Slides - Rede de Difração

Centro: Centro de Ciências Matemáticas e da Natureza (CCMN). Unidade: Instituto de Física. Curso: FÍSICA IV-A Rede de Difração • Difração por três fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Difração por N (>3) fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Resolução de um espectrógrafo: poder de resolução. Rede de Difração • Difração por três fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Difração por N (>3) fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Resolução de um espectrógrafo: poder de resolução. 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Considere válido o regime de Fraunhofer. ✓ <latexit sha1_base64="Wxl9 I98Tzu9Oz/aCfzMT0DKwo=">ACyXicjVHLTsJAFD3UF+ILd emkZi4Iq3B6JLoxsQNJgImQMy0DSl+3UiMSVP+BWf8z4B/oX 3hlLohKj07Q9c+49Z+be60SeSKRlveaMmdm5+YX8YmFpeWV1rb i+0UjCNHZ53Q29ML5wWMI9EfC6FNLjF1HMme94vOkMj1W8ecPjR ITBuRxFvOzfiB6wmWSqEZbDrhkl8WSVb0MqeBnYESslULiy9o o4sQLlL4AgCXtgSOhpwYaFiLgOxsTFhISOc9yjQNqUsjhlMGK H9O3TrpWxAe2VZ6LVLp3i0RuT0sQOaULKiwmr0wdT7WzYn/zHm tPdbcR/Z3MydWYkDsX7pJ5n91qhaJHg51DYJqijSjqnMzl1R3R d3c/FKVJIeIOIW7FI8Ju1o56bOpNYmuXfW6fibzlSs2rtZbop3 dUsasP1znNOgsVe2K+X9s0qpepSNOo8tbGOX5nmAKk5Q528r/ CIJzwbp8a1cWvcfaYauUyziW/LePgA4qWRsQ=</latexit> Considere a vista lateral de um experimento com três fendas idênticas de largura a e espaçadas pela distância d, conforme a figura. d <l ate xit sha 1_b ase 64= "2J nE9D zfk TFZ 4MQ A/u H/y DJd9 SY= ">A AC xHi cjV HLSs NAF D2N r1p fVZ dugk VwV RKp 6LI oiM sW7 ANqk WQ6 raG TB5 mJU Ir+g Fv9 NvE P9C +8M 6ag FtEJ Sc6 ce8 +Zu f6 iQi kcpz Xgr Wwu LS8 Ulw tra1 vbG 6Vt 3fa Ms5 Sxl sFn 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pd V7A Pa Isl 0W ofm xWR SK EX8 Ab f6a +I f6F 94 Z0x BL aIT kp w59 54 zc+ /1 Yl8 kyn Fe c9b C4 tLy Sn 61s La +sb lV 3N5 pJ lEq GW +wy I9 k23 MT7 ou QN5 RQ Pm/ Hk ruB 5/ OWN zr X8d aY y0R E4 Y2a xL wXu MNQ DA RzF VH X3e lt seS UH bPs eV DJQ An Zqk fF3T RR wSG FAE 4Q ijC Pl wk9 HR QgY OY uB6 mx ElC ws Q57 lE gbU pZ nDJ cYk f0 HdK uk 7Eh 7b VnY tS MTv Hp laS 0c UCa iP IkY X2 abe Kpc db sb9 5T 46n vN qG/ l3 kFx Cr cEf uX bpb 5X 52u RW GAU 1O DoJ piw +j qWO aS mq7 om 9tf ql LkE BO ncZ /i kjA zy lmf ba NJT O26 t6Jv 5l Mze o9 y3J Tv Otb 0o ArP 8c 5D5 pH 5Uq 1f HxV LdX Os lHn sY d9H NI 8T1 DD Jep ok PcA j3 jCs 3V hBZ ay xp+ pVi 7T 7OL bs h4+ AH V/j +4 =</ la tex it > O efeito da difração é obtido com a interferência resultante de todas as ondas que saem de uma mesma fenda. O efeito da interferência é o b t i d o c o m a interferência resultante e n t r e o s t r ê s r a i o s resultantes que saem de cada uma das fendas. 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Cr cEf uX bpb 5X 52u RW GAU 1O DoJ piw +j qWO aS mq7 om 9tf ql LkE BO ncZ /i kjA zy lmf ba NJT O26 t6Jv 5l Mze o9 y3J Tv Otb 0o ArP 8c 5D5 pH 5Uq 1f HxV LdX Os lHn sY d9H NI 8T1 DD Jep ok PcA j3 jCs 3V hBZ ay xp+ pVi 7T 7OL bs h4+ AH V/j +4 =</ la tex it > d <l ate xit sha 1_b ase 64= "2J nE9D zfk TFZ 4MQ A/u H/y DJd9 SY= ">A AC xHi cjV HLSs NAF D2N r1p fVZ dugk VwV RKp 6LI oiM sW7 ANqk WQ6 raG TB5 mJU Ir+g Fv9 NvE P9C +8M 6ag FtEJ Sc6 ce8 +Zu f6 iQi kcpz Xgr Wwu LS8 Ulw tra1 vbG 6Vt 3fa Ms5 Sxl sFn Ha9 T3J RD xlg qU4N 0k5 V7o C97 x+ c63 rnjq Qzi 6Ep NEt 4Pv VEU DAPm KaK ag5 tyx ak6 Ztnz wM1 Bf lqx OUX XGO AGAw ZQn BEU IQF PEh 6en DhIC Guj ylx KaH AxD nuUS JtR lmc Mjx ix/ Qd0 a6Xs xHt tac 0ak anC HpTU to4 IE1 MeS lhf Zpt 4plx 1ux v3l Pjq e82 ob+ fe4X EKt wS+ 5du lvl fna5 FY hTU 0NA NSW G0d Wx3C UzX dE3 t79 Upc ghIU 7jA cVT wsw oZ3 2j Ua2 nVv PRN /M5 ma1 XuW 52Z4 17e kAb s/x zkP 2kdV t1Y 9bt Yq9 bN8 1EX sYR+ HNM 8T1 HGJ Blr G+xF PeL YuL GFJ K/t MtQ q5Zh fl vXw AT4 0j3 E=< /lat exi t> Considere válido o regime de Fraunhofer. ✓ <latexit sha1_base64="Wxl9 I98Tzu9Oz/aCfzMT0DKwo=">ACyXicjVHLTsJAFD3UF+ILd emkZi4Iq3B6JLoxsQNJgImQMy0DSl+3UiMSVP+BWf8z4B/oX 3hlLohKj07Q9c+49Z+be60SeSKRlveaMmdm5+YX8YmFpeWV1rb i+0UjCNHZ53Q29ML5wWMI9EfC6FNLjF1HMme94vOkMj1W8ecPjR ITBuRxFvOzfiB6wmWSqEZbDrhkl8WSVb0MqeBnYESslULiy9o o4sQLlL4AgCXtgSOhpwYaFiLgOxsTFhISOc9yjQNqUsjhlMGK H9O3TrpWxAe2VZ6LVLp3i0RuT0sQOaULKiwmr0wdT7WzYn/zHm tPdbcR/Z3MydWYkDsX7pJ5n91qhaJHg51DYJqijSjqnMzl1R3R d3c/FKVJIeIOIW7FI8Ju1o56bOpNYmuXfW6fibzlSs2rtZbop3 dUsasP1znNOgsVe2K+X9s0qpepSNOo8tbGOX5nmAKk5Q528r/ CIJzwbp8a1cWvcfaYauUyziW/LePgA4qWRsQ=</latexit> Considere a vista lateral de um experimento com três fendas idênticas de largura a e espaçadas pela distância d, conforme a figura. d <l ate xit sha 1_b ase 64= "2J nE9D zfk TFZ 4MQ A/u H/y DJd9 SY= ">A AC xHi cjV HLSs NAF D2N r1p fVZ dugk VwV RKp 6LI oiM sW7 ANqk WQ6 raG TB5 mJU Ir+g Fv9 NvE P9C +8M 6ag FtEJ Sc6 ce8 +Zu f6 iQi kcpz Xgr Wwu LS8 Ulw tra1 vbG 6Vt 3fa Ms5 Sxl sFn 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No regime de Fraunhofer, o campo resultante na tela é obtido pela multiplicação da função (máscara) da difração centrada na origem pelo resultado da interferência de três ondas. 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MT7 ou QN5 RQ Pm/ Hk ruB 5/ OWN zr X8d aY y0R E4 Y2a xL wXu MNQ DA RzF VH X3e lt seS UH bPs eV DJQ An Zqk fF3T RR wSG FAE 4Q ijC Pl wk9 HR QgY OY uB6 mx ElC ws Q57 lE gbU pZ nDJ cYk f0 HdK uk 7Eh 7b VnY tS MTv Hp laS 0c UCa iP IkY X2 abe Kpc db sb9 5T 46n vN qG/ l3 kFx Cr cEf uX bpb 5X 52u RW GAU 1O DoJ piw +j qWO aS mq7 om 9tf ql LkE BO ncZ /i kjA zy lmf ba NJT O26 t6Jv 5l Mze o9 y3J Tv Otb 0o ArP 8c 5D5 pH 5Uq 1f HxV LdX Os lHn sY d9H NI 8T1 DD Jep ok PcA j3 jCs 3V hBZ ay xp+ pVi 7T 7OL bs h4+ AH V/j +4 =</ la tex it > Para simplificar, podemos considerar apenas o raio resultante de cada venda que sai do ponto médio de cada uma. Rede de Difração { <latexi t sha1_base64 ="Ea9rONy/Fm9 TFBd5e15HfgLp mA=">ACxXic jVHLSsNAFD2Nr 1pfVZdugkVwV Kp6LoQpdV7AP aIsl0WofmxWRSK EX8Abf6a+If6F 94Z0xBLaITkpw 5954zc+/1Yl8k ynFec9bC4tLyS n61sLa+sblV3N 5pJlEqGW+wyI9k 23MT7ouQN5RQP m/HkruB5/OWNz rX8daYy0RE4Y2 axLwXuMNQDARz FVHX3eltseSUHb PseVDJQAnZqkf F3TRwSGFAE4 QijCPlwk9HRQg YOYuB6mxElCws Q57lEgbUpZnDJc Ykf0HdKuk7Eh7 bVnYtSMTvHpla S0cUCaiPIkYX2 abeKpcdbsb95T 46nvNqG/l3kFx CrcEfuXbpb5X52 uRWGAU1ODoJpi w+jqWOaSmq7om 9tfqlLkEBOncZ /ikjAzylmfbaN JTO26t6Jv5lMz eo9y3JTvOtb0o ArP8c5D5pH5Uq 1fHxVLdXOslHn sYd9HNI8T1DJ epokPcAj3jCs3 VhBZayxp+pVi7T 7OLbsh4+AHV/j +4=</latexit>a <latexi t sha1_base64 ="Yf7V05o/yi IDak5XNsqY/MB 9vaI=">ACxH icjVHLSsNAFD 2Nr1pfVZdugkV wVRKp6LIoiMsW 7ANqkSd1tDJg 8xEKEV/wK1+m /gH+hfeGaegFt EJSc6ce8+Zuf 6KQ+FdJzXgrW wuLS8Ulwtra1v bG6Vt3faIsmzg LWChCdZ1/cE42 HMWjKUnHXTjH mRz1nH5+reOe OZSJM4is5SVk/ 8kZxOAwDTxLV 9G7KFafq6GXPA 9eACsxqJOUXG OABAFyRGCIQl zeBD09ODCQUp cH1PiMkKhjPc o0TanLIYZXjEj uk7ol3PsDHtl afQ6oBO4fRmpL 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Cr cEf uX bpb 5X 52u RW GAU 1O DoJ piw +j qWO aS mq7 om 9tf ql LkE BO ncZ /i kjA zy lmf ba NJT O26 t6Jv 5l Mze o9 y3J Tv Otb 0o ArP 8c 5D5 pH 5Uq 1f HxV LdX Os lHn sY d9H NI 8T1 DD Jep ok PcA j3 jCs 3V hBZ ay xp+ pVi 7T 7OL bs h4+ AH V/j +4 =</ la tex it > d <l ate xit sha 1_b ase 64= "2J nE9D zfk TFZ 4MQ A/u H/y DJd9 SY= ">A AC xHi cjV HLSs NAF D2N r1p fVZ dugk VwV RKp 6LI oiM sW7 ANqk WQ6 raG TB5 mJU Ir+g Fv9 NvE P9C +8M 6ag FtEJ Sc6 ce8 +Zu f6 iQi kcpz Xgr Wwu LS8 Ulw tra1 vbG 6Vt 3fa Ms5 Sxl sFn Ha9 T3J RD xlg