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R. Diamant, M. Fernández-Guasti / Optics Communications 294 (2013) 64–72 67Fig. 3. Amplitude equation solutions for (a) ‘‘stepþ’’, n i ¼ 1, n t ¼ 1:5 and (b) ‘‘step’’, n i ¼ 1:5, n t ¼ 1. The junction is at z¼0.Fig. 4. (a) Plot of the profile in Eq. (17) ‘‘lin&tanh’’, it is constructed by joining a constant function with a hyperbolic tangent. It models a gradual interface with D ¼ 2l andone discontinuity in the first derivative at z¼0, where this derivative is increasing. The amplitude equation solution A d for ‘‘lin&tanhþ’’ profile is also shown and (b) acloser look at the solution in the oscillating region.these waves point in opposite directions; if there is a maximum, itmeans the fields point in the same direction. If maxima occur atL max ¼ ð ð1=4Þ ðm=2ÞÞl and minima at L min ¼ ðm=2Þl, form ¼ 0,1,2,3 ..., the electric field waves must have phase differenceof d ¼ p between them at z¼0 (the discontinuity site), thusrevealing a phase change of p due to reflection at the junctionplane. If maxima and minima positions are interchanged, wavesmust be in phase at z¼0, revealing no phase change due toreflection at the junction plane. For the former interpretation tobe valid, the refractive index n(z) must be reasonably constantalong the interval where the local extrema are found.The location of the critical points for the ‘‘stepþ’’ case, showsan approximate phase change of p upon reflection at z¼0. Thereis a mean offset of ð0:004Þp from the exact d ¼ p value with verysmall deviation (sop=1000). For ‘‘step ’’, maxima fall close toL max ¼ ðm=2Þl and minima at L min ¼ ð ð1=4Þ m=2Þl, revealingan almost null phase change due to reflection at the junction.However, there is a mean offset of ð0:004Þp from the exact d ¼ 0value with standard deviation of s ¼ð0:002Þp. These resultsmatch very well with Fresnel predictions.4. Gradual and continuous interfaces4.1. Type C 0 profilesA type C 0 profile with only one piecewise junction can beconstructed by joining a constant function with a hyperbolictangent at z¼0:(n lin&tanh þ ðzÞ¼ n a zo0n a þðn g n a Þ tanh a 1Dz ð17ÞzZ0,Herea 1 ¼ 1 1:95ln ln 1:05 ,2 0:05 0:95so that D is the interface thickness, the distance needed for n tochange 90% of the total Dn ¼ n g n a . The function n lin&tanh þ ðzÞ iscontinuous, at the junction it is increasing while its derivativedn=dz is discontinuous and is also increasing. Interface thicknessis chosen to be reasonably soft D ¼ 2l to discard any significantsource of reflection except at the junction. Fig. 4(a) shows a plot ofthis profile. Light is always assumed to be incident from the left.The numerical solution of the corresponding amplitude equation(8) is also plotted in Fig. 4.In the case of ‘‘lin&tanhþ’’ profile, applying Eq. (11) for zo0,the reflectivity is found to be R lin&tanh þ ¼ 3:15565 10 2 %, with astandard deviation of s R ¼ 1:8 10 6 %. This reflectivity is muchsmaller than the abrupt interface result 4%, since it is related to avery gradual interface (D ¼ 2l). Yet, it is still far greater than thereflectivity computed for the analytic ‘‘tanh’’ profile stated in Eq.(13), with the same interface thickness, R tanh ¼ 3:16840 10 10 %.The first derivative discontinuity in n(z), thus enhances thereflectivity substantially.To analyze phase relationship between incident and reflectedwaves let us take as reference the reflection plane (always drawnat z¼0 in the figures). In this case, it lies where the first orderderivative is discontinuous, although the refractive index itself iscontinuous. The phase between incident and reflected waves maybe evaluated at any such region with constant refractive index.However, it is easier to consider a plane located at a multiple ofthe half-integer wavelength, L ¼ mðl=2Þ where m is an integer.From any of these planes, the incident wave travels to thereflection plane and then the reflected part travels back an integer
