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The Fresnel equations ki kr Ei Bi Er qi qr Br ni Interface qt Et Bt nt kt Augustin Fresnel Augustin Fresnel (1788-1827) did experiments to establish the wave theory and derived expressions for the fraction of a light wave reflected and transmitted by a flat interface between two media with different refractive indices. These expressions now carry his name. Augustin Fresnel “S” and “P” polarizations “S” polarization is the perpendicular polarization, and it sticks up out of the plane of incidence y Plane of incidence (z = 0) is the plane that contains the incident and reflected k-vectors. z x Plane of the interface (x = 0) (perpendicular to page) “P” polarization is the parallel polarization, and it lies parallel to the plane of incidence. Fresnel equations – perpendicular polarized light (s) 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. Fresnel was the first to do this calculation. Here we do a somewhat simpler derivation than given in the text. ki kr Ei Bi Er qi qr Br Interface Beam geometry for light with its electric field perpendicular to the plane of incidence (i.e., out of the page) y ni z qt Et Bt nt kt x Boundary condition for the electric field As you will learn or remember from Physics 320: The tangential component of E to the interface is always continuous, and the normal component of E to the interface is only continuous in the absence of a surface charge. ki In other words, for our configuration the total Efield is continuous because all E-fields are in the zdirection, which is in the plane of the interface (xz). Bi Ei Er qi qr y x z kr Br ni Interface qt Et Bt Hence: nt kt Ei ( x, y 0, z, t ) Er ( x, y 0, z, t ) Et ( x, y 0, z, t ) Ignoring the rapidly varying parts of the light wave and keeping only the complex amplitudes: E0t E0i E0r Boundary conditions for the magnetic field y Also: The normal component of B to the interface is always continuous and the tangential component of B to the interface is continuous in the absence of a surface current and assuming i = r = t = 0. Hence, in the absence of a surface current, the total Bfield in the plane of the interface is continuous. ki qi Bi qi qi qr Interface Here, all B-fields are in the xy-plane, so we take the x-components: Er Ei z kr Br qt Et Bt x ni nt kt Bi ( x, y 0, z, t ) cos qi Br ( x, y 0, z, t ) cos qr Bt ( x, y 0, z, t ) cos qt Fresnel equations - perpendicular polarized light (s) Ignoring the rapidly varying parts of the light wave and keeping only the complex amplitudes: B0i cos qi B0r cos qr B0t cosqt but: qi qr B0r B0i cosqi B0t cosqt E E E n ni ( E0 r E0i ) cos qi nt E0t cos qt v c/n c Substitute for E0t E0i E0r ni ( E0r E0i ) cosqi nt ( E0r E0i ) cosqt also: B Rearranging: E0r ni cosqi nt cosqt E0i ni cosqi nt cosqt Solving for E0 r / E0i yields the reflection coefficient r Fresnel equations for perpendicular polarized light E0 r ni cos qi nt cos qt r E0i ni cos qi nt cos qt Amplitude reflection coefficient Similarly, from previous slide: ni ( E0 r E0i ) cos qi nt E0t cos qt Substitute for E0r E0t E0i ni ( E0t E0i E0i ) cos qi nt E0t cos qt Solving for E0t / E0i yields the transmission coefficient t E0t 2ni cos qi t E0i ni cos qi nt cos qt Amplitude transmission coefficient Fresnel equations for parallel polarized light (p) In a similar fashion the Fresnel equations for parallel polarized light can be derived. Here we give only the result: E0 r ni cos qt nt cos qi r|| E0i ni cos qt nt cos qi Amplitude reflection coefficient E0t 2ni cos qi t|| E 0i ni cos qt nt cos qi Amplitude transmission coefficient Notational simplification using Snell's law Using Snell’s law: ni sin qi nt sin qt , the Fresnel's equations can be written as: sin qi qt r sin qi qt 2sin qt cos qi t sin qi qt tan qi qt r tan qi qt 2sin qt cos qi t sin qi qt cos qi qt A note of caution: the literature is not standardized and all possible sign variations can be found. In case you are unsure, relate back to the specific field direction from which the equations were derived. Reflection & transmission coefficients for an air-to-glass interface nair 1 < nglass 1.5 (external reflection: ni < nt) Note: • Total reflection at qi = 90° for both polarizations • Zero reflection for parallel polarization at qP=56.3° “Brewster's angle” This polarization by reflection is exploited in numerous optical devices. 1 0 1 0 45 qi degrees q P 56.3o 90 Near glancing incidence, the floor looks mirrorlike Try this with your copy of Hecht. Hold the book horizontally at the level of the middle of your eye and face a bright source you will see a nice reflection of the source Reflection coefficients for a glass-to-air interface nglass 1.5 > nair 1 (internal reflection: nt < ni) 1 Total internal reflection above the "critical angle" here: qt = p/2 and all light is 0 reflected back into the medium (The sine in Snell's Law can't be > 1!). nt qc arcsin ni o 41.8 This phenomenon of total internal reflection has many practical applications in optics. 1 0 45 qi degrees Zero reflection for parallel polarization at Brewster's angle: q'P=33.7°. Here the reflected light is linearly polarized in a plane perpendicular to the incident plane. 