qU4N 0k5 V7o C97 x+ c63 rnjq Qzi 6Ep NEt 4Pv VEU DAPm KaK ag5 tyx ak6 Ztnz wM1 Bf lqx OUX XGO AGAw ZQn BEU IQF PEh 6en DhIC Guj ylx KaH AxD nuUS JtR lmc Mjx ix/ Qd0 a6Xs xHt tac 0ak anC HpTU to4 IE1 MeS lhf Zpt 4plx 1ux v3l Pjq e82 ob+ fe4X EKt wS+ 5du lvl fna5 FY hTU 0NA NSW G0d Wx3C UzX dE3 t79 Upc ghIU 7jA cVT wsw oZ3 2j Ua2 nVv PRN /M5 ma1 XuW 52Z4 17e kAb s/x zkP 2kdV t1Y 9bt Yq9 bN8 1EX sYR+ HNM 8T1 HGJ Blr G+xF PeL YuL GFJ K/t MtQ q5Zh fl vXw AT4 0j3 E=< /lat exi t> A onda resultante da interferência dos três raios, dentro da máscara da difração, vale: R✓ <latexit sha1_ba se64="/LZiEDMqiqQEtZ4NLbXQGUmAGh 4=">ACy3icjVHLTsJAFD3UF+ILdem mkZi4Iq3 B6JLoxo0JGkESIKQtA0zoK9O pCaJLf8Ct/pfxD/QvDOWRCVGp2l75t xz7sy91419nkjLes0Zc/MLi0v5cLK6 tr6RnFzq5FEqfBY3Yv8SDRdJ2E+D1ldc umzZiyYE7g+u3ZHpyp+fcNEwqPwSo5j 1gmcQcj73HMkUc3LblsOmXS6 xZJVtvQy Z4GdgRKyVYuKL2ijhwgeUgRgCEJ+3C Q0NOCDQsxcR1MiBOEuI4z3KNA3pRUjBQ OsSP6DmjXytiQ9ipnot0eneLTK8hpYo 8EekEYXWaqeOpzqzY3JPdE51tzH93 SxXQKzEkNi/fFPlf32qFok+jnUNnGqKN aOq87Isq e6Kurn5pSpJGWLiFO5RXBD2 tHPaZ1N7El276q2j429aqVi19zJtind1 Sxqw/XOcs6BxULYr5cOLSql6ko06jx3 sYp/meYQqzlBDXc/xEU94Ns6NxLg17j6 lRi7zbOPbMh4+ANcxknY=</latexit> R✓ − kdsen(✓) <latexit sha1_ base64="9dJcY8mI768E1ERSi MYkaor9m0=">AC4nicjVHLSs NAFD3G9zvqUoRgEerCkpFl6Ibl ypWC7a UZDJqaF7MTIRSXLlzJ27 9Abf6MeIf6F94Z4ygFtEJSc6ce 8+Zuf6WRK5bovA9bg0PDI6Nj4 xOTU9MysPTd/LNcMF5naZSKhu 9JHoUJr6tQRbyRCe7FfsRP/M6u jp9ciHDNDlS3Yy3Yu8Cc9C5im i2vbSYbupLrjynLVO0GuK2JE8u Sp/cKtu+RWXLOcflA tQAnF2k/ tZzQRIAVDjhgcCRThCB4kPaeowk VGXAs94gSh0MQ5rjB2pyOGV4 xHboe06704JNaK89pVEzOiWiV5 DSwQpUsoThPVpjonxlmzv3n3j Ke+W5f+fuEVE6twQexfus/M/+p 0LQpn2DI1hFRTZhdHStctMVf XPnS1WKHDLiNA4oLgzo/zs2M 0tSu e+uZ+KvJ1KzesyI3x5u+JQ 24+nOc/eB4vVKtVTYOaqXtnWLU Y1jEMso0z01sYw/7qJP3NR7wiC crsG6sW+vuI9UaKDQL+Las+3fPj prx</latexit> R✓ − 2kdsen(✓) <latexit sha1 _base64="01k05TdbOnH4GrbA exXzQwUG5U=">AC43icjVHL SgMxFD2Or/quhRksAi6sExF0W XRjUsVq4ItJZPGdui8yGQEKe7 cuRO3/oBb/RfxD/QvIkR1CKa YWZOzr3nJPdePw2DTHney4AzOD Q8MloYG5+YnJqeKc7OHWdJLrm o8SRM5KnPMhEGsaipQIXiNJWCR X4oTvzuro6fXAiZBUl8pC5T0Y hYOw7OA84Uc3i4mGzrjpCMXdt vdvq1WXkZiK+WvkgV5vFklf2z HL7QcWCEuzaT4rPqKOFBw5Igj EUIRDMGT0nKECDylxDfSIk4QC Exe4wjhpc8oSlMGI7dK3Tbszy 8a0156ZUXM6JaRXktLFMmkSypO E9WmuiefGWbO/efeMp7bJf19 6xURq9Ah9i/dZ+Z/dboWhXNsmx oCqik1jK6OW5fcdEXf3P1SlSK HlDiNWxSXhLlRfvbZNZrM1K57y 0z81WRqVu+5zc3xpm9JA678HG c/OF4vVzbKmwcbpeqOHXUBC1jC Cs1zC1XsYR818r7GAx7x5Ajnx rl17j5SnQGrmce35dy/A3Oamy 0=</latexit> E(✓) = E0sen h kasen(✓) 2 i kasen(✓) 2 [cos ('1) + cos ('2) + cos ('3)] <latexit sha1_b ase64="WUxgEfETmdiZoTUoBCKJc6 hy30=">ADjnicjVHbtQwEJ0UE qhJdBHXixWSFtV2iZLoX2pqECV+lgk tq20WSLH6921NjfZTqUqygfyBah/A H/B2HUQsOLiKMnxOXPGM560yoTSYXj rfn37q8/2Hi4+ejx1vaT4OmzC1XW kvERK7NSXqVU8UwUfKSFzvhVJTnN04 xfpsv3Rr+85lKJsviobyo+yem8EDP BqEYqCT6f9mO94JruHsfTmaSsOU3CJ pY5Ubxo4zP9Di2/JL+oDtL2wzbWI r5Qk/a5q9RXSZWKov68TWV1UIk0Z1 /l+ytasNOI3tkVX3Vqa6CJOiFg9Aus goiB3rg1nkZfIEYplACgxpy4FCARp wBYXPGCIoUJuAg1yEpGwOocWNtFb YxTHCIrsEr9z3I0dW+De5FTWzfCUD F+JTgIv0VNinERsTiNWr21mw/4pd2N zmtpu8J+6XDmyGhbI/svXRf6vz/Si YQZHtgeBPVWMd0xl6W2t2IqJz91pT FDhZzBU9QlYmad3T0T61G2d3O31Op fbaRhzZ652Bq+mSpxwNHv41wF8NBd DB4/eGgd/LOjXoDnsML6OM8D+Ezu AcRsC8fW/kfISP/Df+Mf+27vQNc95 duCX5Z9B2y72b8=</latexit> Considere a vista lateral de um experimento com três fendas idênticas de largura a e espaçadas pela distância d, conforme a figura. d <l ate xit sha 1_b ase 64= "2J nE9D zfk TFZ 4MQ A/u H/y DJd9 SY= ">A AC xHi cjV HLSs NAF D2N r1p fVZ dugk VwV RKp 6LI oiM sW7 ANqk WQ6 raG TB5 mJU Ir+g Fv9 NvE P9C +8M 6ag FtEJ Sc6 ce8 +Zu f6 iQi kcpz Xgr Wwu LS8 Ulw tra1 vbG 6Vt 3fa Ms5 Sxl sFn Ha9 T3J RD xlg qU4N 0k5 V7o C97 x+ c63 rnjq Qzi 6Ep NEt 4Pv VEU DAPm KaK ag5 tyx ak6 Ztnz wM1 Bf lqx OUX XGO AGAw ZQn BEU IQF PEh 6en DhIC Guj ylx KaH AxD nuUS JtR lmc Mjx ix/ Qd0 a6Xs xHt tac 0ak anC HpTU to4 IE1 MeS lhf Zpt 4plx 1ux v3l Pjq e82 ob+ fe4X EKt wS+ 5du lvl fna5 FY hTU 0NA NSW G0d Wx3C UzX dE3 t79 Upc ghIU 7jA cVT wsw oZ3 2j Ua2 nVv PRN /M5 ma1 XuW 52Z4 17e kAb s/x zkP 2kdV t1Y 9bt Yq9 bN8 1EX sYR+ HNM 8T1 HGJ Blr G+xF PeL YuL GFJ K/t MtQ q5Zh fl vXw AT4 0j3 E=< /lat t>exi y <latexit sha1_base64="q2eCfEkXjhGZ0qFnP6LvTpWDLtY=" >ACxHicjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LIoiMsW7ANqkWQ6rUMnD5KJUIr+gFv9NvEP9C+8M6agFtEJSc6ce8+Zuf6sRSpc pzXgrWwuLS8Ulwtra1vbG6Vt3faZQljLdYJKOk63splyLkLSWU5N04V7gS97x+c63rnjSqi8EpNYt4PvFEohoJ5iqjm5KZcaq OWfY8cHNQb4aUfkF1xgAkOGABwhFGEJDyk9PbhwEBPXx5S4hJAwcY57lEibURanDI/YMX1HtOvlbEh7ZkaNaNTJL0JKW0ckCai 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GAU 1O DoJ piw +j qWO aS mq7 om 9tf ql LkE BO ncZ /i kjA zy lmf ba NJT O26 t6Jv 5l Mze o9 y3J Tv Otb 0o ArP 8c 5D5 pH 5Uq 1f HxV LdX Os lHn sY d9H NI 8T1 DD Jep ok PcA j3 jCs 3V hBZ ay xp+ pVi 7T 7OL bs h4+ AH V/j +4 =</ la tex it > O objetivo é entender na prática a interferência com três ondas. O método de fasores ajuda nessa análise. Rede de Difração { <latexi t sha1_base64 ="Ea9rONy/Fm9 TFBd5e15HfgLp mA=">ACxXic jVHLSsNAFD2Nr 1pfVZdugkVwV Kp6LoQpdV7AP aIsl0WofmxWRSK EX8Abf6a+If6F 94Z0xBLaITkpw 5954zc+/1Yl8k ynFec9bC4tLyS n61sLa+sblV3N 5pJlEqGW+wyI9k 23MT7ouQN5RQP m/HkruB5/OWNz rX8daYy0RE4Y2 axLwXuMNQDARz FVHX3eltseSUHb PseVDJQAnZqkf F3TRwSGFAE4 QijCPlwk9HRQg YOYuB6mxElCws Q57lEgbUpZnDJc Ykf0HdKuk7Eh7 bVnYtSMTvHpla S0cUCaiPIkYX2 abeKpcdbsb95T 46nvNqG/l3kFx CrcEfuXbpb5X52 uRWGAU1ODoJpi w+jqWOaSmq7om 9tfqlLkEBOncZ /ikjAzylmfbaN JTO26t6Jv5lMz eo9y3JTvOtb0o ArP8c5D5pH5Uq 1fHxVLdXOslHn sYd9HNI8T1DJ epokPcAj3jCs3 VhBZayxp+pVi7T 7OLbsh4+AHV/j +4=</latexit>a <latexi t sha1_base64 ="Yf7V05o/yi IDak5XNsqY/MB 9vaI=">ACxH icjVHLSsNAFD 2Nr1pfVZdugkV wVRKp6LIoiMsW 7ANqkSd1tDJg 8xEKEV/wK1+m /gH+hfeGaegFt EJSc6ce8+Zuf 6KQ+FdJzXgrW wuLS8Ulwtra1v bG6Vt3faIsmzg LWChCdZ1/cE42 HMWjKUnHXTjH mRz1nH5+reOe OZSJM4is5SVk/ 8kZxOAwDTxLV 9G7KFafq6GXPA 9eACsxqJOUXG OABAFyRGCIQl zeBD09ODCQUp cH1PiMkKhjPc o0TanLIYZXjEj uk7ol3PsDHtl afQ6oBO4fRmpL 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Cr cEf uX bpb 5X 52u RW GAU 1O DoJ piw +j qWO aS mq7 om 9tf ql LkE BO ncZ /i kjA zy lmf ba NJT O26 t6Jv 5l Mze o9 y3J Tv Otb 0o ArP 8c 5D5 pH 5Uq 1f HxV LdX Os lHn sY d9H NI 8T1 DD Jep ok PcA j3 jCs 3V hBZ ay xp+ pVi 7T 7OL bs h4+ AH V/j +4 =</ la tex it > d <l ate xit sha 1_b ase 64= "2J nE9D zfk TFZ 4MQ A/u H/y DJd9 SY= ">A AC xHi cjV HLSs NAF D2N r1p fVZ dugk VwV RKp 6LI oiM sW7 ANqk WQ6 raG TB5 mJU Ir+g Fv9 NvE P9C +8M 6ag FtEJ Sc6 ce8 +Zu f6 iQi kcpz Xgr Wwu LS8 Ulw tra1 vbG 6Vt 3fa Ms5 Sxl sFn Ha9 T3J RD xlg qU4N 0k5 V7o C97 x+ c63 rnjq Qzi 6Ep NEt 4Pv VEU DAPm KaK ag5 tyx ak6 Ztnz wM1 Bf lqx OUX XGO AGAw ZQn BEU IQF PEh 6en DhIC Guj ylx KaH AxD nuUS JtR lmc Mjx ix/ Qd0 a6Xs xHt tac 0ak anC HpTU to4 IE1 MeS lhf Zpt 4plx 1ux v3l Pjq e82 ob+ fe4X EKt wS+ 5du lvl fna5 FY hTU 0NA NSW G0d Wx3C UzX dE3 t79 Upc ghIU 7jA cVT wsw oZ3 2j Ua2 nVv PRN /M5 ma1 XuW 52Z4 17e kAb s/x zkP 2kdV t1Y 9bt Yq9 bN8 1EX sYR+ HNM 8T1 HGJ Blr G+xF PeL YuL GFJ K/t MtQ q5Zh fl vXw AT4 0j3 E=< /lat exi t> A onda resultante da interferência dos três raios, dentro da máscara da difração, vale E(✓) = E0sen h kasen(✓) 2 i kasen(✓) 2 [cos ('1) + cos ('2) + cos ('3)] <latexit sha1_b ase64="WUxgEfETmdiZoTUoBCKJc6 hy30=">ADjnicjVHbtQwEJ0UE qhJdBHXixWSFtV2iZLoX2pqECV+lgk