23. Fresnel Equations • EM Waves at boundaries • Fresnel Equations: Reflection and Transmission Coefficients • Brewster’s Angle • Total Internal Reflection (TIR) • Evanescent Waves • The Complex Refractive Index • Reflection from Metals We will derive the Fresnel equations r : reflection coefficient rTE Er cos θ − n2 − sin 2 θ = = E cos θ + n2 − sin 2 θ rTM Er − n2 cos θ + n2 − sin 2 θ = = E n2 cos θ + n2 − sin 2 θ E θ Et 2n cos θ = E n2 cos θ + n2 − sin 2 θ θr n1 n2 θt t : transmission coefficient E 2cos θ tTE = t = E cos θ + n2 − sin 2 θ tTM = Er Et n≡ ntransmitted n2 = nincident n1 EM Waves at an Interface r r r r Ei = Eoi exp ⎡i ki ⋅ r − ωi t ⎤ ⎣ ⎦ r r r r Reflected beam : Er = Eor exp ⎡i kr ⋅ r − ωr t ⎤ ⎣ ⎦ r r r r Transmitted beam : Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤ ⎣ ⎦ ( Incident beam : ) ( ) ( TE mode ) r Eoi r Ei r ki = n1k0 r kr = n1k0 r kt = n2 k0 r ki n1 n2 r Et r n2 θi n1 r kr r Eot r kt TM mode n2 r Er n1 Note the definition of the positive E-field directions in both cases. r Eor EM Waves at an Interface r r r r Ei = Eoi exp ⎡i ki ⋅ r − ωi t ⎤ ⎣ ⎦ r r r r Reflected beam : Er = Eor exp ⎡i kr ⋅ r − ωr t ⎤ ⎣ ⎦ r r r r Transmitted beam : Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤ ⎣ ⎦ ( Incident beam : ) ( ) ( ) At the boundary between the two media (the x − y plane), all waves must exist simultaneously, and the tangential component must be equal on both sides of the interface. r Therefore, for all time t and for all boundary points r on the interface, ) r ) r ) r n × Ei + n × Er = n × Et r r r r r r ) r ) r ) r ⎡ ⎤ ⎡ ⎤ ⎡ n × Eoi exp i ki ⋅ r − ωi t + n × Eor exp i kr ⋅ r − ωr t = n × Eot exp i kt ⋅ r − ωt t ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ( ) ( ) ( ) Assuming that the wave amplitudes are constant , the only way that this can be true over the entire interface and for all t is if : r r r r r r ⇒ ki ⋅ r − ωi t = kr ⋅ r − ωr t = kt ⋅ r − ωt t : Phase matching at the boundary! ( ) ( ) ( ) r Eoi r Ei r ki n1 n2 r Er n̂ r kr r Eor r rr Eot r kt r Et EM Waves at an Interface Phase matching condition: r r r r r r ki ⋅ r − ωi t = kr ⋅ r − ωr t = kt ⋅ r − ωt t ( ) ( ) ( r Ei ) r At r = 0, this results in ωi t = ωr t = ωt t ⇒ r ki n1 n2 ωi = ωr = ωt (Frequency does not change at the boundary!) At t = 0, this results in r r r r r r ⇒ ki ⋅ r = k r ⋅ r = kt ⋅ r r Er r kr r ki = n1k0 r kr = n1k0 r kt = n2 k0 r r r kt r Et n2 k0 n1k0 r kr r ki (Phases on the boundary does not change!) r r ⇒ ki ,r ,t ⋅ r = constant r r → the equation for a plane perpendicular to ki ,r ,t and r . r kt Normal r r r ⇒ ki , kr , and kt are coplanar in the plane of incidence. r r x EM Waves at an Interface r Ei At t = 0, r r r r r