68R. Diamant, M. Fernández-Guasti / Optics Communications 294 (2013) 64–72number of wavelengths, and thus no phase needs to be added dueto the extra path traveled by this wave. If no phase change occursby reflection at the derivative discontinuity (L ¼ 0), the reflectedwave must be in phase with the incident wave after m wavelengthstravel. However, consider the plane L ¼ 1:0l in Fig. 4(b).The incident and reflected waves are not in phase. In the middlebetween one amplitude maximum and a contiguous minimum,these waves must have a relative phase difference of d ¼ 7ðp=2Þ.The numerical solution shown at L ¼ 1:0l in Fig. 4(b), reveals aphase difference of p=2. Therefore, there must be a phase changeof d ¼ p=2 introduced by the reflection at the plane where thederivative is discontinuous. A detailed analysis of the numericalsolution shows that there is only a small offset ofð 0:02470:004Þp=2 from the exact d ¼ p=2 value. This is certainlya very surprising result, since we know from Fresnel equations,that a phase change of p is introduced between two transparentdielectric media when the refractive index is discontinuous andincreasing. The present result shows that a phase change of p=2can be introduced between two transparent dielectrics whoserefractive index is equal at the interface but its first derivative isdiscontinuous. The former ‘‘lin&tanh’’ profile is increasing at thejunction plane as well as the derivative dn=dz, to see what wouldbe the case for an increasing profile n(z) but decreasing firstderivative dn=dz let us devise another C 0 type profile. Let us againpiecewise join a hyperbolic tangent with a constant function atz¼0, but this time placing the hyperbolic tangent first:(n tanh&lin þ ðzÞ¼ n g þðn g n a Þ tanh a 1Dz zo0ð18ÞzZ0:n gOnce more, the interface thickness is D ¼ 2l to discard anysignificant source of reflection except for the junction.Fig. 5(a) shows a plot of this profile. Again, light is assumed tobe incident from the left. Plots for the amplitude equation (8)numerical solution are shown in Fig. 5.In the case of ‘‘tanh&linþ’’, Eq. (11) must be applied tooscillations at the far left, for Lt 4l, since the medium is notsufficiently homogeneous closer to L ¼ 0. Oscillations are distortedwhen the refractive index varies considerably as a functionof position. Computed reflectivity is R tanh&lin þ ¼ 6:27498 10 3 %with a standard deviation of s R ¼ 7:115 10 5 . The maxima andminima are approximately located at L max ¼ð1=8 m=2Þl andL min ¼ð 1=8 m=2Þl, as seen by looking at the inset in Fig. 5(b).The phase difference between incident and reflected waves is veryclose to d ¼ p=2 atL ¼ m=2 revealing an approximate phasechange of p=2 due to reflection at the junction plane. There is anoffset of ð 0:01070:002Þp from the exact d ¼ p=2 value. Thesign in d ¼ 7ðp=2Þ depends on whether the discontinuous derivativeincreases or decreases, rather than on the refractive indexvalues at the reflection plane.By interchanging the n a and n b refractive indices for the‘‘lin&tanhþ’’ and ‘‘tanh&linþ’’ profiles, one can generate anothertwo C 0 type profiles:(n lin&tanh ðzÞ¼ n g zo0n g þðn a n g Þ tanh a 1Dz ð19ÞzZ0(n tanh&lin ðzÞ¼ n a þðn a n g Þ tanh a 1Dz n azo0zZ0:ð20ÞWhile the refractive index profile is decreasing for both profiles,the first derivative is increasing in one case and decreasing in theother. Performing the same analysis to the correspondingFig. 5. (a) Plot of the profile in Eq. (18) ‘‘tanh&lin’’, it is constructed by joining a hyperbolic tangent with a constant function. It models a gradual interface with D ¼ 2l andone discontinuity in the first derivative at z¼0, where this derivative is decreasing. The amplitude equation solution A d for ‘‘tanh&linþ’’ profile is also shown and (b) acloser look at the solution in the oscillating region.Table 2Calculated reflectivity and approximate phase difference d between incident and reflected waves at the reflection plane, z¼0, related to the type C 0 profiles. The symbol s Rstands for the reflectivity standard deviation.C 0 function Reflectivity (%) R Reflectivity relative standarddeviation (%) s R 100Rlin&tanhþ 3:15565 10 2 0.0057 p2tanh&linþ 6:27498 10 3 1.1 p2lin&tanh 6:24063 10 3 0.013 p2tanh&lin 3:16123 10 2 0.48 p2Approximate phasechange at reflection plane z¼0Offset from the exact p 2 orpvalue in p units20:02470:0030:01470:0030:00570:0060:01970:005
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