tq20WSLH6921NjfZTqUqygfyBah/A H/B2HUQsOLiKMnxOXPGM560yoTSYXj rfn37q8/2Hi4+ejx1vaT4OmzC1XW kvERK7NSXqVU8UwUfKSFzvhVJTnN04 xfpsv3Rr+85lKJsviobyo+yem8EDP BqEYqCT6f9mO94JruHsfTmaSsOU3CJ pY5Ubxo4zP9Di2/JL+oDtL2wzbWI r5Qk/a5q9RXSZWKov68TWV1UIk0Z1 /l+ytasNOI3tkVX3Vqa6CJOiFg9Aus goiB3rg1nkZfIEYplACgxpy4FCARp wBYXPGCIoUJuAg1yEpGwOocWNtFb YxTHCIrsEr9z3I0dW+De5FTWzfCUD F+JTgIv0VNinERsTiNWr21mw/4pd2N zmtpu8J+6XDmyGhbI/svXRf6vz/Si YQZHtgeBPVWMd0xl6W2t2IqJz91pT FDhZzBU9QlYmad3T0T61G2d3O31Op fbaRhzZ652Bq+mSpxwNHv41wF8NBd DB4/eGgd/LOjXoDnsML6OM8D+Ezu AcRsC8fW/kfISP/Df+Mf+27vQNc95 duCX5Z9B2y72b8=</latexit> R✓ <latexit sha1_ba se64="/LZiEDMqiqQEtZ4NLbXQGUmAGh 4=">ACy3icjVHLTsJAFD3UF+ILdem mkZi4Iq3 B6JLoxo0JGkESIKQtA0zoK9O pCaJLf8Ct/pfxD/QvDOWRCVGp2l75t xz7sy91419nkjLes0Zc/MLi0v5cLK6 tr6RnFzq5FEqfBY3Yv8SDRdJ2E+D1ldc umzZiyYE7g+u3ZHpyp+fcNEwqPwSo5j 1gmcQcj73HMkUc3LblsOmXS6 xZJVtvQy Z4GdgRKyVYuKL2ijhwgeUgRgCEJ+3C Q0NOCDQsxcR1MiBOEuI4z3KNA3pRUjBQ OsSP6DmjXytiQ9ipnot0eneLTK8hpYo 8EekEYXWaqeOpzqzY3JPdE51tzH93 SxXQKzEkNi/fFPlf32qFok+jnUNnGqKN aOq87Isq e6Kurn5pSpJGWLiFO5RXBD2 tHPaZ1N7El276q2j429aqVi19zJtind1 Sxqw/XOcs6BxULYr5cOLSql6ko06jx3 sYp/meYQqzlBDXc/xEU94Ns6NxLg17j6 lRi7zbOPbMh4+ANcxknY=</latexit> R✓ − kdsen(✓) <latexit sha1_ base64="9dJcY8mI768E1ERSi MYkaor9m0=">AC4nicjVHLSs NAFD3G9zvqUoRgEerCkpFl6Ibl ypWC7a UZDJqaF7MTIRSXLlzJ27 9Abf6MeIf6F94Z4ygFtEJSc6ce 8+Zuf6WRK5bovA9bg0PDI6Nj4 xOTU9MysPTd/LNcMF5naZSKhu 9JHoUJr6tQRbyRCe7FfsRP/M6u jp9ciHDNDlS3Yy3Yu8Cc9C5im i2vbSYbupLrjynLVO0GuK2JE8u Sp/cKtu+RWXLOcflA tQAnF2k/ tZzQRIAVDjhgcCRThCB4kPaeowk VGXAs94gSh0MQ5rjB2pyOGV4 xHboe06704JNaK89pVEzOiWiV5 DSwQpUsoThPVpjonxlmzv3n3j Ke+W5f+fuEVE6twQexfus/M/+p 0LQpn2DI1hFRTZhdHStctMVf XPnS1WKHDLiNA4oLgzo/zs2M 0tSu e+uZ+KvJ1KzesyI3x5u+JQ 24+nOc/eB4vVKtVTYOaqXtnWLU Y1jEMso0z01sYw/7qJP3NR7wiC crsG6sW+vuI9UaKDQL+Las+3fPj prx</latexit> R✓ − 2kdsen(✓) <latexit sha1 _base64="01k05TdbOnH4GrbA exXzQwUG5U=">AC43icjVHL SgMxFD2Or/quhRksAi6sExF0W XRjUsVq4ItJZPGdui8yGQEKe7 cuRO3/oBb/RfxD/QvIkR1CKa YWZOzr3nJPdePw2DTHney4AzOD Q8MloYG5+YnJqeKc7OHWdJLrm o8SRM5KnPMhEGsaipQIXiNJWCR X4oTvzuro6fXAiZBUl8pC5T0Y hYOw7OA84Uc3i4mGzrjpCMXdt vdvq1WXkZiK+WvkgV5vFklf2z HL7QcWCEuzaT4rPqKOFBw5Igj EUIRDMGT0nKECDylxDfSIk4QC Exe4wjhpc8oSlMGI7dK3Tbszy 8a0156ZUXM6JaRXktLFMmkSypO E9WmuiefGWbO/efeMp7bJf19 6xURq9Ah9i/dZ+Z/dboWhXNsmx oCqik1jK6OW5fcdEXf3P1SlSK HlDiNWxSXhLlRfvbZNZrM1K57y 0z81WRqVu+5zc3xpm9JA678HG c/OF4vVzbKmwcbpeqOHXUBC1jC Cs1zC1XsYR818r7GAx7x5Ajnx rl17j5SnQGrmce35dy/A3Oamy 0=</latexit> Sem o efeito da difração, as ondas resultantes de cada fenda, em função da distância da última fenda ao ponto considerado e em função do tempo, fica Rede de Difração E(φ) = cos(kR✓ − !t) + cos(kR✓ − !t − φ) + cos(kR✓ − !t − 2φ) <lat exit sh a1_base 64="0PD 3j+ZP +IMjVtL 7Korj1N 3z08="> ADMni cjVHLSs NAFL1Nf dR31aW bwSK0iC UtFd0IR RFcqtgH GCmT6bQ dmRCMh Gk+Ff+i QvRnbh R8AfceW eM4AMfE 5KcOfec M3Nn3NA TsbLt24 yVHRufm MxNTc/ Mzs0v5B eXmrFMI sYbTHoy ars05p4 IeEMJ5f F2GHqu x5vucM9 XW+d8y gWMjhRF yE/82k/ ED3BqEK qk/f3i0 4ECWyQ xwm4+Lw uOoAV eUbBH+ rxPiSqR 9Z+LGpq A3zVI+ rkC3bZN oN8B5U FCAdhz J/Aw50Q QKDBHzg EIBC7AG FGJ9TqI ANIXJnM EIuQiRM ncMlTK M3QRVHB UV2iN8+ zk5TNsC 5zoyNm+ EqHr4RO gmsoUei LkKsVyO mnphkz f6UPTKZ em8X+Hf TLB9ZBQ Nk/K9K /r070o 6MG26UF gT6Fhd HcsTUnM qeidkw9 dKUwIkd O4i/UIM TPO93Mm xhOb3vX ZUlN/N ErN6jlL tQk86V3 iBVe+Xu d30KyWK 7Xy5lGt UN9Nrzo HK7AKRb zPLajD ARxCA7O v4SVjZb LWlXVn3 VsPb1Ir k3qW4dO wnl8B9i Sz9g= </latex it> Esse campo varia com o tempo. As fases que diferenciam cada campo são phi (segundo termo da soma) e 2phi (terceiro termo da soma). Sem o efeito da difração, as ondas resultantes de cada fenda, em função da distância da última fenda ao ponto considerado e em função do tempo, fica Rede de Difração E(φ) = cos(kR✓ − !t) + cos(kR✓ − !t − φ) + cos(kR✓ − !t − 2φ) <lat exit sh a1_base 64="0PD 3j+ZP +IMjVtL 7Korj1N 3z08="> ADMni cjVHLSs NAFL1Nf dR31aW bwSK0iC UtFd0IR RFcqtgH GCmT6bQ dmRCMh Gk+Ff+i QvRnbh R8AfceW eM4AMfE 5KcOfec M3Nn3NA TsbLt24 yVHRufm MxNTc/ Mzs0v5B eXmrFMI sYbTHoy ars05p4 IeEMJ5f F2GHqu x5vucM9 XW+d8y gWMjhRF yE/82k/ ED3BqEK qk/f3i0 4ECWyQ xwm4+Lw uOoAV eUbBH+ rxPiSqR 9Z+LGpq A3zVI+ rkC3bZN oN8B5U FCAdhz J/Aw50Q QKDBHzg EIBC7AG FGJ9TqI ANIXJnM EIuQiRM ncMlTK M3QRVHB UV2iN8+ zk5TNsC 5zoyNm+ EqHr4RO gmsoUei LkKsVyO mnphkz f6UPTKZ em8X+Hf TLB9ZBQ Nk/K9K /r070o 6MG26UF gT6Fhd HcsTUnM qeidkw9 dKUwIkd O4i/UIM TPO93Mm xhOb3vX ZUlN/N ErN6jlL tQk86V3 iBVe+Xu d30KyWK 7Xy5lGt UN9Nrzo HK7AKRb zPLajD ARxCA7O v4SVjZb LWlXVn3 VsPb1Ir k3qW4dO wnl8B9i Sz9g= </latex it> Esse campo varia com o tempo. As fases que diferenciam cada campo são phi (segundo termo da soma) e 2phi (terceiro termo da soma). O desafio é determinar a amplitude de E. Ou seja, uma expressão constante que seja independente da variável temporal t, que constitui um campo que oscila com frequência omega. Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária: Rede de Difração O método de fasores simplifica essa soma simétrica de funções harmônicas em função de um cosseno de omega. O método consiste em associar cada termo com um vetor de amplitude unitária em um espaço abstrato (x,y) pelas relações: ~v0 = [1, 0] <lat exit s ha1_ba se64="5 HIO4pb bPdzl5 Yhf5AM6 u8QNJ+ g=">A C1nicj VHLSsN AFD2Nr1 pfqS7d BIvgQk oiFd0IR TcuK9g H1FKS6 bSGpklI JpVS6k 7c+gNu 9ZPEP9C /8M6Yg lpEJyQ5 c+49Z+ be64Se GwvTfM1 oc/MLi 0vZ5dz K6tr6hp 7frMVB EjFeZY EXRA3Hj rn+rw qXOHxR hxe+B4 vO70z2 S8PuR7 Ab+pRi FvDWwe 7bdZkt iGr+a shZ+Ph pG0aJ0b T2jdb b1gFk2 1jFlgpa CAdFUC /QVX6C AQ4IBO HwIwh5 sxPQ0Yc FESFwL Y+IiQq 6Kc0yQI 21CWZw ybGL79 O3Rrpmy Pu2lZ6 zUjE7x 6I1IaWC XNAHlR YTlaYa KJ8pZsr 95j5Wn vNuI/k7 qNSBW4 JrYv3T TzP/qZC 0CXRyr GlyqKV SMrI6lL onqiry 58aUqQ Q4hcRJ3 KB4RZk o57bOh NLGqXfb WVvE3l SlZuWdp boJ3eU sasPVz nLOgdlC 0SsXDi 1KhfJq Oot7G CP5nmE Ms5RQZ W8b/CIJ zxrDe1 Wu9PuP 1O1TKrZ wrelPX wAJbeVN w=</l atexit > ~vφ = [cos(φ), sen(φ)] <lat exit s ha1_ba se64="P aycnm2 2mb7Rl sBvzwlp GI1sP3 Q=">A C8nicj VHLSsQ wFD3W93 vUpZvi ICjo0B FN4Lox qWCo8J 0GNoYN dgXSTog Zb7CnT tx6w+4 1Y8Q/0D /wptYw QeiKW3P Pfek9 zcMIuE 0p73OP 09vUPD A4Nj4y OjU9MVq amD1Wa S8YbLI 1SeRwGi kci4Q0 tdMSPM 8mDOIz4 UXixY/ JHS6VS JMDfZn xVhycJ eJUsEAT 1a4s+x 3Oik63 XfjZuei 6m27TZ 6laMNH iUuHL2F U86b7H rXal6t U8u9yfo F6CKsq 1l1ae4O MEKRhy xOBIoA lHCKDoa aIODxl xLRTES ULC5jm6 GCFtTl WcKgJi L+h7RlG zZBOKj aeyaka 7RPRKUr qYJ01K dZKw2c2 1+dw6G /Y378J 6mrNd0j 8svWJi Nc6J/U v3Uflfn elF4xQ btgdBP WMd2x 0iW3t2 JO7n7q SpNDRpz BJ5SXh JlVftyz azXK9m 7uNrD5 F1tpWBO zsjbHq zklDbj +fZw/we FKrb5a W9tfrW 5tl6Mew izmsED zXMcWd rGHBnlf 4R4PeH S0c+3cO LfvpU5 PqZnBl +XcvQH0 w6Fp</ latexi t> ~v2φ = [cos(2φ), sen(2φ)] <l ate xit sha 1_b ase 64= "CE Ml3M boY +SU LZG e6L hVu fS8m Ds= ">A AC 9Xi cjV HLSh xBF D12 TOI jiZ 24dN M4C BMI Q8 wYj YBi RuXC o4K 08P QXZ Yzh f2iq nqC NPM b2W UXs vUH 3Jpf CP6 B/o W3y hJM hqD VdPe 59 5zq m7d pEy F0mF 4Pe e9m H/5 6vX C4t Lym7 fvV vz3 Hw5 VU nGe6 xIC 3mc xIq nIu c9L XTKj 0vJ 4yx J+V Fyt mPy RxMu lSj yA3 1e8 kEW j3Jx Kli siR r6Y Th rJ5 Mh3U nKs diG nwJ +hE rVN OGHz /Vk cwC xfO pIw ZDvx G2Q ruC WdB 2oA G39 gr/D yKc oAB DhQ wcO ThF DEU PX2 0Ea Ikb oCa OElI 2Dz HFE ukr aiK U0V M7Bl 9Rx T1H ZtT bDy VTP aJa VXk jLA Bmk Kqp OEzW 6Bz VfW 2bD /86 6tpz nbO