2)The amplitude decays rapidly in the y-direction within a few wavelengths. 3)Energy circulates back and forth across the interface, but averaged in a zero net energy flow through the boundary. Properties of evanescent waves:

35 Frustrated total internal reflection (FTIR): When an evanescent wave extends into a medium with higher index of refraction, energy may flow across the boundary (similar to tunneling in quantum mechanics). FTIR beam splitter d

36 Read: Ch4: 7 Homework: Ch4: 59,60,61,69,76 (Optional) Due: October 17