r ki ⋅ r = kr ⋅ r = kt ⋅ r = constant r ki θι Considering the relation for the incident and reflected beams, r r r r ki ⋅ r = k r ⋅ r ki r sin θi = kr r sin θ r ⇒ Since the incident and reflected beams are in the same medium, nω sin θi = sin θ r ki = k r = i θi = θ r : law of reflection ⇒ ⇒ c θr r Er r kr n 1 θt x n2 r kt r Et n2 k0 n1k0 Considering the relation for the incident and transmitted beams, r r r r ⇒ ki ⋅ r = kt ⋅ r ki r sin θi = kt r sin θt But the incident and transmitted beams are in different media, nω nω ki = i kt = t ni sin θi = nt sin θt : law of refraction ⇒ c c r kr θr θi θt r ki r kt Normal r r x Development of the Fresnel Equations From Maxwell ' s EM field theory, we have the boundary conditions at the interface TE-case for the TE case : Ei + Er = Et Bi cos θi − Br cos θ r = Bt cos θ t The above conditions imply that the tangential r r components of both E and B are equal on both sides of the interface. We have also assumed that μi ≅ μt ≅ μ0 , as is true for most dielectric materials. For the TM mode : Ei cos θi + Er cos θ r = Et cos θ t − Bi + Br = − Bt TM-case Development of the Fresnel Equations ⎛c⎞ Recall that E = v B = ⎜ ⎟ B ⇒ ⎝n⎠ B= nE c n2 TE-case Let n1 = refractive index of incident medium n2 = refractive index of refracting medium n1 For the TE mode : Ei + Er = Et n1 Ei cos θi − n1 Er cos θ r = n2 Et cos θ t TM-case n2 For the TM mode : Ei cos θi + Er cos θ r = Et cos θt − n1 Ei + n1 Er = − n2 Et n1 Development of the Fresnel Equations Eliminating Et from each set of equations n2 and solving for the reflection coefficient we obtain : TE case : rTE = TM case : rTM = where n = Er cos θi − n cos θt = Ei cos θi + n cos θt TE-case n1 Er −n cos θi + cos θt = n cos θi + cos θt Ei n2 n1 TM-case n2 We know that sin θi = n sin θt sin 2 θi n cos θt = n 1 − sin θt = n 1 − = 2 n 2 n 2 − sin 2 θi n1 Now we have derived the Fresnel Equations Substituting we obtain the Fresnel equations for reflection coefficients r : TE case : rTE TM case : rTM cos θi − E = r = Ei cos θi + n 2 − sin 2 θi n2 n − sin θi 2 2 2 2 2 Er − n cos θi + n − sin θi = = Ei n 2 cos θi + n 2 − sin 2 θi n≡ TE-case n2 n1 n1 For the transmission coefficient t : 2 cos θi E TE case : tTE = t = Ei cos θi + n 2 − sin 2 θi TM case : tTM = TE : 2n cos θi Et = Ei n 2 cos θi + n 2 − sin 2 θi tTE = rTE + 1 TM : ntTM = 1 − rTM TM-case n2 These just mean the boundary conditions. For the TE case : Ei + Er = Et For the TM mode : − Bi + Br = − Bt n1 Power : Reflectance (R) and Transmittance (T) The quantities r and t are ratios of electric field amplitudes. The ratios R and T are the ratios of reflected and transmitted powers, respectively, to the incident power : P P R= r T = t Pi