f0T 5U Rqz Em9 indQ +Vz daY XjV N8t j0I 6qm0 jOm OZ fK3 o5 efCo K0 OJX EGn 1Be EmZ W+XD Pgd Uo2 7u5 29j mb2y lYU 3MX G2F W3N KGn D73 HOg sNO q91 tbe 53G 9tf3 agX sIZ 1NG meW 9jGL vbQ I+/ vuM QVf nvf vB/e T+/ Xfa k35 zSr +Gt5 F3f zsa Id< /la tex it> e(φ) = cos (!t) + cos (!t + φ) + cos (!t + 2φ) <latexit sha1 _base64="v3tvPIEYxdkiA5U3z oO+BCslUJg=">ADKXicjVHP a9swGH31tjbLtjZdj72IhUFCI DilY70UwnbpMYUlDdSl2IqSiNi WkeVCf2L+p/0tN+XHfZbr12 p3SHMiajk3G9tP73nvSJ0VZLH Pj+1/WvEePn6xvVJ5Wnz1/sbl V2345yFWhuehzFSs9jMJcxDIVf SNLIaZFmESxeIkmr239ZMLoX Op0g/mMhNnSThJ5Vjy0B1Xhu JRpBNZfMw4CoPYjE2jUAlYhIyE 2g5mZoma62WtazqLMWe0Cxty Q5r9X9tu8GWwWdEtRjp6qfUKA ERQ4CiQSGEIxwiR03OKDnxkx J1hTpwmJF1d4ApV8hakEqQIiZ3 Rd0Kz05JNaW4zc+fmtEpMryYn w2vyKNJpwnY15uqFS7bs37LnL tPu7ZL+UZmVEGswJfZfvoXyf32 2F4MxDlwPknrKHGO742VK4U7F 7pwtdWUoISPO4hHVNWHunItzZs 6Tu97t2Yau/t0pLWvnvNQW+GF 3SRfcuX+dq2Cw1+7st98c79e7 8qrmAXr9Cg+3yLo7Q5+yb3 CLO/z0r2P3mfv62+pt1Z6dvD H8L79ArKctog=</latexit> Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária: Rede de Difração O método de fasores simplifica essa soma simétrica de funções harmônicas em função de um cosseno de omega. O método consiste em associar cada termo com um vetor de amplitude unitária em um espaço abstrato (x,y) pelas relações: ~v0 = [1, 0] <lat exit s ha1_ba se64="5 HIO4pb bPdzl5 Yhf5AM6 u8QNJ+ g=">A C1nicj VHLSsN AFD2Nr1 pfqS7d BIvgQk oiFd0IR TcuK9g H1FKS6 bSGpklI JpVS6k 7c+gNu 9ZPEP9C /8M6Yg lpEJyQ5 c+49Z+ be64Se GwvTfM1 oc/MLi 0vZ5dz K6tr6hp 7frMVB EjFeZY EXRA3Hj rn+rw qXOHxR hxe+B4 vO70z2 S8PuR7 Ab+pRi FvDWwe 7bdZkt iGr+a shZ+Ph pG0aJ0b T2jdb b1gFk2 1jFlgpa CAdFUC /QVX6C AQ4IBO HwIwh5 sxPQ0Yc FESFwL Y+IiQq 6Kc0yQI 21CWZw ybGL79 O3Rrpmy Pu2lZ6 zUjE7x 6I1IaWC XNAHlR YTlaYa KJ8pZsr 95j5Wn vNuI/k7 qNSBW4 JrYv3T TzP/qZC 0CXRyr GlyqKV SMrI6lL onqiry 58aUqQ Q4hcRJ3 KB4RZk o57bOh NLGqXfb WVvE3l SlZuWdp boJ3eU sasPVz nLOgdlC 0SsXDi 1KhfJq Oot7G CP5nmE Ms5RQZ W8b/CIJ zxrDe1 Wu9PuP 1O1TKrZ wrelPX wAJbeVN w=</l >atexit A amplitude do vetor resultante (soma dos três vetores) é, precisamente, a amplitude da soma que se busca determinar. ~vφ = [cos(φ), sen(φ)] <lat exit s ha1_ba se64="P aycnm2 2mb7Rl sBvzwlp GI1sP3 Q=">A C8nicj VHLSsQ wFD3W93 vUpZvi ICjo0B FN4Lox qWCo8J 0GNoYN dgXSTog Zb7CnT tx6w+4 1Y8Q/0D /wptYw QeiKW3P Pfek9 zcMIuE 0p73OP 09vUPD A4Nj4y OjU9MVq amD1Wa S8YbLI 1SeRwGi kci4Q0 tdMSPM 8mDOIz4 UXixY/ JHS6VS JMDfZn xVhycJ eJUsEAT 1a4s+x 3Oik63 XfjZuei 6m27TZ 6laMNH iUuHL2F U86b7H rXal6t U8u9yfo F6CKsq 1l1ae4O MEKRhy xOBIoA lHCKDoa aIODxl xLRTES ULC5jm6 GCFtTl WcKgJi L+h7RlG zZBOKj aeyaka 7RPRKUr qYJ01K dZKw2c2 1+dw6G /Y378J 6mrNd0j 8svWJi Nc6J/U v3Uflfn elF4xQ btgdBP WMd2x 0iW3t2 JO7n7q SpNDRpz BJ5SXh JlVftyz azXK9m 7uNrD5 F1tpWBO zsjbHq zklDbj +fZw/we FKrb5a W9tfrW 5tl6Mew izmsED zXMcWd rGHBnlf 4R4PeH S0c+3cO LfvpU5 PqZnBl +XcvQH0 w6Fp</ latexi t> ~v2φ = [cos(2φ), sen(2φ)] <l ate xit sha 1_b ase 64= "CE Ml3M boY +SU LZG e6L hVu fS8m Ds= ">A AC 9Xi cjV HLSh xBF D12 TOI jiZ 24dN M4C BMI Q8 wYj YBi RuXC o4K 08P QXZ Yzh f2iq nqC NPM b2W UXs vUH 3Jpf CP6 B/o W3y hJM hqD VdPe 59 5zq m7d pEy F0mF 4Pe e9m H/5 6vX C4t Lym7 fvV vz3 Hw5 VU nGe6 xIC 3mc xIq nIu c9L XTKj 0vJ 4yx J+V Fyt mPy RxMu lSj yA3 1e8 kEW j3Jx Kli siR r6Y Th rJ5 Mh3U nKs diG nwJ +hE rVN OGHz /Vk cwC xfO pIw ZDvx G2Q ruC WdB 2oA G39 gr/D yKc oAB DhQ wcO ThF DEU PX2 0Ea Ikb oCa OElI 2Dz HFE ukr aiK U0V M7Bl 9Rx T1H ZtT bDy VTP aJa VXk jLA Bmk Kqp OEzW 6Bz VfW 2bD /86 6tpz nbO f0T 5U Rqz Em9 indQ +Vz daY XjV N8t j0I 6qm0 jOm OZ fK3 o5 efCo K0 OJX EGn 1Be EmZ W+XD Pgd Uo2 7u5 29j mb2y lYU 3MX G2F W3N KGn D73 HOg sNO q91 tbe 53G 9tf3 agX sIZ 1NG meW 9jGL vbQ I+/ vuM QVf nvf vB/e T+/ Xfa k35 zSr +Gt5 F3f zsa Id< /la tex it> e(φ) = cos (!t) + cos (!t + φ) + cos (!t + 2φ) <latexit sha1 _base64="v3tvPIEYxdkiA5U3z oO+BCslUJg=">ADKXicjVHP a9swGH31tjbLtjZdj72IhUFCI DilY70UwnbpMYUlDdSl2IqSiNi WkeVCf2L+p/0tN+XHfZbr12 p3SHMiajk3G9tP73nvSJ0VZLH Pj+1/WvEePn6xvVJ5Wnz1/sbl V2345yFWhuehzFSs9jMJcxDIVf SNLIaZFmESxeIkmr239ZMLoX Op0g/mMhNnSThJ5Vjy0B1Xhu JRpBNZfMw4CoPYjE2jUAlYhIyE 2g5mZoma62WtazqLMWe0Cxty Q5r9X9tu8GWwWdEtRjp6qfUKA ERQ4CiQSGEIxwiR03OKDnxkx J1hTpwmJF1d4ApV8hakEqQIiZ3 Rd0Kz05JNaW4zc+fmtEpMryYn w2vyKNJpwnY15uqFS7bs37LnL tPu7ZL+UZmVEGswJfZfvoXyf32 2F4MxDlwPknrKHGO742VK4U7F 7pwtdWUoISPO4hHVNWHunItzZs 6Tu97t2Yau/t0pLWvnvNQW+GF 3SRfcuX+dq2Cw1+7st98c79e7 8qrmAXr9Cg+3yLo7Q5+yb3 CLO/z0r2P3mfv62+pt1Z6dvD H8L79ArKctog=</latexit> Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária: Rede de Difração O método de fasores simplifica essa soma simétrica de funções harmônicas em função de um cosseno de omega. O método consiste em associar cada termo com um vetor de amplitude unitária em um espaço abstrato (x,y) pelas relações: ~v0 = [1, 0] <lat exit s ha1_ba se64="5 HIO4pb bPdzl5 Yhf5AM6 u8QNJ+ g=">A C1nicj VHLSsN AFD2Nr1 pfqS7d BIvgQk oiFd0IR TcuK9g H1FKS6 bSGpklI JpVS6k 7c+gNu 9ZPEP9C /8M6Yg lpEJyQ5 c+49Z+ be64Se GwvTfM1 oc/MLi 0vZ5dz K6tr6hp 7frMVB EjFeZY EXRA3Hj rn+rw qXOHxR hxe+B4 vO70z2 S8PuR7 Ab+pRi FvDWwe 7bdZkt iGr+a shZ+Ph pG0aJ0b T2jdb b1gFk2 1jFlgpa CAdFUC /QVX6C AQ4IBO HwIwh5 sxPQ0Yc FESFwL Y+IiQq 6Kc0yQI 21CWZw ybGL79 O3Rrpmy Pu2lZ6 zUjE7x 6I1IaWC XNAHlR YTlaYa KJ8pZsr 95j5Wn vNuI/k7 qNSBW4 JrYv3T TzP/qZC 0CXRyr GlyqKV SMrI6lL onqiry 58aUqQ Q4hcRJ3 KB4RZk o57bOh NLGqXfb WVvE3l SlZuWdp boJ3eU sasPVz nLOgdlC 0SsXDi 1KhfJq Oot7G CP5nmE Ms5RQZ W8b/CIJ zxrDe1 Wu9PuP 1O1TKrZ wrelPX wAJbeVN w=</l >atexit A amplitude do vetor resultante (soma dos três vetores) é, precisamente, a amplitude da soma que se busca determinar. O método funciona porque somar vetores simétricos de forma gráfica é mais simples que uma simplificação direta que depende de omega! ~vφ = [cos(φ), sen(φ)] <lat exit s ha1_ba se64="P aycnm2 2mb7Rl sBvzwlp GI1sP3 Q=">A C8nicj VHLSsQ wFD3W93 vUpZvi ICjo0B FN4Lox qWCo8J 0GNoYN dgXSTog Zb7CnT tx6w+4 1Y8Q/0D /wptYw QeiKW3P Pfek9 zcMIuE 0p73OP 09vUPD A4Nj4y OjU9MVq amD1Wa S8YbLI 1SeRwGi kci4Q0 tdMSPM 8mDOIz4 UXixY/ JHS6VS JMDfZn xVhycJ eJUsEAT 1a4s+x 3Oik63 XfjZuei 6m27TZ 6laMNH iUuHL2F U86b7H rXal6t U8u9yfo F6CKsq 1l1ae4O MEKRhy xOBIoA lHCKDoa aIODxl xLRTES ULC5jm6 GCFtTl WcKgJi L+h7RlG zZBOKj aeyaka 7RPRKUr qYJ01K dZKw2c2 1+dw6G /Y378J 6mrNd0j 8svWJi Nc6J/U v3Uflfn elF4xQ btgdBP WMd2x 0iW3t2 JO7n7q SpNDRpz BJ5SXh JlVftyz azXK9m 7uNrD5 F1tpWBO zsjbHq zklDbj +fZw/we FKrb5a W9tfrW 5tl6Mew izmsED zXMcWd rGHBnlf 4R4PeH S0c+3cO LfvpU5 PqZnBl +XcvQH0 w6Fp</ latexi t> ~v2φ = [cos(2φ), sen(2φ)] <l ate xit sha 1_b ase 64= "CE Ml3M boY +SU LZG e6L hVu fS8m Ds= ">A AC 9Xi cjV HLSh xBF D12 TOI jiZ 24dN M4C BMI Q8 wYj YBi RuXC o4K 08P QXZ Yzh f2iq nqC NPM b2W UXs vUH 3Jpf CP6 B/o W3y hJM hqD VdPe 59 5zq m7d pEy F0mF 4Pe e9m H/5 6vX C4t Lym7 fvV vz3 Hw5 VU nGe6 xIC 3mc xIq nIu c9L XTKj 0vJ 4yx J+V Fyt mPy RxMu lSj yA3 1e8 kEW j3Jx Kli siR r6Y Th rJ5 Mh3U nKs diG nwJ +hE rVN OGHz /Vk cwC xfO pIw ZDvx G2Q ruC WdB 2oA G39 gr/D yKc oAB DhQ wcO ThF DEU PX2 0Ea Ikb oCa OElI 2Dz HFE ukr aiK U0V M7Bl 9Rx T1H ZtT bDy VTP aJa VXk jLA Bmk Kqp OEzW 6Bz VfW 2bD /86 6tpz nbO f0T 5U Rqz Em9 indQ +Vz daY XjV N8t j0I 6qm0 jOm OZ fK3 o5 efCo K0 OJX EGn 1Be EmZ W+XD Pgd Uo2 7u5 29j mb2y lYU 3MX G2F W3N KGn D73 HOg sNO q91 tbe 53G 9tf3 agX sIZ 1NG meW 9jGL vbQ I+/ vuM QVf nvf vB/e