Dosya:Spherical Lens.gif
Spherical_Lens.gif (543 × 543 piksel, dosya boyutu: 6,8 MB, MIME türü: image/gif, döngüye girdi, 92 kare, 9,2 sn)
Not: Teknik sınırlamalar nedeniyle, bu gibi yüksek çözünürlüklü GIF resimlerinin küçük resimlerinde animasyon yoktur.
Bu dosya Wikimedia Commons deposunda bulunmaktadır ve diğer projeler tarafından kullanılıyor olabilir. Aşağıda dosya açıklama sayfasındaki açıklama gösteriliyor.
Özet
AçıklamaSpherical Lens.gif |
English: Visualization of light through a spherical lens, as a function of the radii of curvature of the two facets. Notice that we are far from the "thin lens" approximation. |
Tarih | |
Kaynak | https://twitter.com/j_bertolotti/status/1392058658520027138 |
Yazar | Jacopo Bertolotti |
İzin (Bu dosyanın tekrar kullanımı) |
https://twitter.com/j_bertolotti/status/1030470604418428929 |
Mathematica 12.0 code
\[Lambda]0 = 1.; k0 = N[(2 \[Pi])/\[Lambda]0]; (*The wavelength in vacuum is set to 1, so all lengths are now in units of wavelengths*)
\[Delta] = \[Lambda]0/15; \[CapitalDelta] = 40*\[Lambda]0; (*Parameters for the grid*)
\[Sigma] = 10 \[Lambda]0; (*width of the gaussian beam*)
sourcef[x_, y_] := E^(-(x^2/(2 \[Sigma]^2))) E^(-((y + \[CapitalDelta]/2)^2/(2 (\[Lambda]0/2)^2))) E^(I k0 y);
\[Phi]in = Table[Chop[sourcef[x, y]], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/ 2, \[Delta]}]; (*Discretized source*)
d = \[Lambda]0/2; (*typical scale of the absorbing layer*)
imn = Table[ Chop[5 (E^-((x + \[CapitalDelta]/2)/d) + E^((x - \[CapitalDelta]/2)/d) + E^-((y + \[CapitalDelta]/2)/d) + E^((y - \[CapitalDelta]/2)/d))], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}]; (*Imaginary part of the refractive index (used to emulate absorbing boundaries)*)
dim = Dimensions[\[Phi]in][[1]];
L = -1/\[Delta]^2*KirchhoffMatrix[GridGraph[{dim, dim}]]; (*Discretized Laplacian*)
ycenter = Map[y0 /. # &, FullSimplify[Solve[(x1)^2 + (y1 - y0)^2 == r^2, {y0}]][[All, 1, All]] ];
surface2[x_] := Evaluate[Evaluate[((Sqrt[r^2 - (x)^2] + y0) /. {y0 -> ycenter[[1]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> 100 \[CapitalDelta]} ] ];
surface1[x_] := Evaluate[((-Sqrt[r^2 - (x)^2] + y0 - 1) /. {y0 -> ycenter[[2]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> 100 \[CapitalDelta]}];
frames1 = Table[
ren = Table[ If[y < Re@Evaluate[surface2[x]] && y > Re@surface1[x], n0, 1], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}];
n = ren + I imn;
b = -(Flatten[n]^2 - 1) k0^2 Flatten[\[Phi]in]; (*Right-hand side of the equation we want to solve*)
M = L + DiagonalMatrix[ SparseArray[Flatten[n]^2 k0^2]]; (*Operator on the left-hand side of the equation we want to solve*)
\[Phi]s = Partition[LinearSolve[M, b], dim]; (*Solve the linear system*)
ImageAdd[
ArrayPlot[ Transpose[(Abs[\[Phi]in + \[Phi]s]/Max[Abs[\[Phi]in + \[Phi]s]])^2][[(4 d)/\[Delta] ;; (-4 d)/\[Delta], (4 d)/\[Delta] ;; (-4 d)/\[Delta]]], ColorFunction -> "AvocadoColors" , DataReversed -> True, Frame -> False, PlotRange -> {0, 1}],
ArrayPlot[Transpose@Re[(n - 1)/5] , DataReversed -> True , ColorFunctionScaling -> False, ColorFunction -> GrayLevel, Frame -> False]
](*Plot everything*)
, {n0, 1, 2.5, 0.24}];
surface2[x_] := Evaluate[Evaluate[((Sqrt[r^2 - (x)^2] + y0) /. {y0 -> ycenter[[1]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> (c^2 + (c \[CapitalDelta])/2 + (5 \[CapitalDelta]^2)/16)/(