Pi A From conservation of energy : ⇒ 1= R +T Pi = Pr + Pt We can express the power in each of the fields in terms of the product of an irradiance and area : Pi = I i Ai Pr = I r Ar Pt = I t At ⇒ I i Ai = I r Ar + I t At I i A cos θi = I r A cos θ r + I t A cos θt I i cos θi = I r cos θ r + I t cos θt But I = ⇒ 2 I out cos θ out ⎛ nout Eout cos θ out Power _ ratio = =⎜ I in cos θin ⎜⎝ nin Ein 2 cos θin 1 1 1 n1ε 0 cE02i cos θi = n1ε 0 cE02r cos θ r + n2ε 0 cE02t cos θ t 2 2 2 2 2 2 ⎛ cos θt ⎞ E02t E0 r n2 E0t cos θt E0 r ⇒ 1= 2 + = 2 + n⎜ ⎟ 2 = R+T co s E0i n1 E02i cos θi E0i θ i ⎠ E0 i ⎝ 1 n ε 0 cE02 ⇒ 2 E02r R = 2 = r2 E0i ⎛ cos θt ⎞ E02t ⎛ cos θt ⎞ 2 T = n⎜ ⎟t ⎟ 2 = n⎜ ⎝ cos θi ⎠ E0i ⎝ cos θi ⎠ R = rr* = r 2 ⎛ cos θ t T = ⎜⎜ n ⎝ cos θ i ⎛ cos θ t ⎞ ⎟⎟tt* = ⎜⎜ n ⎝ cos θ i ⎠ ⎞ 2 ⎟⎟ t ⎠ ⎞ ⎟ ⎟ ⎠ 23-2. External and Internal Reflection rTE = cos θ i − n 2 − sin 2 θ i co s θ i + n 2 − sin 2 θ i rTM = − n 2 cos θ i + t n 2 − sin 2 θ i n 2 c os θ i + n 2 − sin 2 θ i tTE ,TM > 0 rTE ,TM > 0 rTM External Reflection [ n = n2 / n1 > 1 ] ⇒ n2 > n1 ⇒ n = n2 / n1 > 1 ⇒ ( n 2 − sin 2 θ ) ≥ 0 rTE rTE ,TM < 0 n=1.50 ⇒ rTE ,TM are always real ⇒ If rTE ,TM > 0 then there are no phase changes after reflection. ⇒ If rTE ,TM < 0 then there are always π (= 180o ) phase changes. → rTE ,TM = − rTE ,TM = eiπ rTE ,TM Note for the TM case : ⇒ rTM (θ = θ p ) = 0 when θ p = tan n −1 Brewster’s angle (or, polarizing angle) (No reflection of TM mode) Internal Reflection [ n = n2 / n1 < 1 ] rTE = cos θ i − n 2 − sin 2 θ i co s θ i + n 2 − sin 2 θ i rTM = − n 2 cos θ i + n 2 c os θ i + rTE ,TM > 0 n 2 − sin 2 θ i n 2 − sin 2 θ i TIR region rTE ,TM < 0 n1 > n2 ⇒ n = n2 / n1 < 1 ⇒ ( n 2 − sin 2 θ ) > 0, or , ( n 2 − sin 2 θ ) < 0 ⇒ If ( n 2 − sin 2 θ ) > 0, rTE ,TM are always real → If rTE ,TM > 0 then there are no phase changes after reflection. → If rTE ,TM < 0 then there are π (= 180o ) phase changes. ⇒ If ( n 2 − sin 2 θ ) = 0, rTE ,TM =1 → sin θ c = n = (n2 / n1 ) critical angle ⇒ If ( n 2 − sin 2 θ ) < 0, rTE ,TM =1, BUT rTE ,TM are complex ! → rTE ,TM =1 Note Brewster's angle (θ p = tan −1 n ) for the TM case : rTM = 0 Total internal reflection (TIR) when θ > θc → rTE ,TM = rTE ,TM eiφ = eiφ → φ (-π ~ +π ) phase change may occur after reflection Derivation of Brewster’s Angle θc Brewster's angle θ p ( for polarizing angle) : rTM (θ p ) = ⇒ − n 2 cos θ p + n 2 − sin 2 θ p n cos θ p + n − sin θ p 2 2 2 θp =0 R n cos θ p = n − sin θ p 4 2 2 2 n 4 cos 2 θ p − n 2 + sin 2 θ p = (n − 1) ⎡⎣ n c os θ p − sin θ p ⎤⎦ = 0 2 2 θp 2 2 TE TM internal reflection ⇒ θ p = tan −1 n For n = 1.50, θ p = 56.31° Brewster ‘s angle : tan θ p = n : n > 1 or n < 1 Æ External & Internal reflections, but