T+/ Xfa k35 zSr +Gt5 F3f zsa Id< /la tex it> e(φ) = cos (!t) + cos (!t + φ) + cos (!t + 2φ) <latexit sha1 _base64="v3tvPIEYxdkiA5U3z oO+BCslUJg=">ADKXicjVHP a9swGH31tjbLtjZdj72IhUFCI DilY70UwnbpMYUlDdSl2IqSiNi WkeVCf2L+p/0tN+XHfZbr12 p3SHMiajk3G9tP73nvSJ0VZLH Pj+1/WvEePn6xvVJ5Wnz1/sbl V2345yFWhuehzFSs9jMJcxDIVf SNLIaZFmESxeIkmr239ZMLoX Op0g/mMhNnSThJ5Vjy0B1Xhu JRpBNZfMw4CoPYjE2jUAlYhIyE 2g5mZoma62WtazqLMWe0Cxty Q5r9X9tu8GWwWdEtRjp6qfUKA ERQ4CiQSGEIxwiR03OKDnxkx J1hTpwmJF1d4ApV8hakEqQIiZ3 Rd0Kz05JNaW4zc+fmtEpMryYn w2vyKNJpwnY15uqFS7bs37LnL tPu7ZL+UZmVEGswJfZfvoXyf32 2F4MxDlwPknrKHGO742VK4U7F 7pwtdWUoISPO4hHVNWHunItzZs 6Tu97t2Yau/t0pLWvnvNQW+GF 3SRfcuX+dq2Cw1+7st98c79e7 8qrmAXr9Cg+3yLo7Q5+yb3 CLO/z0r2P3mfv62+pt1Z6dvD H8L79ArKctog=</latexit> 0 1 2 3 -2 -1 0 1 2 x y Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração No gráfico ao lado, para um ângulo beta pequeno, o vetor resultante é dado pela regra do paralelogramo. ~v0 = [1, 0] <latexi t sha1_base64 ="5HIO4pbPd zl5Yhf5AM6u8Q NJ+g=">AC1n icjVHLSsNAFD 2Nr1pfqS7dBIv gQkoiFd0IRTcu K9gH1FKS6bSGp klIJpVS6k7c+ gNu9ZPEP9C/8M 6YglpEJyQ5c+4 9Z+be64SeGwv TfM1oc/MLi0vZ 5dzK6tr6hp7fr MVBEjFeZYEXRA 3Hjrn+rwqXO HxRhxe+B4vO7 0z2S8PuR7Ab+ pRiFvDWwe7b dZktiGr+ashZ +PhpG0aJ0bT2j db1gFk21jFl gpaCAdFUC/QV X6CAQ4IBOHwI wh5sxPQ0YcFES FwLY+IiQq6Kc 0yQI21CWZwybG L79O3RrpmyPu2 lZ6zUjE7x6I1I aWCXNAHlRYTl aYaKJ8pZsr95j 5WnvNuI/k7qNS BW4JrYv3TzP /qZC0CXRyrGly qKVSMrI6lLonq iry58aUqQ4hc RJ3KB4RZko57 bOhNLGqXfbWVv E3lSlZuWdpboJ 3eUsasPVznLO gdlC0SsXDi1Kh fJqOot7GCP5 nmEMs5RQZW8b/ CIJzxrDe1Wu9 PuP1O1TKrZwre lPXwAJbeVNw= </latexit> ~vφ = [cos(φ), sen(φ)] <latexi t sha1_base64 ="Paycnm2mb 7RlsBvzwlpGI1 sP3Q=">AC8n icjVHLSsQwFD 3W93vUpZviICj o0BFN4LoxqWC o8J0GNoYNdgXS TogZb7CnTtx6 w+41Y8Q/0D/wp tYwQeiKW3Pfe ek9zcMIuE0p7 3OP09vUPDA4N j4yOjU9MVqamD 1WaS8YbLI1SeR wGikci4Q0tdM SPM8mDOIz4UXi xY/JHS6VSJMD fZnxVhycJeJU sEAT1a4s+x3Oi k63XfjZuei6m2 7TZ6laMNHiUuH L2FU86b7HrXa l6tU8u9yfoF6C Ksq1l1ae4OMEK RhyxOBIoAlHC KDoaIODxlxLR TESULC5jm6GCF tTlWcKgJiL+h7 RlGzZBOKjaey aka7RPRKUrqYJ 01KdZKw2c21+d w6G/Y378J6mr Nd0j8svWJiNc6 J/Uv3UflfnelF 4xQbtgdBPWM d2x0iW3t2JO7 n7qSpNDRpzBJ5 SXhJlVftyzazX K9m7uNrD5F1t pWBOzsjbHqzkl Dbj+fZw/weFKr b5aW9tfrW5tl6 MewizmsEDzXM cWdrGHBnlf4R4 PeHS0c+3cOLfv pU5PqZnBl+Xc 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bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> 0 1 2 3 -2 -1 0 1 2 x y Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração No gráfico ao lado, para um ângulo beta pequeno, o vetor resultante é dado pela regra do paralelogramo. ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh 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<latexi t sha1_base64 ="Paycnm2mb 7RlsBvzwlpGI1 sP3Q=">AC8n icjVHLSsQwFD 3W93vUpZviICj o0BFN4LoxqWC o8J0GNoYNdgXS TogZb7CnTtx6 w+41Y8Q/0D/wp tYwQeiKW3Pfe ek9zcMIuE0p7 3OP09vUPDA4N j4yOjU9MVqamD 1WaS8YbLI1SeR wGikci4Q0tdM SPM8mDOIz4UXi xY/JHS6VSJMD fZnxVhycJeJU sEAT1a4s+x3Oi k63XfjZuei6m2 7TZ6laMNHiUuH L2FU86b7HrXa l6tU8u9yfoF6C Ksq1l1ae4OMEK RhyxOBIoAlHC KDoaIODxlxLR TESULC5jm6GCF tTlWcKgJiL+h7 RlGzZBOKjaey aka7RPRKUrqYJ 01KdZKw2c21+d w6G/Y378J6mr Nd0j8svWJiNc6 J/Uv3UflfnelF 4xQbtgdBPWM d2x0iW3t2JO7 n7qSpNDRpzBJ5 SXhJlVftyzazX K9m7uNrD5F1t pWBOzsjbHqzkl Dbj+fZw/weFKr b5aW9tfrW5tl6 MewizmsEDzXM cWdrGHBnlf4R4 PeHS0c+3cOLfv pU5PqZnBl+Xc vQH0w6Fp</lat exit> ~v2φ = [cos(2φ), sen(2φ)] <latexi t sha1_base64 ="CEMl3MboY+ SULZGe6LhVufS 8mDs=">AC9X icjVHLShxBFD 12TOIjiZ24dNM 4CBMIQ8wYjYB iRuXCo4K08PQX ZYzhf2iqnqCN PMb2WUXsvUH3J pfCP6B/oW3yhJ MhqDVdPe595 zqm7dpEyF0mF4 Pe9mH/56vXC4 tLym7fvVvz3Hw 5VUnGe6xIC3 mcxIqnIuc9LXT Kj0vJ4yxJ+VFy tmPyRxMulSjy A31e8kEWj3JxK lisiRr6YThrJ 5Mh3UnKsdiGnw J+hErVNOGHz/ VkcwCxfOpIwZD vxG2QruCWdB2o AG39gr/DyKco ABDhQwcOThFD EUPX20EaIkboC aOElI2DzHFEuk raiKU0VM7Bl9 RxT1HZtTbDyV TPaJaVXkjLABm kKqpOEzW6BzV fW2bD/86tpzn bOf0T5URqzEm 9indQ+VzdaYXj VN8tj0I6qm0j OmOZfK3o5ef CoK0OJXEGn1B eEmZW+XDPgdU o27u529jmb2yl YU3MXG2FW3NKG nD73HOgsNOq9 1tbe53G9tf3a gXsIZ1NGmeW9j GLvbQI+/vuMQV fnvfvB/eT+/X fak35zSr+Gt5F 3fzsaId</late xit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> 0 1 2 3 -2 -1 0 1 2 x y Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração No gráfico ao lado, para um ângulo beta pequeno, o vetor resultante é dado pela regra do paralelogramo. ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 o1NceiNS6jgiTUB5EWF5mq7iXKW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6ljmkqiuyJvrX6oS5BASJ/GA4h TLOPb0t7+AsR5Qq</latexit>FhpTzPutKE6vaZW8tFX9TmZKVe5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcYUaGuQd4hFPeNbq2q12p91/pmq5 ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr gkzIxy3GfbaFJTu+6tZ+KvJlOzes+K3Axv+pY0YPf7OH+Ci52qu1vdO92t1A6LUZewgU1s0z3UcMxTlAnb4kHPOLJOrOG1q195Fq TRSadXxZ1v07meKViw=</latexit> ~v0 = [1, 0] <latexi t sha1_base64 ="5HIO4pbPd zl5Yhf5AM6u8Q NJ+g=">AC1n icjVHLSsNAFD 2Nr1pfqS7dBIv gQkoiFd0IRTcu K9gH1FKS6bSGp klIJpVS6k7c+ gNu9ZPEP9C/8M 6YglpEJyQ5c+4 9Z+be64SeGwv TfM1oc/MLi0vZ 5dzK6tr6hp7fr MVBEjFeZYEXRA 3Hjrn+rwqXO HxRhxe+B4vO7 0z2S8PuR7Ab+ pRiFvDWwe7b dZktiGr+ashZ +PhpG0aJ0bT2j db1gFk21jFl gpaCAdFUC/QV X6CAQ4IBOHwI wh5sxPQ0YcFES FwLY+IiQq6Kc 0yQI21CWZwybG L79O3RrpmyPu2 lZ6zUjE7x6I1I aWCXNAHlRYTl aYaKJ8pZsr95j 5WnvNuI/k7qNS BW4JrYv3TzP 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iRuXCo4K08PQX ZYzhf2iqnqCN PMb2WUXsvUH3J pfCP6B/oW3yhJ MhqDVdPe595 zqm7dpEyF0mF4 Pe9mH/56vXC4 tLym7fvVvz3Hw 5VUnGe6xIC3 mcxIqnIuc9LXT Kj0vJ4yxJ+VFy tmPyRxMulSjy A31e8kEWj3JxK lisiRr6YThrJ 5Mh3UnKsdiGnw J+hErVNOGHz/ VkcwCxfOpIwZD vxG2QruCWdB2o AG39gr/DyKco ABDhQwcOThFD EUPX20EaIkboC aOElI2DzHFEuk raiKU0VM7Bl9 RxT1HZtTbDyV TPaJaVXkjLABm kKqpOEzW6BzV fW2bD/86tpzn bOf0T5URqzEm 9indQ+VzdaYXj VN8tj0I6qm0j OmOZfK3o5ef CoK0OJXEGn1B eEmZW+XDPgdU o27u529jmb2yl YU3MXG2FW3NKG nD73HOgsNOq9 1tbe53G9tf3a gXsIZ1NGmeW9j GLvbQI+/vuMQV fnvfvB/eT+/X fak35zSr+Gt5F 3fzsaId</late xit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> 0 1 2 3 -2 -1 0 1 2 x y Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração No gráfico ao lado, para um ângulo beta pequeno, o vetor resultante é dado pela regra do paralelogramo. ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 o1NceiNS6jgiTUB5EWF5mq7iXKW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6ljmkqiuyJvrX6oS5BASJ/GA4h 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<latexi t sha1_base64 ="Paycnm2mb 7RlsBvzwlpGI1 sP3Q=">AC8n icjVHLSsQwFD 3W93vUpZviICj o0BFN4LoxqWC o8J0GNoYNdgXS TogZb7CnTtx6 w+41Y8Q/0D/wp tYwQeiKW3Pfe ek9zcMIuE0p7 3OP09vUPDA4N j4yOjU9MVqamD 1WaS8YbLI1SeR wGikci4Q0tdM SPM8mDOIz4UXi xY/JHS6VSJMD fZnxVhycJeJU sEAT1a4s+x3Oi k63XfjZuei6m2 7TZ6laMNHiUuH L2FU86b7HrXa l6tU8u9yfoF6C Ksq1l1ae4OMEK RhyxOBIoAlHC KDoaIODxlxLR TESULC5jm6GCF tTlWcKgJiL+h7 RlGzZBOKjaey aka7RPRKUrqYJ 01KdZKw2c21+d w6G/Y378J6mr Nd0j8svWJiNc6 J/Uv3UflfnelF 4xQbtgdBPWM d2x0iW3t2JO7 n7qSpNDRpzBJ5 SXhJlVftyzazX K9m7uNrD5F1t pWBOzsjbHqzkl Dbj+fZw/weFKr b5aW9tfrW5tl6 MewizmsEDzXM cWdrGHBnlf4R4 PeHS0c+3cOLfv pU5PqZnBl+Xc vQH0w6Fp</lat exit> ~v2φ = [cos(2φ), sen(2φ)] <latexi t sha1_base64 ="CEMl3MboY+ SULZGe6LhVufS 8mDs=">AC9X icjVHLShxBFD 12TOIjiZ24dNM 4CBMIQ8wYjYB iRuXCo4K08PQX ZYzhf2iqnqCN PMb2WUXsvUH3J pfCP6B/oW3yhJ MhqDVdPe595 zqm7dpEyF0mF4 Pe9mH/56vXC4 