2 (c + \[CapitalDelta]/4))} ] ];
surface1[x_] := Evaluate[((-Sqrt[r^2 - (x)^2] + y0 - 1) /. {y0 -> ycenter[[2]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> 100 \[CapitalDelta]}];
frames2 = Table[
ren = Table[ If[y < Re@Evaluate[surface2[x]] && y > Re@surface1[x], 2.5, 1], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}];
n = ren + I imn;
b = -(Flatten[n]^2 - 1) k0^2 Flatten[\[Phi]in]; (*Right-hand side of the equation we want to solve*)
M = L + DiagonalMatrix[ SparseArray[Flatten[n]^2 k0^2]]; (*Operator on the left-hand side of the equation we want to solve*)
\[Phi]s = Partition[LinearSolve[M, b], dim]; (*Solve the linear system*)
ImageAdd[
ArrayPlot[
Transpose[(Abs[\[Phi]in + \[Phi]s]/Max[Abs[\[Phi]in + \[Phi]s]])^2][[(4 d)/\[Delta] ;; (-4 d)/\[Delta], (4 d)/\[Delta] ;; (-4 d)/\[Delta]]], ColorFunction -> "AvocadoColors" , DataReversed -> True,
Frame -> False, PlotRange -> {0, 1}],ArrayPlot[Transpose@Re[(n - 1)/5] , DataReversed -> True , ColorFunctionScaling -> False, ColorFunction -> GrayLevel, Frame -> False]
](*Plot everything*)
, {c, -(\[CapitalDelta]/4) + 0.01, 0, \[CapitalDelta]/(20*3)}];
surface2[x_] := Evaluate[Evaluate[((Sqrt[r^2 - (x)^2] + y0) /. {y0 -> ycenter[[1]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> 5/8 \[CapitalDelta]} ] ];
surface1[x_] := Evaluate[((-Sqrt[r^2 - (x)^2] + y0 - 1) /. {y0 -> ycenter[[2]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> (c^2 + (c \[CapitalDelta])/2 + (5 \[CapitalDelta]^2)/16)/(2 (c + \[CapitalDelta]/4))}];
frames3 = Table[
ren = Table[If[y < Re@Evaluate[surface2[x]] && y > Re@surface1[x], 2.5, 1], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}];
n = ren + I imn;
b = -(Flatten[n]^2 - 1) k0^2 Flatten[\[Phi]in]; (*Right-hand side of the equation we want to solve*)
M = L + DiagonalMatrix[SparseArray[Flatten[n]^2 k0^2]]; (*Operator on the left-hand side of the equation we want to solve*)
\[Phi]s = Partition[LinearSolve[M, b], dim]; (*Solve the linear system*)
ImageAdd[
ArrayPlot[Transpose[(Abs[\[Phi]in + \[Phi]s]/Max[Abs[\[Phi]in + \[Phi]s]])^2][[(4 d)/\[Delta] ;; (-4 d)/\[Delta], (4 d)/\[Delta] ;; (-4 d)/\[Delta]]], ColorFunction -> "AvocadoColors" , DataReversed -> True, Frame -> False, PlotRange -> {0, 1}],
ArrayPlot[Transpose@Re[(n - 1)/5] , DataReversed -> True , ColorFunctionScaling -> False, ColorFunction -> GrayLevel, Frame -> False]
](*Plot everything*)
, {c, -(\[CapitalDelta]/4) + 0.01, -\[CapitalDelta]/10, \[CapitalDelta]/(20*3)}];
surface2[x_] := Evaluate[Evaluate[((Sqrt[r^2 - (x)^2] + y0) /. {y0 -> ycenter[[1]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> (c^2 + (c \[CapitalDelta])/2 + (5 \[CapitalDelta]^2)/16)/(2 (c + \[CapitalDelta]/4))} ] ];
surface1[x_] := Evaluate[((-Sqrt[r^2 - (x)^2] + y0 - 1) /. {y0 -> ycenter[[2]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> 109/120 \[CapitalDelta]}];
frames4 = Table[
ren = Table[If[y < Re@Evaluate[surface2[x]] && y > Re@surface1[x], 2.5, 1], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}];
n = ren + I imn;
b = -(Flatten[n]^2 - 1) k0^2 Flatten[\[Phi]in]; (*Right-hand side of the equation we want to solve*)