TM-polarization only Critical angle : sin θ c = n :n < 1 Æ TE & TM polarizations, but Internal reflection only external reflection Total Internal Reflection Internal reflection : n = n2 <1 n1 (TIR) R θc R=1 For θ ≥ θ c = sin −1 n , called total internal reflection(TIR), ⇒ r = 1 and R = rr* = 1 for both (TE and TM) cases. ⇒ r is a complex number rTE rTM internal reflection 2 2 Er cos θi − i sin θi − n = = Ei cos θi + i sin 2 θi − n 2 2 2 2 Er − n cos θi + i sin θi − n = = Ei n 2 cos θi + i sin 2 θi − n 2 r Complex value 23-3. Phase changes on reflection External reflection tTE ,TM > 0 Phase shift after External Reflection rTE ,TM > 0 rTM rTE ,TM is always a real number for external reflection, then the phase shift is 0° for rTE ,TM > 0, and the phase shift is 180°(= π ) for rTE ,TM < 0. rTE ,TM < 0 rTE n=1.50 External Reflection TE For TE case, π phase shift for all incident angles External Reflection TM For TM case, π phase shift for θ < θp No phase shift for θ > θp Phase shift after Internal Reflection Internal reflection ⇒ rTE > 0 for θ < θ c = sin −1 n ⇒ rTE is complex in TIR region where θ > θ c Complex value → rTE = rTE eiφTE = eiφTE In TIR region ⇒ rTM > 0 for θ < θ p = tan −1 n : θ > θc ⇒ rTM < 0 for θ p < θ < θ c → rTM = − rTM = eiφTM rTM → φTM = π ⇒ rTM is complex in TIR region where θ > θ c → rTM = rTM eiφTM = eiφTM TIR For TE case, no phase shift for θ < θc φTE(θ) phase shift for θ > θc TIR For TM case, no phase shift for θ < θp π phase shift for θp < θ < θc φTM(θ) phase shift for θ > θc Phase shifts on total Internal Reflection for both TE- and TM-cases When θ ≥ θ c (TIR case) then r is complex and for both the TE and TM cases has the form : a − ib cos α − i sin α e − iα sin α b r= = = + iα = e − i 2α = eiφ ⇒ tan α = = φ = − 2α a + ib cos α + i sin α e a cos α φ is the phase shift on total internal reflection(TIR ). TE case : rTE a = cos θi ⇒ φTE 2 2 Er cos θi − i sin θ i − n = = Ei cos θi + i sin 2 θi − n 2 b = sin 2 θ i − n 2 Internal reflection sin 2 θi − n 2 ⎛ φTE ⎞ tan α = tan ⎜ − ⎟= cos θ ⎝ 2 ⎠ ⎛ sin 2 θ − n 2 i = − 2 tan ⎜ ⎜ cos θi ⎝ −1 ⎞ ⎟ ⎟ ⎠ : θi > θ c A similar analysis for the TM case gives : φTM ⎛ sin 2 θ − n 2 i = π − 2 tan ⎜ 2 ⎜ n cos θ i ⎝ −1 ⎞ ⎟ ⎟ ⎠ : θi > θ c TIR (Complex r ) Therefore, rTE ,TM after TIR is ……….. Internal reflection rTE ,TM For TIR case ( θ incident > θ c ) rTE rTM 2 2 Er cos θi − i sin θi − n = = Ei cos θi + i sin 2 θi − n 2 Complex value 2 2 2 Er −n cos θi + i sin θi − n = = Ei n 2 cos θi + i sin 2 θi − n 2 φTM ⎛ sin 2 θ − n 2 i = π − 2 tan ⎜ 2 ⎜ n cos θ ⎝ φTE ⎛ sin 2 θ − n 2 i = − 2 tan ⎜ ⎜ cos θ ⎝ −1 −1 ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ φTE ,TM Summary of Phase Shifts on Internal Reflection