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7. Fresnel's Equations for Reflection and Refraction

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7. Fresnel's Equations for Reflection and Refraction Fresnel's Equations for Reflection and Refraction Incident, transmitted, and reflected beams at interfaces Reflection and transmission coefficients The Fresnel Equations Brewster's Angle Total internal reflection Power reflectance and transmittance Phase shifts in reflection The mysterious evanescent wave Reflection and transmission for an arbitrary angle of incidence at one (1) interface • Only Maxwell+Boundary conditions need. Gives Fresnels equations Maxwell’s eqns. r ∇ ⋅D ρ =ext r ∇ ⋅B 0 = r r ∂B ∇ ×D = − dt r r ∂D r ∇ ×H = + j dt Boundary onditionsc of EM wave • Tangential components of the: - E and H fields (from Gauss’ theorem) • Normal components of - D and B fields (from Stoke’s theorem) r (r 2 )r ( 1 ) n×12( E − E ) = 0 r (r 2 ) r ( 1 ) 1 r 2 n⋅12( B − B ) = 0 n12 Definitions: Planes of Incidence and the Interface and the polarizations Perpendicular (“S”) Incident medium polarization sticks out of or into the plane of r r incidence. ki kr E Plane of incidence Ei r ni (here the xy plane) is the plane that θi θr contains the incident InterfacePlane of the interface (here the and reflected k- yz plane)y (perpendicular to page) vectors. θt x nt z E r Parallel (“P”) t k polarization lies t parallel to the plane of incidence. Transmitting medium Shorthand notation for the polarizations Perpendicular (S) polarization sticks up out of the plane of incidence. Parallel (P) polarization lies parallel to the plane of incidence. rEErEE== // Fresnel Equations For example, ⊥ 0000riri tEEtEE== // ⊥ 0000titi We would like to compute the fraction of a light wave reflected and transmitted by a flat interface between two media with different refrac- tive indices. Fresnel was the first to do this calculation. It proceeds by considering r r the boundary ki kr conditions at Er Ei ni the interface Bi θ θ Br for the electric i r y Interface and magnetic Beam geometry fields of the for light with its z x θ light waves. electric field per- t E pendicular to the t nt We’ll do the plane of incidence r (i.e., out of the page) Bt perpendicular kt polarization first. Boundary Condition for the Electric y Field at an Interface The Tangential Electric Field is Continuous z x In other words: r r The total E-field in ki kr Er the plane of the Ei ni interface is B B i θi θr r continuous. Interface Here, all E-fields are θt in the z-direction, E t nt which is in the plane B r of the interface (xz), t kt so: Ei(x, y = 0, z, t) + Er(x, y = 0, z, t) = Et(x, y = 0, z, t) Boundary Condition for the Magnetic y Field at an Interface z x The Tangential Magnetic Field* is Continuous In other words: r r k The total B-field in i kr Er the plane of the Ei ni interface is θi B B θ θi θr r continuous. i i Interface Here, all B-fields are θ in the xy-plane, so t Et n we take the t B r x-components: t kt –Bi(x, y=0, z, t) cos(θi) + Br(x, y=0, z, t) cos(θr) = –Bt(x, y=0, z, t) cos(θt) *It's really the tangential B/μ, but we're using μ = μ0 Reflection and Transmission for Perpendicularly (S) Polarized Light Canceling the rapidly varying parts of the light wave and keeping only the complex amplitudes: EE00ir+ = E 0 t −+BB00iirrcos(θ ) cos(θθ ) =− B 0 tt cos( ) But B= E/( c00 / n )== nE / c and θriθ : nEir()cos()cos()00− E iθ i=− nE tt0 θ t Substituting forE0000tirt using EE+ = E: nEiriitrit()cos()()cos()00− Eθ =− nE00 + E θ Reflection & Transmission Coefficients for Perpendicularly Polarized Light Rearranging nEiriitrit()cos()()cos()00− Eθ =− nE00 + E θ yields: En00ri[]cos(θ i )+= n t cos(θθθ t ) En ii[] cos( i ) − n t cos( t ) Solving for EE00ri/ yields the reflection coefficient : rEE⊥ ==00riiittiitt/[ n cos(θ ) − n cos(θθθ )] /[ n cos( )+ n cos( )] Analogously, the transmission coefficient, EE00ti/ , is tEE⊥ ==00tiiiiitt/ 2 n cos(θ ) /[ n cos(θθ )+ n cos( )] These equations are called the Fresnel Equations for perpendicularly polarized light. Simpler expressions for r┴ and t┴ Recall the magnification at an interface, m: θi wi ni w cos(θ ) nt m ==tt w θ t wiicos(θ ) t Also let ρ be the ratio of the refractive indices, nt / ni. Dividing numerator and denominator of r and t by ni cos(θi): rn⊥ =−[ iittiittcos(θ ) n cos(θθθρρ )] /[ n cos( ) +=−+ n cos( )] [ 1 m] /[ 1 m] tn⊥ =+=2iiiitt cos(θ ) /[ n cos(θθ ) n