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Chapter 4 Propagation of Light September 12,14 Scattering and interference 4.1 Introduction

+. -. +. B. q. E. Chapter 4 Propagation of Light September 12,14 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.

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Chapter 4 Propagation of Light September 12,14 Scattering and interference 4.1 Introduction

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  1. + - + B q E Chapter 4 Propagation of Light September 12,14 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?

  2. 4.2.1 Scattering and interference Scattering in tenuous media (d >l): Lateral scattering:random phase qm Summation is a random walk (no interference). Primary light P Forward scattering Lateral scattering P Forward scattering:in phase Summation is a constructive interference.

  3. 4.2.2 The transmission of light through dense media Scattering in dense media (d<<l): Lateral scattering: destructive interference Wherever there is an atom A, there will be an atom A'l/2 away laterally. There is little or no light scattered laterally or backwards in a dense homogenous media. Forward scattering: in phase, constructive interference. Primary light P Forward scattering Lateral scattering Tyndall and Mie scattering: When the size of the molecules gradually increases, destructive interference starts for the scatter of short wavelength. Scattering of the longer wavelengths increases proportionally. Why is the cloud (or milk) white? P

  4. + - Primary + secondary Primary 4.2.3 Transmission and the index of refraction Why does n >1 lead to v < c? (Photons are always moving with speed of light c…) Both the primary and the secondary waves propagate with speed c in the space between atoms. They overlap and generate the transmitted wave. • Phase lagof the oscillation of the damped oscillator, compared to the driving field: • (It is a little easier to use the phase factor eiwt when discussing phase lag or phase lead.)

  5. 2) Additional phase lagof p/2 for emitting the secondary wave from the oscillator (Problem 4.5): Dx P Note that the “+g ” comes from our initial choice ofIf we instead use then x Additional phase Damped oscillator phase

  6. Re(n) a b=a +p /2 Total phase lagb = a + p/2, p/2<b < 3p /2. Note that a phase lag of b > p is equivalent to a phase lead of 2p-b. Transmitted = primary + secondary Successive phase shift at each atom is tantamount to a difference (slowing down) in phase velocity. Primary Secondary Transmitted

  7. Read: Ch4: 1-2 Homework: Ch4: 2,4,5 Due: September 21

  8. September 17 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 l/2 deep on the surface. P External reflection: nincident < ntransmitting Internal reflection: nincident > ntransmitting

  9. D B qr qi C A 4.3.1 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. The law of reflection: 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: Specular reflectionDiffuse reflection

  10. B viDt D qi qt A vtDt C 4.4 Refraction Refraction:Forward scattering on an interface + primary beam. 4.4.1 The law of refraction Refraction of light from a point source: Snell’s law: 1) 2) The incident, reflected, and refracted rays all lie in the plane of incidence. Example 4.2 Water P S

  11. S S' S" S S' S vt S' • 4.4.2 Huygens’s principle • Huygens’s principle (1690):How can we determine a new wavefront S' from a known wavefront S? • 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 widely used in anisotropic media

  12. Read: Ch4: 3-4 Homework: Ch4: 6,8,15,26,27,29 Due: September 28

  13. S qi h ni O nt x b qt P a ni September 19 Fermat’s principle 4.5 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 Optical path length (OPL): Fermat’s principle (restatement): Light traverses the route that has the smallest optical path length.

  14. Superior mirage (sea, airplane) Inferior mirage (road, desert) Optical miragesin the atmosphere:

  15. P S S P OPL×w/c=f 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.

  16. Read: Ch4: 5 Homework: Ch4: 36,38 Due: September 28

  17. y x E1tx E2tx September 21,24 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.  4.6.1 Waves at an interface Boundary conditions: 1) The tangent component of the electric field E is continuous on the interface. 2) The tangent component of H (= B/m) is continuous on the interface.

  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 kiz=0. The three participating waves are: y ki 1) Incident wave: 2) Reflected wave: 3) Refracted wave: Here E0i, E0r, E0t are complex amplitudes. The boundary condition (at the interface y = 0) is now kr qr qi x y=0 qt kt

  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 wr=wt=wi. • Similarly, we have krx=ktx=kixby fixing (z=0,t=0) and varying x on the interface. • Similarly, we have krz=ktz (=kiz)=0 by fixing (x=0,t=0) and varying z on the interface. • The laws of reflection and refraction: • 1)krx=kix, krz=kiz =0  • a) The incident beam, the reflected beam and the normal are in one plane; • b) kisinqi = krsinqr. Since ki= kr, we have qi = qr. • 2)ktx=kix, ktz=kiz =0  • a) The incident beam, the refracted beam and the normal are in one plane; • b) kisinqi = ktsinqt. Since ki/ni= kt/nt, we have ni sinqi = nt sinqt.