M = L + DiagonalMatrix[SparseArray[Flatten[n]^2 k0^2]]; (*Operator on the left-hand side of the equation we want to solve*)
\[Phi]s = Partition[LinearSolve[M, b], dim]; (*Solve the linear system*)
ImageAdd[
ArrayPlot[Transpose[(Abs[\[Phi]in + \[Phi]s]/Max[Abs[\[Phi]in + \[Phi]s]])^2][[(4 d)/\[Delta] ;; (-4 d)/\[Delta], (4 d)/\[Delta] ;; (-4 d)/\[Delta]]], ColorFunction -> "AvocadoColors" , DataReversed -> True, Frame -> False, PlotRange -> {0, 1}], ArrayPlot[Transpose@Re[(n - 1)/5] , DataReversed -> True , ColorFunctionScaling -> False, ColorFunction -> GrayLevel, Frame -> False]
](*Plot everything*)
, {c, 0, -(\[CapitalDelta]/4) + 0.01, -(\[CapitalDelta]/(20*3))}];
surface2[x_] := Evaluate[Evaluate[((Sqrt[r^2 - (x)^2] + y0) /. {y0 -> ycenter[[1]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> 100 \[CapitalDelta]} ] ];
surface1[x_] := Evaluate[((-Sqrt[r^2 - (x)^2] + y0 - 1) /. {y0 -> ycenter[[2]]}) /. {y1 -> -(\[CapitalDelta]/4), x1 -> \[CapitalDelta]/2, r -> (c^2 + (c \[CapitalDelta])/2 + (5 \[CapitalDelta]^2)/16)/(2 (c + \[CapitalDelta]/4))}];
frames5 = Table[
ren = Table[If[y < Re@Evaluate[surface2[x]] && y > Re@surface1[x], 2.5, 1], {x, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}, {y, -\[CapitalDelta]/2, \[CapitalDelta]/2, \[Delta]}];
n = ren + I imn;
b = -(Flatten[n]^2 - 1) k0^2 Flatten[\[Phi]in]; (*Right-hand side of the equation we want to solve*)
M = L + DiagonalMatrix[SparseArray[Flatten[n]^2 k0^2]]; (*Operator on the left-hand side of the equation we want to solve*)
\[Phi]s = Partition[LinearSolve[M, b], dim]; (*Solve the linear system*)
ImageAdd[
ArrayPlot[Transpose[(Abs[\[Phi]in + \[Phi]s]/Max[Abs[\[Phi]in + \[Phi]s]])^2][[(4 d)/\[Delta] ;; (-4 d)/\[Delta], (4 d)/\[Delta] ;; (-4 d)/\[Delta]]], ColorFunction -> "AvocadoColors" , DataReversed -> True, Frame -> False, PlotRange -> {0, 1}], ArrayPlot[Transpose@Re[(n - 1)/5] , DataReversed -> True , ColorFunctionScaling -> False, ColorFunction -> GrayLevel, Frame -> False]
](*Plot everything*)
, {c, -\[CapitalDelta]/10, -(\[CapitalDelta]/4) + 0.01, -(\[CapitalDelta]/(20*3))}];
ListAnimate[ Flatten[ Join[Table[frames1[[1]], {5}], frames1, Table[frames2[[1]], {5}], frames2, Table[frames3[[1]], {5}], frames3, Table[frames4[[1]], {5}], frames4, Table[frames5[[1]], {5}], frames5, Table[frames1[[-1]], {5}], Reverse@frames1]] ]
Lisanslama
Bu dosya Creative Commons Evrensel Kamu Malı İthafı altındadır. | |
Bu çalışmayı oluşturan kişi bu senet ile eser hakkında tüm dünya çapında telif hakkı yasaları kapsamında, yasalar tarafından izin verilen ölçülerde ve diğer benzer tüm haklarından feragat etmiş ve kamu malı olarak nitelendirmiştir. Siz bu çalışmayı ve eseri hiç bir izin almadan ticari amaçlar da dahil olmak üzere kopyalayabilir, değiştirebilir ve serbestçe dağıtabilirsiniz.
http://creativecommons.org/publicdomain/zero/1.0/deed.enCC0Creative Commons Zero, Public Domain Dedicationfalsefalse |
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betimlenen
Vikiveri ögesi olmayan bir değer
11 Mayıs 2021
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Tarih/Saat | Küçük resim | Boyutlar | Kullanıcı | Yorum | |
---|---|---|---|---|---|
güncel | 11.19, 12 Mayıs 2021 | 543 × 543 (6,8 MB) | wikimediacommons>Berto | Uploaded own work with UploadWizard |
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