College 9 Lit.=Literatuur K=Kennen T/A=Toepassen/Afleiden

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1 College 9 Lit.=Literatuur K=Kennen T/A=Toepassen/Afleiden
Hecht H6 staat in dit college centraal Werkcollegeopgaves: zie blackboard Thema Lit. K T/A Opmerking Breking, Huygens, Principe van Fermat (deels herhaling) Hecht H6.4 X Concept kunnen uitleggen/toepassen. Vergelijkingen kunnen reproduceren Fresnel coeeficienten H6.6 Afleiding kunnen produceren Evanescent wave H6.7 QED, Phasor approach to Feynman’s summation of probability amplitudes H6.11.1 Figuur 6.68 T Concept achter figuur 6.68 kunnen uitleggen/toepassen.

2 Trillingen, Golven en Optica (TGO)
TGO in dagelijks leven: geluid, licht en mechanische trillingen. TGO in de natuurkunde: belangrijke basiskennis om verschijnselen uit de Quantum Mechanica en Electromagnetisme te beschrijven en te begrijpen.                   Dr. Marcel Vreeswijk Nikhef Prof. Dr. Johannes F. de Boer Laserlab VU Januari Trillingen & Golven Feb-Maart Golven & Optica

3 Fermat’s principe Het pad tussen twee punten dat afgelegd wordt door een lichtbundel is dat pad dat afgelegd wordt in de kortste tijd. → Snell’s Wet Afleiding op p. 107 van Hecht

4 Snell’s wet zorgt er voor dat dingen gekromd lijken in water.
Picture borrowed from

5 Snell's wet verklaart vele allledaagse verschijnselen
De brekingsindex neemt toe met dichtheid. Astronomen speculeren of licht om een planeet kan draaien ergens in het universum… Het effect kan een rol spelen in fata morgana’s.

6 Snell’s wet verklaart waarom de zon platter wordt als hij onder gaat.
Lichtstralen dichter bij de horizon buigen meer af dan stralen verder weg.

7 Breking en Reflectie, volgens Maxwell
We kunnen de hoeken wel uitrekenen, maar hoe zit het met de grootte van de velden? Je verwacht dat er een soort behoud van E-veld en B-veld zal gelden. Echter, de atoomtrillingen (‘gebonden oppervlaktelading’) dragen ook bij aan deze velden. Gevolg (m.b.v. Maxwell vgl): de tangentiele componenten van E en B/m zijn continue op een grensvlak. (m= magnetische permeabiliteit voor ieder materiaal) Inkomend (i) Reflected (r) Transmitted (t) De Maxwell theorie wordt zo nodig op tentamen gegeven  dus niet uit hoofd leren. Je moet de informatie wel kunnen gebruiken /toepassen/ Interpreteren. Ik probeer consistent groene kaders te gebruiken

8 Breking en Reflectie De eerste stap: waar wijst het E veld naar toe? Twee extreme mogelijkheden: 1. “S” polarisatie. E veld steekt uit de plane of incidence. (senk) 2. “P” polarisatie ligt in de plane of incidence (parallel) t r i van overgang Vlak P-Polarisatie invalsvlak orientatie E vectoren uit Maxwell theorie. van overgang Vlak S-Polarisatie invalsvlak i r Het vlak van de overgang (interface) steekt uit papier. Het invalsvlak is het papier. (plane of incidence) t

9 Fresnel Equations Ei Er Et
Aim: We would like to compute the fraction of a light wave reflected and transmitted by a flat interface between two media with different refractive indices. for the perpendicular polarization for the parallel polarization where E0i, E0r, and E0t are the field complex amplitudes. We consider the boundary conditions at the interface for the electric and magnetic fields of the light waves. We’ll do the perpendicular polarization first.

10 Boundary Condition for the Electric Field at an Interface
The Tangential Electric Field is Continuous In other words: The total E-field in the plane of the interface is continuous. Here, all E-fields are in the z-direction, which is in the plane of the interface (xz), so: Ei(x, y = 0, z, t) + Er(x, y = 0, z, t) = Et(x, y = 0, z, t) Er Ei ni Bi Br qi qr x y z Interface qt Et nt Bt

11 Ruimte voor Afleiding ni nt qi qr qt Ei Bi Er Br Et Bt x y z Interface
Waarom kijken we alleen naar de grootte van E veld. Immers: ni nt qi qr qt Ei Bi Er Br Et Bt Interface x y z The Tangential Electric Field is Continuous

12 Boundary Condition for the Magnetic Field at an Interface
The Tangential Magnetic Field* is Continuous In other words: The total B-field in the plane of the interface is continuous. Here, all B-fields are in the xy-plane, so we take the x-components: –Bi(x, y=0, z, t) cos(qi) + Br(x, y=0, z, t) cos(qr) = –Bt(x, y=0, z, t) cos(qt) *It's really the tangential B/m, but we're using m = m0 Er Ei Bi ni qi qi qr Br qi x y z Interface qt Et nt Bt