  20. Eis kr Ei Ei Ers Bi Bis ki ki ki Bi qi qi qi qr Brs x x x ni nt Ets qt Bts kt 4.6.2 The Fresnel equations What are E0r and E0t for a given E0i? Answer: Solving the two boundary conditions. s polarization (TE mode) and p polarization (TM mode): p polarization 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. s polarization y y y Erp Boundary condition 1, continuity of Etan: Eip Brp Boundary condition 2, continuity of Btan/m : Bip Solution for the p-polarization: Etp Therefore, s-polarized light won’t produce any p-polarized reflection or refraction, and vise versa. Btp

  21. kr Ei Er ki Br Bi qi qr x ni nt qt Et kt Bt I) E perpendicular to the plane-of-incidence (s polarization): Boundary condition 1, continuity of Etan: Boundary condition 2, continuity of Btan/m : y Solving these two equations for E0r and E0t (using Cramer’s rule), we have For dielectrics with mi ≈mt ≈ m0, we have the Amplitude reflection (and transmission) coefficients:

  22. Er kr Ei ki Bi Br qi qr ni x nt qt Et Bt kt II) E parallel to the plane-of-incidence ( p polarization): y Boundary condition 1, continuity of Etan: Boundary condition 2, continuity of Btan/m : Solving these two equations for E0r and E0t (using Cramer’s rule), we have For nonmagnetic dielectrics, the amplitude reflection (and transmission) coefficients:

  23. Finally using Snell’s law we have the Fresnel equations: Note that the negative sign comes from our definition of the positive E-field direction of each wave. Example 4.4

  24. Read: Ch4: 6 Homework: Ch4: 43,44,45 Due: September 28

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

  26. September 26 Reflectance and transmittance 4.6.3 Interpretation of the Fresnel equations I. Amplitude coefficients External reflection Internal reflection

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

  28. II. Phase shift: Note: The phase shift of 0 or p depends on our choice of the positive E-field direction of the reflected or refracted light wave. Phase shift of internal reflection will be discussed later in the part of total internal reflection. External reflection

  29. Acosqi A III. Reflectance and transmittance: Internal reflection External reflection

  30. Read: Ch4: 6 Homework: Ch4: 49,58,62(Optional),71,78(Optional),83,85 Due: October 5

  31. September 28 Total internal reflection 4.7 Total internal reflection Total internal reflection: For internal reflection, when qi ≥qc= arcsin(nt/ni), all the incoming energy is reflected back into the incident medium. I) Reflection 1) Reflectance: total reflectance. 2) Phase: Negative d means a phase lead. means p-wave leads the incident wave more than s-wave does. Therefore s lags p in phase.

  32. Relative phase difference between r┴ and r//: Maximum relative phase differencedm:

  33. * (Reading) Generating circularly polarized light: 1) Letdm =p/2, we have nti= 0.414, or ni= 2.414 if nt= 1 (air). This means ni≥ 2.414 is desired. These materials are not easily attainable. 2) Let us try two reflections. Letdm =p/4, we have ni= 1.497. For a glass of n =1.51, qi = 48º37' and 54º37' will make d =p/4. Fresnel’s rhomb: p p s s Linear polarization Circular polarization 54º37' Total internal reflection introduces a phase shift which is equivalent to a retarder with its slow axis aligned parallel to the s-polarization.

  34. y kt qt ktcosqt x ktsinqt qi II) Transmission: the evanescent wave Properties of evanescent waves: • The wave propagates in the x-direction. • The amplitude decays rapidly in the y-direction within a few wavelengths. • Energy circulates back and forth across the interface, but averaged in a zero net energy flow through the boundary.

  35. d FTIR beam splitter 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).

  36. Read: Ch4: 7 Homework: Ch4: 86,87,91 Due: October 5

  37. La nature ne s'est pas embarrassée des difficultés d'analyse. Nature is not embarrassed by difficulties of analysis. Augustin Fresnel

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