tLym7fvVvz3Hw 5VUnGe6xIC3 mcxIqnIuc9LXT Kj0vJ4yxJ+VFy tmPyRxMulSjy A31e8kEWj3JxK lisiRr6YThrJ 5Mh3UnKsdiGnw 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cWdrGHBnlf4R4 PeHS0c+3cOLfv pU5PqZnBl+Xc vQH0w6Fp</lat exit> ~v2φ = [cos(2φ), sen(2φ)] <latexi t sha1_base64 ="CEMl3MboY+ SULZGe6LhVufS 8mDs=">AC9X icjVHLShxBFD 12TOIjiZ24dNM 4CBMIQ8wYjYB iRuXCo4K08PQX ZYzhf2iqnqCN PMb2WUXsvUH3J pfCP6B/oW3yhJ MhqDVdPe595 zqm7dpEyF0mF4 Pe9mH/56vXC4 tLym7fvVvz3Hw 5VUnGe6xIC3 mcxIqnIuc9LXT Kj0vJ4yxJ+VFy tmPyRxMulSjy A31e8kEWj3JxK lisiRr6YThrJ 5Mh3UnKsdiGnw J+hErVNOGHz/ VkcwCxfOpIwZD vxG2QruCWdB2o AG39gr/DyKco ABDhQwcOThFD EUPX20EaIkboC aOElI2DzHFEuk raiKU0VM7Bl9 RxT1HZtTbDyV TPaJaVXkjLABm kKqpOEzW6BzV fW2bD/86tpzn bOf0T5URqzEm 9indQ+VzdaYXj VN8tj0I6qm0j OmOZfK3o5ef CoK0OJXEGn1B eEmZW+XDPgdU o27u529jmb2yl YU3MXG2FW3NKG nD73HOgsNOq9 1tbe53G9tf3a gXsIZ1NGmeW9j GLvbQI+/vuMQV fnvfvB/eT+/X fak35zSr+Gt5F 3fzsaId</late xit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> 0 1 2 3 -2 -1 0 1 2 x y Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração No gráfico ao lado, para um ângulo beta pequeno, o vetor resultante é dado pela regra do paralelogramo. ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 o1NceiNS6jgiTUB5EWF5mq7iXKW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6ljmkqiuyJvrX6oS5BASJ/GA4h FhpTzPutKE6vaZW8tFX9TmZKVe5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcYUaGuQd4hFPeNbq2q12p91/pmq5 TLOPb0t7+AsR5Qq</latexit> ~vres <latexit sha1_base64="C7acy1tGvKyomPY6vEH3tU2XA=" >AC13icjVHLSgMxFD2Or1pfVZduBovgqkxF0WXRjcsK9iFWykxMdXBeJmilOJO3PoDbvWPxD/Qv/AmpuAD0Qwzc3LuPSe59wZF ErleS9jzvjE5NR0YaY4Oze/sFhaWm7KNBeMN1gapaId+JHYcIbKlQRb2eC+3EQ8VZwua/jrT4XMkyTI3Wd8dPYP0/CXsh8RVS3tNz pczboD7uDjohdweXQ7ZbKXsUzy/0JqhaUYVc9LT2jgzOkYMgRgyOBIhzBh6TnBFV4yIg7xYA4QSg0cY4hiqTNKYtThk/sJX3PaXdi 2YT2lMaNaNTInoFKV2skyalPEFYn+aeG6cNfub98B46rtd0z+wXjGxChfE/qUbZf5Xp2tR6GHX1BSTZlhdHXMuSmK/rm7qeqFD lkxGl8RnFBmBnlqM+u0UhTu+6tb+KvJlOzes9sbo43fUsacPX7OH+C5malulXZPtwq1/bsqAtYxRo2aJ47qOEAdTI+woPeMSTc+zc OLfO3UeqM2Y1K/iynPt3nF2XDA=</latexit> O tamanho do raio resultante pode ser determinado graficamente. ~v2φ <latexit sha1_base64="7zrFDcmPH6CrVmY4Rx81Ysb6Ub0=" >AC1HicjVHLTsJAFD3UF+ID1KWbRmLihSC0SXRjUtM5JEAIe0wITSNu2UhCAr49YfcKvfZPwD/QvjCVRidFp2p4595w7c+91A ldE0rJeU8bK6tr6Rnozs7W9s5vN7e3XIz8OGa8x3/XDpmNH3BUer0khXd4MQm6PHZc3nNGlijcmPIyE793IacA7Y3vgib5gtiSqm8u 2J5zNJvPurNQOhmLezeWtgqWXuQyKCcgjWVU/94I2evDBEGMDg+SsAsbET0tFGEhIK6DGXEhIaHjHNkyBuTipPCJnZE3wHtWgnr 0V7ljLSb0SkuvSE5TRyTxydSFidZup4rDMr9rfcM51T3W1KfyfJNSZWYkjsX76F8r8+VYtEH+e6BkE1BZpR1bEkS6y7om5ufqlKUo aAOIV7FA8JM+1c9NnUnkjXrnpr6/ibVipW7VmijfGubkDLv4c5zKolwrFcuH0upyvXCSjTuMQRziheZ6hgitUdMzf8QTno26cWvc GfefUiOVeA7wbRkPH83Alfg=</latexit> ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr gkzIxy3GfbaFJTu+6tZ+KvJlOzes+K3Axv+pY0YPf7OH+Ci52qu1vdO92t1A6LUZewgU1s0z3UcMxTlAnb4kHPOLJOrOG1q195Fq TRSadXxZ1v07meKViw=</latexit> ~v0 = [1, 0] <latexi t sha1_base64 ="5HIO4pbPd zl5Yhf5AM6u8Q NJ+g=">AC1n icjVHLSsNAFD 2Nr1pfqS7dBIv gQkoiFd0IRTcu K9gH1FKS6bSGp klIJpVS6k7c+ gNu9ZPEP9C/8M 6YglpEJyQ5c+4 9Z+be64SeGwv TfM1oc/MLi0vZ 5dzK6tr6hp7fr MVBEjFeZYEXRA 3Hjrn+rwqXO HxRhxe+B4vO7 0z2S8PuR7Ab+ pRiFvDWwe7b dZktiGr+ashZ +PhpG0aJ0bT2j db1gFk21jFl gpaCAdFUC/QV X6CAQ4IBOHwI wh5sxPQ0YcFES FwLY+IiQq6Kc 0yQI21CWZwybG L79O3RrpmyPu2 lZ6zUjE7x6I1I aWCXNAHlRYTl aYaKJ8pZsr95j 5WnvNuI/k7qNS BW4JrYv3TzP /qZC0CXRyrGly qKVSMrI6lLonq iry58aUqQ4hc RJ3KB4RZko57 bOhNLGqXfbWVv E3lSlZuWdpboJ 3eUsasPVznLO gdlC0SsXDi1Kh fJqOot7GCP5 nmEMs5RQZW8b/ CIJzxrDe1Wu9 PuP1O1TKrZwre lPXwAJbeVNw= </latexit> ~vφ = [cos(φ), sen(φ)] <latexi t sha1_base64 ="Paycnm2mb 7RlsBvzwlpGI1 sP3Q=">AC8n icjVHLSsQwFD 3W93vUpZviICj o0BFN4LoxqWC 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RxT1HZtTbDyV TPaJaVXkjLABm kKqpOEzW6BzV fW2bD/86tpzn bOf0T5URqzEm 9indQ+VzdaYXj VN8tj0I6qm0j OmOZfK3o5ef CoK0OJXEGn1B eEmZW+XDPgdU o27u529jmb2yl YU3MXG2FW3NKG nD73HOgsNOq9 1tbe53G9tf3a gXsIZ1NGmeW9j GLvbQI+/vuMQV fnvfvB/eT+/X fak35zSr+Gt5F 3fzsaId</late xit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 o1NceiNS6jgiTUB5EWF5mq7iXKW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6ljmkqiuyJvrX6oS5BASJ/GA4h FhpTzPutKE6vaZW8tFX9TmZKVe5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcYUaGuQd4hFPeNbq2q12p91/pmq5 TLOPb0t7+AsR5Qq</latexit> ~vres <latexit sha1_base64="C7acy1tGvKyomPY6vEH3tU2XA=" >AC13icjVHLSgMxFD2Or1pfVZduBovgqkxF0WXRjcsK9iFWykxMdXBeJmilOJO3PoDbvWPxD/Qv/AmpuAD0Qwzc3LuPSe59wZF ErleS9jzvjE5NR0YaY4Oze/sFhaWm7KNBeMN1gapaId+JHYcIbKlQRb2eC+3EQ8VZwua/jrT4XMkyTI3Wd8dPYP0/CXsh8RVS3tNz pczboD7uDjohdweXQ7ZbKXsUzy/0JqhaUYVc9LT2jgzOkYMgRgyOBIhzBh6TnBFV4yIg7xYA4QSg0cY4hiqTNKYtThk/sJX3PaXdi 2YT2lMaNaNTInoFKV2skyalPEFYn+aeG6cNfub98B46rtd0z+wXjGxChfE/qUbZf5Xp2tR6GHX1BSTZlhdHXMuSmK/rm7qeqFD lkxGl8RnFBmBnlqM+u0UhTu+6tb+KvJlOzes9sbo43fUsacPX7OH+C5malulXZPtwq1/bsqAtYxRo2aJ47qOEAdTI+woPeMSTc+zc OLfO3UeqM2Y1K/iynPt3nF2XDA=</latexit> A figura ampliada para o vetor resultantes é dada ao O t a m a n h o d o v e t o r resultante é obtido com a soma de três termos, a t r a v é s d e r e t a s perpendiculares ao vetor v_phi. ~v2φ <latexit sha1_base64="7zrFDcmPH6CrVmY4Rx81Ysb6Ub0=" >AC1HicjVHLTsJAFD3UF+ID1KWbRmLihSC0SXRjUtM5JEAIe0wITSNu2UhCAr49YfcKvfZPwD/QvjCVRidFp2p4595w7c+91A ldE0rJeU8bK6tr6Rnozs7W9s5vN7e3XIz8OGa8x3/XDpmNH3BUer0khXd4MQm6PHZc3nNGlijcmPIyE793IacA7Y3vgib5gtiSqm8u 2J5zNJvPurNQOhmLezeWtgqWXuQyKCcgjWVU/94I2evDBEGMDg+SsAsbET0tFGEhIK6DGXEhIaHjHNkyBuTipPCJnZE3wHtWgnr 0V7ljLSb0SkuvSE5TRyTxydSFidZup4rDMr9rfcM51T3W1KfyfJNSZWYkjsX76F8r8+VYtEH+e6BkE1BZpR1bEkS6y7om5ufqlKUo aAOIV7FA8JM+1c9NnUnkjXrnpr6/ibVipW7VmijfGubkDLv4c5zKolwrFcuH0upyvXCSjTuMQRziheZ6hgitUdMzf8QTno26cWvc GfefUiOVeA7wbRkPH83Alfg=</latexit> ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr gkzIxy3GfbaFJTu+6tZ+KvJlOzes+K3Axv+pY0YPf7OH+Ci52qu1vdO92t1A6LUZewgU1s0z3UcMxTlAnb4kHPOLJOrOG1q195Fq TRSadXxZ1v07meKViw=</latexit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 o1NceiNS6jgiTUB5EWF5mq7iXKW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6ljmkqiuyJvrX6oS5BASJ/GA4h FhpTzPutKE6vaZW8tFX9TmZKVe5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcYUaGuQd4hFPeNbq2q12p91/pmq5 TLOPb0t7+AsR5Qq</latexit> ~vres <latexit sha1_base64="C7acy1tGvKyomPY6vEH3tU2XA=" >AC13icjVHLSgMxFD2Or1pfVZduBovgqkxF0WXRjcsK9iFWykxMdXBeJmilOJO3PoDbvWPxD/Qv/AmpuAD0Qwzc3LuPSe59wZF ErleS9jzvjE5NR0YaY4Oze/sFhaWm7KNBeMN1gapaId+JHYcIbKlQRb2eC+3EQ8VZwua/jrT4XMkyTI3Wd8dPYP0/CXsh8RVS3tNz pczboD7uDjohdweXQ7ZbKXsUzy/0JqhaUYVc9LT2jgzOkYMgRgyOBIhzBh6TnBFV4yIg7xYA4QSg0cY4hiqTNKYtThk/sJX3PaXdi 2YT2lMaNaNTInoFKV2skyalPEFYn+aeG6cNfub98B46rtd0z+wXjGxChfE/qUbZf5Xp2tR6GHX1BSTZlhdHXMuSmK/rm7qeqFD lkxGl8RnFBmBnlqM+u0UhTu+6tb+KvJlOzes9sbo43fUsacPX7OH+C5malulXZPtwq1/bsqAtYxRo2aJ47qOEAdTI+woPeMSTc+zc OLfO3UeqM2Y1K/iynPt3nF2XDA=</latexit> A figura ampliada para o vetor resultantes é dada ao O t a m a n h o d o v e t o r resultante é obtido com a soma de três termos, a t r a v é s d e r e t a s perpendiculares ao vetor v_phi. ~v2φ <latexit sha1_base64="7zrFDcmPH6CrVmY4Rx81Ysb6Ub0=" >AC1HicjVHLTsJAFD3UF+ID1KWbRmLihSC0SXRjUtM5JEAIe0wITSNu2UhCAr49YfcKvfZPwD/QvjCVRidFp2p4595w7c+91A ldE0rJeU8bK6tr6Rnozs7W9s5vN7e3XIz8OGa8x3/XDpmNH3BUer0khXd4MQm6PHZc3nNGlijcmPIyE793IacA7Y3vgib5gtiSqm8u 