13 Reflection and Transmission for Perpendicularly (S) Polarized Light
Canceling the rapidly varying parts of the light wave and keeping only the complex amplitudes:

14 Reflection & Transmission Coefficients for Perpendicularly Polarized Light
These equations are called the Fresnel Equations for perpendicular polarized light

15 Simpler expressions for r┴ and t┴
Recall the magnification at an interface, m: θt qi wi wt ni nt Also let ρ be the ratio of the refractive indices, nt / ni. Dividing numerator and denominator of r and t by ni cos(qi):

16 Fresnel Equations—Parallel electric field
x y z This B-field points into the page. Ei Br Bi qi qr Er ni Interface Beam geometry for light with its electric field parallel to the plane of incidence (i.e., in the page) Note that Hecht uses a different notation for the reflected field, which is confusing! Ours is better! qt Et nt Bt Note that the reflected magnetic field must point into the screen to achieve The x means “into the screen.”

17 Reflection & Transmission Coefficients for Parallel (P) Polarized Light
For parallel polarized light, B0i - B0r = B0t and E0icos(qi) + E0rcos(qr) = E0tcos(qt) Solving for E0r / E0i yields the reflection coefficient, r||: Analogously, the transmission coefficient, t|| = E0t / E0i, is These equations are called the Fresnel Equations for parallel polarized light.

18 Simpler expressions for r║ and t║
Again, use the magnification, m, and the refractive-index ratio, ρ . And again dividing numerator and denominator of r and t by ni cos(qi):

19 Interpretatie van de hoeken
Brewster angle Critical angle Phase shifts

20 Reflection Coefficients for an Air-to-Glass Interface
Incidence angle, qi Reflection coefficient, r 1.0 .5 -.5 -1.0 r|| r ┴ 0° ° ° ° Brewster’s angle r||=0! nair ~1 nglass ~1.5 Note that: Total reflection at q = 90° for both polarizations Zero reflection for parallel polarization at Brewster's angle (56.3° for these values of ni and nt). qBrewster = arctan(nt /ni)

21 Brewster Angle (I) Fresnel sin⁡( 𝜃 𝑡 ) sin⁡( 𝜃 𝑖 ) = 𝑛 𝑖 𝑛 𝑡 Snell

22 Reflectie-coefficienten (Amplitude)

23 Afleiding Brewster Angle (II)
Snell sin⁡( 𝜃 𝑡 ) sin⁡( 𝜃 𝑖 ) = 𝑛 𝑖 𝑛 𝑡

24 Reflection Coefficients for a Glass-to-Air Interface
Incidence angle, qi Reflection coefficient, r 1.0 .5 -.5 -1.0 r|| r ┴ 0° ° ° ° Total internal reflection Brewster’s angle Critical angle nglass ≈ 1.5 > nair ≈ 1 Total internal reflection above the critical angle qcrit = arcsin(nt /ni) Sine in Snell’s law cannot be larger that one: sin⁡( 𝜃 𝑡 ) sin⁡( 𝜃 𝑖 ) = 𝑛 𝑖 𝑛 𝑡

25 Interpretatie Fresnel Coefficienten, vervolg
Fase verschuiving reflectie

26 We beginnen met loodrechte inval

27 Vergelijking met een golf door een touw
Randvoorwaardes: Medium 1 Medium 2 start later X=0 Zie eerder college. Dit hebben we afgeleid: Dit kan eventueel naar volgend college: volgende onderwerp (staande golven) moet in week3 algemeen Beroemde r en t coëfficiënten

28 Phase Shift in Reflection (for Perpendicularly Polarized Light)
ni nt qi qr qt Ei Bi Er Br Et Bt Interface So there will be destructive interference between the incident and reflected beams just before the surface. Analogously, if ni > nt (glass to air), > 0, and there will be constructive interference.

29 Phase Shift in Reflection (Parallel Polarized Light)
ni nt qi qr qt Ei Bi Er Br Et Bt Interface This also means destructive interference with incident beam. Analogously, if ni > nt (glass to air), r|| > 0, and we have constructive interference just above the interface. Good that we get the same result as for ; it’s the same problem when qi = 0! Also, the phase is opposite above Brewster’s angle.

30 Phase shifts in reflection (air to glass)
0° ° ° ° Incidence angle π ┴ || Incidence angle, θi Reflection coefficient, r 1.0 -1.0 r|| r ┴ 0° ° ° ° Brewster’s angle 180° phase shift for all angles 180° phase shift for angles below Brewster's angle; 0° for larger angles

31 Phase shifts in reflection (glass to air)
0° ° ° ° Incidence angle π ┴ || Incidence angle, θi Reflection coefficient, r 1.0 -1.0 r|| r ┴ 0° ° ° ° Total internal reflection Brewster’s angle Critical angle Interesting phase above the critical angle. Coefficent becomes complex, see evanescent wave (later).

32 If you slowly turn up a laser intensity incident on a piece of glass, where does damage happen first, the front or the back? The obvious answer is the front of the object, which sees the higher intensity first. But constructive interference happens at the back surface between the incident light and the reflected wave. This yields an irradiance that is 44% higher just inside the back surface!