cos( )] 2 /[ 1+ ρm] 1− ρm 2 r = t = ⊥ 1+ ρm ⊥ 1+ ρm Fresnel Equations—Parallel electric field r r This B-field y ki kr points into E B the page. i r z x Bi E θi θr r ni Interface Beam geometry Note that Hecht for light with its θ uses a different electric field t E notation for the parallel to the t nt reflected field, plane of incidence B which is t r confusing! (i.e., in the page) kt Ours is better! Note that the reflected magnetic field must point into the screen to rr r achieve EB ×∝ k . The x means “into the screen.” Reflection & Transmission Coefficients for Parallel (P) Polarized Light For parallel polarized light, B0i -B0r = B0t and E0icos(θi) + E0rcos(θr) = E0tcos(θt) Solving for E0r / E0i yields the reflection coefficient, r||: rEE||== 0riittiitti/ 0 [ n cos(θ ) − n cos(θθθ )] /[ n cos( )+ n cos( )] Analogously, the transmission coefficient, t|| = E0t / E0i, is tEE||== 0tiiiitti/ 0 2 n cos(θ ) /[ n cos(θθ )+ n cos( )] These equations are called the Fresnel Equations for parallel polarized light. Simpler expressions for r║ and t║ rEE||== 0riittiitti/ 0 [ n cos(θ ) − n cos(θθθ )] /[ n cos( )+ n cos( )] tEE||== 0ti/ 0 2 n i cos(θ i ) /[ n i cos(θθ t )+ n t cos( i )] Again, use the magnification, m, and the refractive-index ratio, ρ . r And again dividing numerator and denominator of r and t by ni cos(θi): rm= −+ρ / mρ r|| [ ] [ ] ttm= 2/ + ρ || [ ] m − ρ 2 rr||= tt = m + ρ || m + ρ Reflection Coefficients for an Air-to-Glass Interface 1.0 nair ≈ 1 < nglass ≈ 1.5 Note that: r Brewster’s angle .5 r =0! Total reflection at θ = 90° || r|| for both polarizations 0 Zero reflection for parallel polarization at Brewster's angle (56.3° for these Reflection coefficient, -.5 r values of ni and nt). ┴ -1.0 0° 30° 60° 90° Incidence angle, θi Condition for rII (rp)=0 θ1 + θ2 = 90° ⎛ n ⎞ −1⎜ t ⎟ θ B = tan ⎜ ⎟ ⎝ ni ⎠ Brewster’s angle [pic from Wikipedia] Brewster application: Sunglasses • Polarized sunglasses use the principle of Brewster's angle to eliminate glare from from the sun. In a large range of angles around Brewster's angle the reflection of p-polarized light is lower than s-polarized light. • Thus, if the sun is low in the sky mostly s-polarized light will reflect from water. Sunglasses made up of polarizers (e.g. polaroid film) aligned to block this light consequently block reflections from the water. To accomplish this, sunglass makers assume people will be upright while viewing the water and thus align the polarizers to block the polarization which oscillates along the line connecting the sunglass ear-pieces (i.e. Horizontal). • Photographers use the same principle to remove reflections from water so that they might photograph objects beneath the surface. In this case, the polarizer filter camera attachment can be rotated to be at the correct angle (see figure). Brewster angle Polarizer aligned in ”s” direction Polarizer aligned in ”p” direction Photographs taken of mudflats with a camera polarizer filter rotated to two different angles. In the first picture, the polarizer is rotated to maximize reflections, and in the second, it is rotated 90° to minimize reflections - almost all reflected sunlight is eliminated [pic from Wikipedia] Reflection Coefficients for a Glass-to-Air Interface 1.0 nglass ≈ 1.5 > nair ≈ 1 Critical r angle .5 r Note that: ┴ Total internal reflection Total internal reflection above the critical angle 0 Brewster’s -1 angle θcrit ≡ sin (nt /ni) Reflection coefficient, -.5 (The sine in Snell's Law Critical r|| can't be > 1!): angle -1.0 0° 30° 60° 90° sin(θcrit) = nt /ni sin(90°) Incidence angle, θi Relation between θB and θC • Note tanθB=sinθC, tanθ>sinθ⇒θC>θB θB θc θi ⎛⎞ε 00c 2 Transmittance (T) In= ⎜⎟ E0 ⎝⎠2 IA T ≡ Transmitted Power / Incident Power = tt A = Area IAii θ Compute the w i i n Awcos(θ ) ratio of the i tt= == tm nt beam areas: Awiicos(θ i ) 1D beam wt expansion θt The beam expands in one dimension on refraction. 2 E0t 2 2 = t ε c 2 ⎛⎞00 E0i ⎜⎟nEtt0 2 IAtt ⎝⎠2 ⎡⎤wt nEtt0 w tnt2 cos(θ t ) Tt== ⎢⎥=2 = IA ε c 2 w n cos(θ ) ii ⎛⎞00 ⎣⎦i nEii0 w i i i ⎜⎟nEii0 ⎝⎠2 ⎡⎤(nttcos(θ )) The Transmittance is also ⇒ Tt==⎢⎥22ρ mt n cos θ called the Transmissivity. ⎣⎦⎢⎥()ii() Reflectance (R) ⎛⎞ε 00c 2 In= ⎜⎟ E0 ⎝⎠2 IA R ≡ Reflected Power / Incident Power = rr A = Area IAii θ θ w i r i wi ni nt 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: R = r 2 The Reflectance is also called the Reflectivity.
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