2J5zNJvPurNQOhmLezeWtgqWXuQyKCcgjWVU/94I2evDBEGMDg+SsAsbET0tFGEhIK6DGXEhIaHjHNkyBuTipPCJnZE3wHtWgnr 0V7ljLSb0SkuvSE5TRyTxydSFidZup4rDMr9rfcM51T3W1KfyfJNSZWYkjsX76F8r8+VYtEH+e6BkE1BZpR1bEkS6y7om5ufqlKUo aAOIV7FA8JM+1c9NnUnkjXrnpr6/ibVipW7VmijfGubkDLv4c5zKolwrFcuH0upyvXCSjTuMQRziheZ6hgitUdMzf8QTno26cWvc GfefUiOVeA7wbRkPH83Alfg=</latexit> ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr 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B/ov/BJnwbaIZpjk5X3fe8mXLy 0zaWwY3iwFyw8erqw2Hq09fvJ 0faO5uTU0qtJcDLjKlD5MEyMyW YiBlTYTh6UWSZ5m4iA9e+fiB+ dCG6mKD3ZaiuM8GRfyVPLEjV q7sYUtTx2fl8NIt1zrQwc8+yP RaxV6wTc2V24nIiX8Zajic20V pdjJqtsB36we6DqAYt1KOvmt8R 4wQKHBVyCBSwhDMkMPQdIUKIk rhjzIjThKSPC8yxRtqKsgRlJMS e0Tym3VHNFrR3nsarOZ2S0a9J yfCNIryNGF3GvPxyjs79m/eM +/p7jalNa29cmItJsT+S7fI/F+ dq8XiFLu+Bk1lZ5x1fHapfKv 4m7O/qjKkNJnMnFNeEuVcu3p l5jfG1u7dNfPyHz3Ss2/M6t8J Pd0tqcHS3nfBsNOu3X+91W7 23d6gae4Tl2qJ9v0MN79DEg7y t8xTW+BZfBx+BT8Pl3arBUa7Z xawRfgGaoaYh</latexit> Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 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ao vetor v_phi. ~v2φ <latexit sha1_base64="7zrFDcmPH6CrVmY4Rx81Ysb6Ub0=" >AC1HicjVHLTsJAFD3UF+ID1KWbRmLihSC0SXRjUtM5JEAIe0wITSNu2UhCAr49YfcKvfZPwD/QvjCVRidFp2p4595w7c+91A ldE0rJeU8bK6tr6Rnozs7W9s5vN7e3XIz8OGa8x3/XDpmNH3BUer0khXd4MQm6PHZc3nNGlijcmPIyE793IacA7Y3vgib5gtiSqm8u 2J5zNJvPurNQOhmLezeWtgqWXuQyKCcgjWVU/94I2evDBEGMDg+SsAsbET0tFGEhIK6DGXEhIaHjHNkyBuTipPCJnZE3wHtWgnr 0V7ljLSb0SkuvSE5TRyTxydSFidZup4rDMr9rfcM51T3W1KfyfJNSZWYkjsX76F8r8+VYtEH+e6BkE1BZpR1bEkS6y7om5ufqlKUo aAOIV7FA8JM+1c9NnUnkjXrnpr6/ibVipW7VmijfGubkDLv4c5zKolwrFcuH0upyvXCSjTuMQRziheZ6hgitUdMzf8QTno26cWvc GfefUiOVeA7wbRkPH83Alfg=</latexit> ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr gkzIxy3GfbaFJTu+6tZ+KvJlOzes+K3Axv+pY0YPf7OH+Ci52qu1vdO92t1A6LUZewgU1s0z3UcMxTlAnb4kHPOLJOrOG1q195Fq TRSadXxZ1v07meKViw=</latexit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> |~vres| = 1 + 2 cos(φ) ! <latexit sha1 _base64="oID8fcM2SfDZNShg0 nQgoOmGKuQ=">AC/XicjVFB axQxGH2dartWbft0UtwESrCM rOs2Eth0YvHLbjbQqcsM2m6Gzo zGZJMy7Is/Se9eROv/gGv9Sr9 B/ov/BJnwbaIZpjk5X3fe8mXLy 0zaWwY3iwFyw8erqw2Hq09fvJ 0faO5uTU0qtJcDLjKlD5MEyMyW YiBlTYTh6UWSZ5m4iA9e+fiB+ dCG6mKD3ZaiuM8GRfyVPLEjV q7sYUtTx2fl8NIt1zrQwc8+yP RaxV6wTc2V24nIiX8Zajic20V pdjJqtsB36we6DqAYt1KOvmt8R 4wQKHBVyCBSwhDMkMPQdIUKIk rhjzIjThKSPC8yxRtqKsgRlJMS e0Tym3VHNFrR3nsarOZ2S0a9J yfCNIryNGF3GvPxyjs79m/eM +/p7jalNa29cmItJsT+S7fI/F+ dq8XiFLu+Bk1lZ5x1fHapfKv 4m7O/qjKkNJnMnFNeEuVcu3p l5jfG1u7dNfPyHz3Ss2/M6t8J Pd0tqcHS3nfBsNOu3X+91W7 23d6gae4Tl2qJ9v0MN79DEg7y xawRfgGaoaYh</latexit>t8xTW+BZfBx+BT8Pl3arBUa7Z |~vres| = sen ⇣ 3φ 2 ⌘ sen ⇣ φ 2 ⌘ <latexit sha1_base64="JOPJmQaUEhnzYMAKZOj/TNh2If4="> ADNnicjVHLSsNAFL2N73fVpZtgEeqmpFrRjSC6calgtWCkJNpOzQvZyaChPyXfyJu3Im68wcE71wjqFV0QiZnzj3nZO6MnwRCac e5K1kjo2PjE5NT0zOzc/ML5cWlUxWnkvEmi4NYtnxP8UBEvKmFDngrkdwL/YCf+YMDUz+74lKJODrR1wm/CL1eJLqCeRqpdvnSxarG iWVXeTtzZWhLrnJi7V3b7XSlxzLiFY9yN+BdXWJ3HSTvsizjdyVotfX6/nPsu+qdrni1Bwa9jCoF6ACxTiKy7fgQgdiYJBCBwi0I gD8EDhcw51cCB7gIy5CQiQXUOUyjN0UVR4WH7ADnHq7OCzbCtclU5Gb4lwBfiU4b1tATo04iNn+zqZ5SsmF/y84o0+ztGr9+kRUi q6GP7F+D+V/faYXDV3YoR4E9pQY7pjRUpKp2J2bn/qSmNCgpzBHaxLxIycH+dsk0dR7+ZsPao/kdKwZs0KbQrPZpd4wfXv1zkMTj dq9UZt67hR2dsvrnoSVmAVqnif27AHh3AETcy+hdfSeGnCurHurQfr8V1qlQrPMnwZ1sbwn2+Hg=</latexit> Considere um campo resultante dado pela soma de três componentes harmônicas de amplitude unitária (t=0): Rede de Difração ~v0 <latexit sha1_base64="vtzByQUyFJhYHNdgiahXtmKa8hc=" >ACz3icjVHLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/4Z0xBbWITkhy5tx7zsy91w5dJ xaG8ZrTlpZXVtfy64WNza3tneLuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHsRP412Ia8p5njXxn6DBLENXtTjh LJ7N+asz0frFklA219EVgZqCEbNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJtQFqcMi9gxfUe062SsT3vpGSs1 o1NceiNS6jgiTUB5EWF5mq7iXKW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6ljmkqiuyJvrX6oS5BASJ/GA4h FhpTzPutKE6vaZW8tFX9TmZKVe5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcYUaGuQd4hFPeNbq2q12p91/pmq5 TLOPb0t7+AsR5Qq</latexit> ~vres <latexit sha1_base64="C7acy1tGvKyomPY6vEH3tU2XA=" >AC13icjVHLSgMxFD2Or1pfVZduBovgqkxF0WXRjcsK9iFWykxMdXBeJmilOJO3PoDbvWPxD/Qv/AmpuAD0Qwzc3LuPSe59wZF ErleS9jzvjE5NR0YaY4Oze/sFhaWm7KNBeMN1gapaId+JHYcIbKlQRb2eC+3EQ8VZwua/jrT4XMkyTI3Wd8dPYP0/CXsh8RVS3tNz pczboD7uDjohdweXQ7ZbKXsUzy/0JqhaUYVc9LT2jgzOkYMgRgyOBIhzBh6TnBFV4yIg7xYA4QSg0cY4hiqTNKYtThk/sJX3PaXdi 2YT2lMaNaNTInoFKV2skyalPEFYn+aeG6cNfub98B46rtd0z+wXjGxChfE/qUbZf5Xp2tR6GHX1BSTZlhdHXMuSmK/rm7qeqFD lkxGl8RnFBmBnlqM+u0UhTu+6tb+KvJlOzes9sbo43fUsacPX7OH+C5malulXZPtwq1/bsqAtYxRo2aJ47qOEAdTI+woPeMSTc+zc OLfO3UeqM2Y1K/iynPt3nF2XDA=</latexit> A figura ampliada para o vetor resultantes é dada ao O t a m a n h o d o v e t o r resultante é obtido com a soma de três termos, a t r a v é s d e r e t a s perpendiculares ao vetor v_phi. ~v2φ <latexit sha1_base64="7zrFDcmPH6CrVmY4Rx81Ysb6Ub0=" >AC1HicjVHLTsJAFD3UF+ID1KWbRmLihSC0SXRjUtM5JEAIe0wITSNu2UhCAr49YfcKvfZPwD/QvjCVRidFp2p4595w7c+91A 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TRSadXxZ1v07meKViw=</latexit> e(φ) = cos (0) + cos (φ) + cos (2φ) <latexit sha1 _base64="Gml8gQNkdreXUIz0P Iu/CbHqI=">ADEHicjVHL SsNAFD3G97vq0k2wC2FkoqiG 0F0I7hRsCo0pSTaTuYF5OJIOJ P+Cfu3Ilbf0Dc6U7/wjtji9E JyQ5c+45Z+bO+EkgUuU4jwPW4N DwyOjY+MTk1PTMbGFu/iNM8l 4ncVBLE98L+WBiHhdCRXwk0RyL /QDfuyf7uj68RmXqYijQ3We8G bodSPREcxTRLUKe7zkJj1R3nR ZnLoB76iS40rR7amyXfngtKZP2 xX7o7DyqdIqFJ2qY4b9E9RyUE Q+9uPCA1y0EYMhQwiOCIpwA8p PQ3U4CAhrokL4iQhYeocl5gb 0YqTgqP2FP6dmnWyNmI5jozNW5 GqwT0SnLaWCZPTDpJWK9m3pm kjX7W/aFydR7O6e/n2eFxCr0i P3L1f+16d7Uehgw/QgqKfEMLo 7lqdk5lT0zu1PXSlKSIjTuE1 SZgZ/+cbeNJTe/6bD1TfzFKze o5y7UZXvUu6YJr36/zJzhaqdZ Wq2sHq8Wt7fyqx7CIJZToPtexh V3so07Z13jAE56tK+vGurXu3q XWQO5ZwJdh3b8BUISsfQ=</l atexit> |~vres| = 1 + 2 cos(φ) ! <latexit sha1 _base64="oID8fcM2SfDZNShg0 nQgoOmGKuQ=">AC/XicjVFB axQxGH2dartWbft0UtwESrCM rOs2Eth0YvHLbjbQqcsM2m6Gzo zGZJMy7Is/Se9eROv/gGv9Sr9 B/ov/BJnwbaIZpjk5X3fe8mXLy 0zaWwY3iwFyw8erqw2Hq09fvJ 0faO5uTU0qtJcDLjKlD5MEyMyW YiBlTYTh6UWSZ5m4iA9e+fiB+ dCG6mKD3ZaiuM8GRfyVPLEjV q7sYUtTx2fl8NIt1zrQwc8+yP RaxV6wTc2V24nIiX8Zajic20V pdjJqtsB36we6DqAYt1KOvmt8R 4wQKHBVyCBSwhDMkMPQdIUKIk rhjzIjThKSPC8yxRtqKsgRlJMS e0Tym3VHNFrR3nsarOZ2S0a9J yfCNIryNGF3GvPxyjs79m/eM +/p7jalNa29cmItJsT+S7fI/F+ dq8XiFLu+Bk1lZ5x1fHapfKv 4m7O/qjKkNJnMnFNeEuVcu3p l5jfG1u7dNfPyHz3Ss2/M6t8J Pd0tqcHS3nfBsNOu3X+91W7 23d6gae4Tl2qJ9v0MN79DEg7y xawRfgGaoaYh</latexit>t8xTW+BZfBx+BT8Pl3arBUa7Z |~vres| = sen ⇣ 3φ 2 ⌘ sen ⇣ φ 2 ⌘ <latexit sha1_base64="JOPJmQaUEhnzYMAKZOj/TNh2If4="> ADNnicjVHLSsNAFL2N73fVpZtgEeqmpFrRjSC6calgtWCkJNpOzQvZyaChPyXfyJu3Im68wcE71wjqFV0QiZnzj3nZO6MnwRCac e5K1kjo2PjE5NT0zOzc/ML5cWlUxWnkvEmi4NYtnxP8UBEvKmFDngrkdwL/YCf+YMDUz+74lKJODrR1wm/CL1eJLqCeRqpdvnSxarG iWVXeTtzZWhLrnJi7V3b7XSlxzLiFY9yN+BdXWJ3HSTvsizjdyVotfX6/nPsu+qdrni1Bwa9jCoF6ACxTiKy7fgQgdiYJBCBwi0I gD8EDhcw51cCB7gIy5CQiQXUOUyjN0UVR4WH7ADnHq7OCzbCtclU5Gb4lwBfiU4b1tATo04iNn+zqZ5SsmF/y84o0+ztGr9+kRUi q6GP7F+D+V/faYXDV3YoR4E9pQY7pjRUpKp2J2bn/qSmNCgpzBHaxLxIycH+dsk0dR7+ZsPao/kdKwZs0KbQrPZpd4wfXv1zkMTj dq9UZt67hR2dsvrnoSVmAVqnif27AHh3AETcy+hdfSeGnCurHurQfr8V1qlQrPMnwZ1sbwn2+Hg=</latexit> φ = kdsen(✓) <latexit