φTM φTE ⎧ 0o ⎪ ⎪ ⎪⎪ = ⎨π (= 180o ) ⎪ ⎛ sin 2 θ − n 2 ⎪ i −1 ⎪ π − 2 tan ⎜⎜ n 2 cos θ ⎝ ⎩⎪ ⎧ 0o ⎪⎪ ⎛ sin 2 θ − n 2 =⎨ i −1 ⎪−2 tan ⎜⎜ cos θ ⎪⎩ ⎝ Δφ = φTM − φTE ⎧ = 0o ⎪ ⎨= π ⎪ o 0 > ⎩ θ <θ p' Internal reflection TIR (Complex r ) θ p' < θ <θ c ⎞ ⎟ ⎟ ⎠ θ <θ c θ <θ c ⎞ ⎟ ⎟ ⎠ θ >θ c θ <θ p θ p < θ <θ c θc < θ φTM φTE Δφ Fresnel Rhomb 3π near θ i = 53o when n = 1.5 4 → After two consequentive TIRs, Note φTM − φTE = → φTM − φTE = 3π 2 → Δφ = φTM − φTE = π φTE 2 → Quarter − wave retarder Linearly polarized light (45o) Circularly Polarized light φTM Δφ Quarter-wave retardation after TIR Note φTM − φTE = π 2 φTM near θ i = 69 when n = ??? → Δφ = φTM − φTE = o π 2 → Quarter − wave retarder Linearly polarized light (45o) φTE Circularly Polarized light n Δφ 23-5. Evanescent Waves at an Interface r r r r Ei = Eoi exp ⎡i ki ⋅ r − ωi t ⎤ ⎣ ⎦ r r r r Reflected beam : Er = Eor exp ⎡i kr ⋅ r − ωr t ⎤ ⎣ ⎦ r r r r Transmitted beam : Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤ ⎣ ⎦ ( Incident beam : ) ( ) ( ) For the transmitted beam : r r Et = Eot exp ⎡i kt ⋅ r − ωt t ⎤ ⎣ ⎦ ( ) r r ) ) ) ) kt ⋅ r = ( kt sin θt x + kt cos θt z ) ⋅ ( x x + zz ) = kt ( x sin θ t + z cos θt ) sin 2 θi But , cos θt = 1 − sin θt = 1 − n 2 When sin θi > n ( total internal reflection), then : sin 2 θ i cos θt = i −1 n ⇒ a purely imaginary number Evanescent Waves at an Interface For the transmitted beam with an TIR condition ( sin θi > n ) , we can write the phase factor as : ⎛ sin θ r r t + iz k t ⋅ r = kt ⎜ x ⎜ n ⎝ ⎡ ⎛ k x sin θt ⎞⎤ Et = E0t exp ⎢i ⎜ t − ωt ⎟ ⎥ exp ( −α z ) n ⎠⎦ ⎣ ⎝ ⎞ sin θi −1 ⎟ ⎟ n ⎠ 2 Defining the coefficient α : α = kt sin θi 2π −1 = λt n 2 z sin θ i −1 n 2 n2 n1 > n2 h We can write the transmitted wave as : ⎡ ⎛ k x sin θt ⎞⎤ − ωt ⎟ ⎥ exp ( −α z ) Et = E0t exp ⎢i ⎜ t n ⎠⎦ ⎣ ⎝ n1 The evanescent wave amplitude will decay rapidly as it penetrates into the lower refractive index medium. ⎛ ⎞ Penetration depth: Et = ⎜ e ⎟ Eot ⇒ h = α = ⎝ ⎠ 1 1 Note that the incident and reflection waves form a standing wave in x direction λ sin 2 θ i −1 2π n2 x Frustrated TIR d Tp = fraction of intensity transmitted across gap n1=n2=1.517 1.65 Zhu et al., “Variable Transmission Output Coupler and Tuner for Ring Laser Systems,” Appl. Opt. 24, 3610-3614 (1985). d/λ Frustrated Total Internal Reflectance Pellin-Broca prism d Zhu et al., “Variable Transmission Output Coupler and Tuner for Ring Laser Systems,” Appl. Opt. 24, 3610-3614 (1985). d = 1 ~ λ: changing the reflectance Rotation: changing the wavelength resonant at θB 23-6. Complex Refractive Index ⎛ σ ⎞ For a material with conductivity (σ ) : n% = 1 + i ⎜ ⎟ = nR + i nI ⎝ ε0 ω ⎠ ⎛ σ ⎞ 2 2 n% 2 = 1 + i ⎜ ⎟ = nR − nI + i 2nR nI ⎝ ε0 ω ⎠ Solving for the real and imaginary components we obtain : nR2 − nI2 = 1 2nR nI = σ ε0 ω ⇒ ⇒ ⎛ σ ⎞ nI4 − nI2 − ⎜ ⎟ =0 2 ε ω ⎝ 0 ⎠ 2 ⇒ ⎛ σ ⎞ 2 ⎜ ⎟ − nI = 1 ⎝ 2 nI ε 0 ω ⎠ nR = σ 2 nI ε 0 ω 2 From the quadratic solution we obtain : 2 ⎛ σ ⎞ ⎛ σ ⎞ + + 1± 1+ 4⎜ 1 1 4 ⎜ ⎟ ⎟ 2 2 ω ε ω ε ⎝ 0 ⎠ ⎝ 0 ⎠ ⇒ nI2 = nI2 = 2 2 We need to take the positive root because nI is a real number. 