90 Brewster windows • Polarization by reflection via Brewster's angle is exploited in numerous optical devices Courtesy Thorlabs Inc. Brewster windows are uncoated substrates which are used at Brewster's angle. When used in a laser cavity, a Brewster window causes the laser output to be polarized and there is no reflection. R = 100% 0% reflection! Laser medium R = 90% 0% reflection! c I n 0 E0 2 Transmittance Transmitted Power I t At T Incident Power I i Ai If the beam has width wi: wi qi ni nt qt wt 2 A = Area At wt cos(qt ) Ai wi cos(qi ) The beam expands in one dimension on refraction. 2 0c nt E0t 2 n E I t At 2 wt nt wt 2 t 0 t wt T t 2 2 I i Ai 0c wi ni E0i wi ni wi n E i 2 0i nt cos qt 2 T t ni cos qi The transmittance is also called the transmissivity since E0t E0i 2 2 t2 0c I n E0 2 Reflectance R Reflected Power I r Ar Incident Power I i Ai wi ni nt qi qr A = Area wi 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. E0t I r Ar R I i Ai E0i 2 2 r2 R r2 The reflectance is also called the reflectivity 2 Energy conservation Ii A cos qi I r A cos qr I t A cos qt 1 2 qi q r , ni nr , and I nc 0 E0 2 ni E02i cos qi ni E02r cos qi nt E02t cos qt 2 E0 r nt cos qt E0t 1 E n cos q E i 0 i 0i i 2 1 R T Neglecting absorption in the medium, you can calculate T from knowledge of R. It can also be shown that this relation holds for the individual polarization components: R T 1 R T 1 Energy conservation So you can find perpendicular and parallel polarization components of T from knowledge of R: 2 E0 r ni cos qi nt cos qt R Ro r E n cos q n cos q i t t 0i i 2 2 E0 r ni cos qt nt cos qi R Ri r E0i ni cos qt nt cos qi 2 2 2 note: in your manual, the numerator is: nt cos qi ni cos qt which doesn’t make a difference due to the square Reflectance and transmittance for an air-to-glass interface Perpendicular polarization 1.0 Parallel polarization 1.0 T T .5 .5 R R 0 0 0° 30° 60° 90° Incidence angle, qi 0° 30° 60° Incidence angle, qi Note that: R T 1 R T 1 qB 90° Reflectance and transmittance for a glass-to-air interface Perpendicular polarization 1.0 Parallel polarization 1.0 R .5 R .5 T T 0 0° 30° 0 60° 90° Incidence angle, qi qC 0° 30° 60° Incidence angle, qi Note that: R T 1 R T 1 qC 90° Reflection and transmission at normal incidence When qi = 0, the incident plane becomes undefined, and any distinction between parallel and perpendicular components vanishes: 2 4 nt ni nt ni R R R and T T T 2 n n n n t i t i 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. It can be significantly reduced using anti reflection (AR) coatings (later). 4% back reflection This back reflection is of concern if you are dealing with complicated lens systems containing 10 or 20 of these interfaces. Polarization by reflection A wave reflecting and refracting at an interface The incoming wave drives the bound electrons, and they in turn re-radiate according to the dipole radiation pattern. Brewster's angle revisited From Fresnel equations, the reflection coefficient for parallelpolarized light goes to zero for Brewster's angle incidence: q p qt 90o qt 90o q p ni qt ni sin(q p ) nt sin(qt ) ni sin(q p ) nt sin(90 q p ) nt cos(q p ) nt tan(q p ) ni qp qp nt Brewster's angle can be found from Snell's law: qp +qt = 90° nt q p tan ni 1 Stick model by Robert W. Wood This behavior can be explained from Fresnel equations, but perhaps an easier method is to employ the stick model first proposed by Robert W. Wood: Consider a stick of wood in place of the electric vector. If the wood impacts a water surface at an angle, the stick slides into the water and is not reflected. However, if the stick lands parallel to the water surface, it can bounce back. Because in nature, horizontal surfaces are almost exclusively encountered, it will be a horizontal vibration that is reflected. To minimize or cut glare, Polaroid sunglasses are designed to remove the horizontal vibrations and transmit the electric vector that is vibrating vertically. Electron oscillators and Brewster’s law We can understand Brewster's angle considering the dipole radiation pattern. q p Brewster's angle the reflected beam although completely polarized is rather weak. http://physics.bu.edu/~duffy/ semester2/c27_brewster.html Electron oscillators vibrate under the influence of the refracted (transmitted) wave. When the reflected beam makes a right angle with the transmitted beam, the reflected beam vanishes due to the scattered dipole emission pattern. Glare is horizontally polarized Puddle reflection viewed through polarizer that transmits only horizontally polarized light Puddle reflection viewed through polarizer that transmits only vertically polarized light Light reflected into our eyes from the puddle reflects at about Brewster's Angle. So parallel (i.e., vertical) polarization sees zero reflection. http://www.colorado.edu/physics/2000/applets/polarized.html Polarizer sunglasses transmit only vertically polarized light. That's why I use them for fly-fishing so can see through the water and spot the fish. Sir David Brewster • Brewster’s law named after Sir David Brewster • Professor of physics at St. Andrews University • Inventor of the kaleidoscope, lenticular stereoscope, binocular camera, polyzonal lens, polarimeter, and lighthouse illuminator. Sir David Brewster (1781-1868) http://www.brewstersociety.com/brewster_bio.html