33 Phase shifts with coated optics
Reflections with different magnitudes can be generated using partial metallization or coatings. We’ll see these later. But the phase shifts on reflection are the same! For near-normal incidence: 180° if low-index-to-high and 0 if high-index-to-low. Example: Laser Mirror Highly reflecting coating on this surface Phase shift of 180°

34 Transmittance and Reflectance
We hebben de Fresnel coëfficiënten afgeleid  verhoudingen Amplitudes. Hoe zit het met het VERMOGEN?

35 Transmittance (T) T Transmitted Power / Incident Power
A = Area T Transmitted Power / Incident Power Compute the ratio of the beam areas: qt qi wi wt ni nt 1D beam expansion The beam expands in one dimension on refraction. The Transmittance is also called the Transmissivity.

36 Reflectance (R) R Reflected Power / Incident Power
A = Area R Reflected Power / Incident Power qi wi ni nt qr Because the angle of incidence = the angle of reflection, the beam area doesn’t change on reflection. Also, n is the same for both incident and reflected beams. So: The Reflectance is also called the Reflectivity.

37 Reflectance and Transmittance for an Air-to-Glass Interface
Perpendicular polarization Incidence angle, qi 1.0 .5 0° ° ° ° R T Parallel polarization Incidence angle, qi 1.0 .5 0° ° ° ° R T Note that R + T = 1

38 Reflectance and Transmittance for a Glass-to-Air Interface
Perpendicular polarization Incidence angle, qi 1.0 .5 0° ° ° ° R T Parallel polarization Incidence angle, qi 1.0 .5 0° ° ° ° R T Note that R + T = 1

39 Reflection at normal incidence
When qi = 0, and For an air-glass interface (ni = 1 and nt = 1.5), R = 4% and T = 96% The values are the same, whichever direction the light travels, from air to glass or from glass to air. The 4% has big implications for photography lenses.

40 Practical Applications of Fresnel’s Equations
Windows look like mirrors at night (when you’re in the brightly lit room) One-way mirrors (used by police to interrogate bad guys) are just partial reflectors (actually, aluminum-coated). Disneyland puts ghosts next to you in the haunted house using partial reflectors (also aluminum-coated). Lasers use Brewster’s angle components to avoid reflective losses: R = 100% R = 90% Laser medium 0% reflection! Optical fibers use total internal reflection. Hollow fibers use high- incidence-angle near-unity reflections.

41 The critical angle and beyond
Incidence angle, θi Reflection coefficient, r 1.0 -1.0 r|| r ┴ 0° ° ° ° Total internal reflection Brewster’s angle Critical angle The critical angle and beyond Totale Interne Reflectie Evanescent Waves

42 Total Internal Reflection occurs when sin(θt) > 1, and no transmitted beam can occur.
Note that the irradiance of the transmitted beam goes to zero (i.e., TIR occurs) as it grazes the surface. Brewster’s angle Total Internal Reflection Total internal reflection is 100% efficient, that is, all the light is reflected.

43 Applications of Total Internal Reflection
Beam steerers Beam steerers used to compress the path inside binoculars

44 Three bounces: The Corner Cube
Corner cubes involve three reflections and also displace the return beam in space. Even better, they always yield a parallel return beam: If the beam propagates in the z direction, it emerges in the –z direction, with each point in the beam (x,y) reflected to the (-x,-y) position. Hollow corner cubes avoid propagation through glass and don’t use TIR.

45 Frustrated Total Internal Reflection
By placing another surface in contact with a totally internally reflecting one, total internal reflection can be frustrated. Total internal reflection Frustrated total internal reflection n=1 n=1 n n n n How close do the prisms have to be before TIR is frustrated? This effect provides evidence for evanescent fields—fields that leak through the TIR surface–and is the basis for a variety of spectroscopic techniques.

46 FTIR and fingerprinting
Picture on right due to Center for Image Processing in Education See TIR from a fingerprint valley and FTIR from a ridge.

47 The Evanescent Wave qi ni y qt nt x
The evanescent wave is the "transmitted wave" when total internal reflection occurs. A mystical quantity! So we'll do a mystical derivation:

48 The Evanescent-Wave k-vector
The evanescent wave k-vector must have x and y components: Along surface: ktx = kt sin(qt) Perpendicular to it: kty = kt cos(qt) Using Snell's Law, sin(qt) = (ni /nt) sin(qi), so ktx is meaningful. And again: cos(qt) = [1 – sin2(qt)]1/2 = [1 – (ni /nt)2 sin2(qi)]1/2 = ± iβ Neglecting the unphysical -ib solution, we have: Et(x,y,t) = E0 exp[–kβ y] exp i [ k (ni /nt) sin(qi) x – w t ] The evanescent wave decays exponentially in the transverse direction. ni nt qi qt x y

49 Fermat’s principle: Light travels shortest path in time
Possible path from S to P reflected from a mirror Amplitude and phase in complex plane for each path

50 The End