2 Complex Refractive Index Substituting our expression for the complex refractive index back into our expression for the electric field we obtain r r r r E = E0 exp ⎡i k ⋅ r − ωt ⎤ ⎣ ⎦ r r ω ⎧ ⎡ ⎤⎫ = E0 exp ⎨i ⎢( nR + i nI ) ( uˆk ⋅ r ) − ωt ⎥ ⎬ c ⎦⎭ ⎩ ⎣ r r r ⎤ ⎧ ⎡n ⎡ n ω ⎤⎫ = E0 exp ⎨i ω ⎢ R ( uˆk ⋅ r ) − t ⎥ ⎬ exp ⎢ − I ( uˆk ⋅ r ) ⎥ c ⎦⎭ ⎣ ⎦ ⎩ ⎣ c ( ) The first exponential term is oscillatory. The EM wave propagates with a velocity of nR / c. The second exponential has a real argument (absorbed). Complex Refractive Index r r r r ⎤ ⎧ ⎡n ⎤⎫ ⎡ n ω E = E0 exp ⎨i ω ⎢ R ( uˆk ⋅ r ) − t ⎥ ⎬ exp ⎢ − I ( uˆk ⋅ r ) ⎥ c ⎣ ⎦ ⎦⎭ ⎩ ⎣ c The second term leads to absorption of the beam in metals due to inducing a current in the medium. This causes the irradiance to decrease as the wave propagates through the medium. r r r* r r * ⎡ 2 nI ω ( uˆk ⋅ r ) ⎤ I ≡ EE = E0 E0 exp ⎢ − ⎥ c ⎣ ⎦ r ⎡ 2 nI ω ( uˆk ⋅ r ) ⎤ r ˆ = I − u ⋅ r α exp I = I 0 exp ⎢ − ⎡ ( k )⎦⎤ ⎥ 0 ⎣ c ⎣ ⎦ The absorption coefficient is defined : α = 2 nI ω 4π nI = c λ 23-7. Reflection from Metals Reflection from metals is analyzed by substituting the complex refractive index n% in the Fresnel equations : TE case : rTE cos θi − E = r = Ei cos θi + TM case : rTM = n% 2 − sin 2 θi n% 2 − sin 2 θi −n% 2 cos θ i + Er = Ei n% 2 cos θi + Substituting n% = nR + i nI Reflectance n% 2 − sin 2 θi n% 2 − sin 2 θi we obtain : Er cos θi − TE case : r = = Ei cos θi + (n (n 2 R − nI2 − sin 2 θi ) + i ( 2nR nI ) 2 R − nI2 − sin 2 θi ) + i ( 2nR nI ) 2 2 Er − ⎡⎣( nR − nI ) + i ( 2nR nI ) ⎤⎦ cos θi + TM case : r = = Ei ⎡( nR2 − nI2 ) + i ( 2nR nI ) ⎤ cos θi + ⎣ ⎦ (n 2 R (n 2 R θi − nI2 − sin 2 θi ) + i ( 2nR nI ) − nI2 − sin 2 θi ) + i ( 2nR nI ) Reflection from Metals at normal incidence (θi=0) At normal incidence, θi = 0° : rTE = rTM = cos θi − n% 2 − sin 2 θi cos θi + n% 2 − sin 2 θi At normal incidence (from Hecht, page 113) = 1 − n% 1 + n% − n% 2 cos θi + n% 2 − sin 2 θi n% 2 cos θi + n% 2 − sin 2 θi ∴ r= 1 − ( nR − i nI 1 + ( nR − i nI = 1 − n% 1 + n% ) ) The power reflectance R is given by R = r r* ⎡1 − ( nR − i nI ) ⎤ ⎡1 − ( nR + i nI ) ⎤ ⎛ 1 − 2nR + nR2 + nI2 ⎞ =⎢ ⎥⎢ ⎥=⎜ 2 2 ⎟ + − + + + + + n i n n i n n n n 1 1 1 2 ( ) ( ) ⎝ R I ⎦⎣ R I ⎦ R R I ⎠ ⎣ ( n − 1) R= R 2 ( nR + 1) 2 + nI2 + nI2 visible λ
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