sha1_base64="68nQxnEIRwOXbgsMl8uctRC0CU=" >AC3XicjVHLSsNAFD3GV62vqBvBTbAIdVNSqehGEN24rGCt0JSpKMdmheTiVCK7tyJW3/Arf6O+Af6F94ZI6hFdEKSM+fec2buv V4S8FTa9suYMT4xOTVdmCnOzs0vLJpLy6dpnAmfNfw4iMWZ56Ys4BFrSC4DdpYI5oZewJpe/1DFm5dMpDyOTuQgYe3QvYj4OfdSVT HXHWSHt/rd4eOCK2URVdlR/aYdDc7Zsmu2HpZo6CagxLyVY/NZzjoIoaPDCEYIkjCAVyk9LRQhY2EuDaGxAlCXMcZrlAkbUZjDJc Yv0vaBdK2cj2ivPVKt9OiWgV5DSwgZpYsoThNVplo5n2lmxv3kPtae624D+Xu4VEivRI/Yv3Wfmf3WqFolz7OoaONWUaEZV5+cume 6Kurn1pSpJDglxCncpLgj7WvnZ0trUl276q2r4686U7Fq7+e5Gd7ULWnA1Z/jHAWnW5VqrbJ9XCvtH+SjLmAN6yjTPHewjyPU0SDv azgEU9Gx7gxbo27j1RjLNes4Nsy7t8BIaSZKQ=</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 2 4 sen ⇣ 3φ(✓) 2 ⌘ sen ⇣ φ(✓) 2 ⌘ 3 5 2 <latexit sha1 _base64="9RVl45Yw3NXF6lgDm mvUNEyp0=">ADo3icjVHL ThsxFL3DUJ4FQrvsZkSEFDbRJ IBg4TaLojYUEQAKQ6Rx3ESw7x keypFo3wnYtN1+Yteu05VHhF4 NDPH5z7GtHeSyUDsMHb86f/7 CwuLS8svpxbX2jsvnpUmWFZLz NsjiT1xFVPBYpb2uhY36dS06TK OZX0d03U7/6yaUSWXqhxznvJn SYioFgVCPVq/xq1YgecU13gqO g1QtJzAe6Q/oDSVlZEpkEiqcTy 9aIJUmE6qlpUjYnRIrhSCOcUX eC7k3z7fBdko/Eq+Gv72WG+t+K vUo1rId2BC9Bw4EquHGWVe6BQ B8yYFBAhxS0IhjoKDw6UADQsi R60KJnEQkbJ3DBFbQW6CKo4Ii e4fIc46jk1xbjKVdTNcJcZXo jOAbfRkqJOIzWqBrRc2bCzsku bafY2xn/kshJkNYyQfcs3Vb7X Z3rRMIBD24PAnLmO6YSynsqZ idB/91pTEhR87gPtYlYmad03M OrEfZ3s3ZUlv/bZWGNXPmtAU8m l3iBTeX+dLcNmsN/bq+z/2qs df3VUvwRfYghre5wEcwmcQRu Y9279ZSn/W3/1D/3L/5K5zn+ QxPht/9Ay2B4Y=</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 2 4 sen ⇣ 3φ(✓) 2 ⌘ sen ⇣ φ(✓) 2 ⌘ 3 5 2 <latexit sha1 _base64="9RVl45Yw3NXF6lgDm mvUNEyp0=">ADo3icjVHL ThsxFL3DUJ4FQrvsZkSEFDbRJ IBg4TaLojYUEQAKQ6Rx3ESw7x keypFo3wnYtN1+Yteu05VHhF4 NDPH5z7GtHeSyUDsMHb86f/7 CwuLS8svpxbX2jsvnpUmWFZLz NsjiT1xFVPBYpb2uhY36dS06TK OZX0d03U7/6yaUSWXqhxznvJn SYioFgVCPVq/xq1YgecU13gqO g1QtJzAe6Q/oDSVlZEpkEiqcTy 9aIJUmE6qlpUjYnRIrhSCOcUX eC7k3z7fBdko/Eq+Gv72WG+t+K vUo1rId2BC9Bw4EquHGWVe6BQ B8yYFBAhxS0IhjoKDw6UADQsi R60KJnEQkbJ3DBFbQW6CKo4Ii e4fIc46jk1xbjKVdTNcJcZXo jOAbfRkqJOIzWqBrRc2bCzsku bafY2xn/kshJkNYyQfcs3Vb7X Z3rRMIBD24PAnLmO6YSynsqZ idB/91pTEhR87gPtYlYmad03M OrEfZ3s3ZUlv/bZWGNXPmtAU8m l3iBTeX+dLcNmsN/bq+z/2qs df3VUvwRfYghre5wEcwmcQRu Y9279ZSn/W3/1D/3L/5K5zn+ QxPht/9Ay2B4Y=</latexit> O fator I0 nas equações representa a intensidade máxima com apenas uma fenda. A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 2 4 sen ⇣ 3φ(✓) 2 ⌘ sen ⇣ φ(✓) 2 ⌘ 3 5 2 <latexit sha1 _base64="9RVl45Yw3NXF6lgDm mvUNEyp0=">ADo3icjVHL ThsxFL3DUJ4FQrvsZkSEFDbRJ IBg4TaLojYUEQAKQ6Rx3ESw7x keypFo3wnYtN1+Yteu05VHhF4 NDPH5z7GtHeSyUDsMHb86f/7 CwuLS8svpxbX2jsvnpUmWFZLz NsjiT1xFVPBYpb2uhY36dS06TK OZX0d03U7/6yaUSWXqhxznvJn SYioFgVCPVq/xq1YgecU13gqO g1QtJzAe6Q/oDSVlZEpkEiqcTy 9aIJUmE6qlpUjYnRIrhSCOcUX eC7k3z7fBdko/Eq+Gv72WG+t+K vUo1rId2BC9Bw4EquHGWVe6BQ B8yYFBAhxS0IhjoKDw6UADQsi R60KJnEQkbJ3DBFbQW6CKo4Ii e4fIc46jk1xbjKVdTNcJcZXo jOAbfRkqJOIzWqBrRc2bCzsku bafY2xn/kshJkNYyQfcs3Vb7X Z3rRMIBD24PAnLmO6YSynsqZ idB/91pTEhR87gPtYlYmad03M OrEfZ3s3ZUlv/bZWGNXPmtAU8m l3iBTeX+dLcNmsN/bq+z/2qs df3VUvwRfYghre5wEcwmcQRu Y9279ZSn/W3/1D/3L/5K5zn+ QxPht/9Ay2B4Y=</latexit> O fator I0 nas equações representa a intensidade máxima com apenas uma fenda. As duas expressões são idênticas. Porém, a segunda possui o formato da solução geral com N fendas (aplicada para N=3). A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> φ(✓) = kdsen(✓) = 2⇡ dsen(✓) λ <latexit sha1_base64="EdDe7zd+vUcUHYvPaF2LdJiD87s=" >ADE3icjVFPTxNBH2soFBqx65TGhI6qXZEoxeTBq4eCKYWErCkmZ2dtpOuv8yO2vSbPox/CbeuBGvng1e4aDfwt+MS4IWArPZ3 Tfv96b+c2EeawK4/sXS96j5ZXHT1bXGk/XN549b754eVRkpRayL7I408chL2SsUtk3ysTyONeSJ2EsB+F039YHn6UuVJZ+MrNcniZ 8nKqREtwQNWweBPlEtQMzkYa/Zu/ZNGJVoBNWyHR+g94JchVEI81FdYtgXgUxLRnx+bDZ8ju+G2wRdGvQj0Os+YPBIiQaBEAokU hnAMjoKeE3ThIyfuFBVxmpBydYk5GuQtSVJwYmd0ndMs5OaTWluMwvnFrRKTK8mJ8M2eTLSacJ2NebqpUu27F3Zlcu0e5vRP6yzEm 2ul0dztvPu62env1Va9iE1to032+RQ8fcIg+ZX/FT1ziyvinXn3re/Um+p9rzCP8P7/gfXma41</latexit>INJsTe57tWPtRnezEY4Z3rQVFPuWNsd6JOKd2p2J2zG10ZSsiJsziuiYsnP6nJnzFK53e7bc1X85pWXtXNTaEr/tLumCu/9f5yI4 β(✓) = kasen(✓) = 2⇡ asen(✓) λ <lat exit s ha1_ba se64="3 +jmVvh 4OdB6g H2DBt8z cew7Uk Y=">A DFHicj VFPSxt BH2uVW NqbapH L0NDwV 7CRiztp RDqpTc jNFwR WYnk2TI /mN2Vp AlX8Nv 4q036bX XIl4V7 Lfob6Yr xH+0s+ zum/d7 7838ZsI sUrnx/ cs5b/7 FwuJSb n+cuXV 6uvGm7 V+nhZay J5Io1Q fhDyXk UpkzygT yYNMSx 6HkdwPJ zu2vn8 ida7S5 Js5zeR zEeJGi rBDVH jd0glIZ vBmZMv /fsM5t wVgY6Zr lMpjP0 VpCpYD DUXJRPC KZlENG aAz49bj T9lu8G ewzaFW iGt208 QsBkg hUCGR AJDOAJH Ts8h2v CREXeE kjhNSLm 6xBR18 hakqT gxE7oO6 LZYcUm NLeZuXM LWiWiV 5OT4R1 5UtJpwn Y15uqF S7bsc9 mly7R7O 6V/WGX FxBqMi f2X7075 vz7bi8 EQn1wP inrKHGO 7E1VK4 U7F7pzN dGUoIS PO4gHV NWHhnHf nzJwnd 73bs+W ufuUlr VzUWkL /La7pA tuP7zOx 6C/1Wp vtz7sb Tc7X6qr rmEDb7 FJ9/kRH XxFz3 KPscVr nHjnXnf vQvx1 +pN1d5 1nFveD/ /AOK8r pQ=</l atexit> As posições para os máximos principais e as posições de interferência destrutiva podem ser obtidas com a expressão da parte da interferência. As posições dos máximos principais são obtidas com a condição de os três raios se interferirem construtivamente. A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> A condição para os três raios interferirem construtivamente é φ(✓m) = 2m⇡ , m = (0, ±1, ±2, ±3, · · · ) <latexit sha1 _base64="tMGah8Stk1Gw/5Oc GexvIPvEO4=">ADBnicjVHL ThsxFD0ZaElDH4Eu7GIkIUR TOBim6Q0rLpMkgEkBgUzTiGWIx nrBlPJYTY8yfsuqu67Q90S9Q/ gL/g2plItKhqPRr7+Nx7jn19Y5 3Iwvj+r5q3sPjs+VL9RWP5av Xb5orqwdFVuZcDHmWZPlRHBUik akYGmkScaRzEak4EYfx+a6NH3 4ReSGzdN9caHGiorNUnkoeGaJ GzY+hnsh2aCbCRCO1wXZYT4Vas rATdlhntqidt8JtQpor1XPzZ s083Fmio1Rs+V3fTfYUxBUoIVq DLmFCHGyMBRQkEghSGcIEJB3 zEC+NDEneCSuJyQdHGBKzRIW1K WoIyI2HOaz2h3XLEp7a1n4dSc Tknoz0nJsE6ajPJywvY05uKlc 7bs37wvnae92wWtceWliDWYEPs v3Tzf3W2FoNTfHA1SKpJO8ZW xyuX0r2KvTl7VJUhB02cxWOK54 S5U87fmTlN4Wq3bxu5+J3LtKz d8yq3xL29JTU4+LOdT8FBrxtsd d/vbX6n6pW1/EOa2hTP7fRx2 cMCTvG/zELabetfV+Z9n6V 6tUrzFr8N78cDlDWk/A=</lat exit> dsen(✓m) = mλ <latexit sha1 _base64="qDC1oFyzsEpGnlCWq WFtqa79a/A=">AC5HicjVHL SsNAFD3GV31XbowWATdlEQqu hGKblwqWC0pUzSqQ3Ni8lEKV Ld+7ErT/gVr9F/AP9C+MEdQi OiHJmXPvOTP3XjcJ/FRa1suYMT 4xOTVdmJmdm19YXCour5yncSY 8XvPiIBZ1l6U8CNek74MeD0Rn IVuwC/c3pGKX1xkfpxdCb7CW +G7DLyO7HJFGt4np74IjQTHk 03HJkl0vWCrfNAzN0AjJps1axZ JUtvcxRYOeghHydxMVnOGgjho cMITgiSMIBGFJ6GrBhISGuiQFx gpCv4xDzJI2oyxOGYzYHn0va dfI2Yj2yjPVao9OCegVpDSxSZq Y8gRhdZqp45l2Vuxv3gPtqe7W p7+be4XESnSJ/Uv3mflfnapFo oN9XYNPNSWaUdV5uUumu6Jubn6 pSpJDQpzCbYoLwp5WfvbZ1JpU 1656y3T8VWcqVu29PDfDm7olDd j+Oc5RcL5Ttivl3dNKqXqYj7q ANWxgi+a5hyqOcYIaeV/jAY94M jrGjXFr3H2kGmO5ZhXflnH/Dp wKm50=</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> A condição para os três raios interferirem construtivamente é φ(✓m) = 2m⇡ , m = (0, ±1, ±2, ±3, · · · ) <latexit sha1 _base64="tMGah8Stk1Gw/5Oc GexvIPvEO4=">ADBnicjVHL ThsxFD0ZaElDH4Eu7GIkIUR TOBim6Q0rLpMkgEkBgUzTiGWIx nrBlPJYTY8yfsuqu67Q90S9Q/ gL/g2plItKhqPRr7+Nx7jn19Y5 3Iwvj+r5q3sPjs+VL9RWP5av Xb5orqwdFVuZcDHmWZPlRHBUik akYGmkScaRzEak4EYfx+a6NH3 4ReSGzdN9caHGiorNUnkoeGaJ GzY+hnsh2aCbCRCO1wXZYT4Vas rATdlhntqidt8JtQpor1XPzZ s083Fmio1Rs+V3fTfYUxBUoIVq DLmFCHGyMBRQkEghSGcIEJB3 zEC+NDEneCSuJyQdHGBKzRIW1K WoIyI2HOaz2h3XLEp7a1n4dSc Tknoz0nJsE6ajPJywvY05uKlc 7bs37wvnae92wWtceWliDWYEPs v3Tzf3W2FoNTfHA1SKpJO8ZW xyuX0r2KvTl7VJUhB02cxWOK54 S5U87fmTlN4Wq3bxu5+J3LtKz d8yq3xL29JTU4+LOdT8FBrxtsd d/vbX6n6pW1/EOa2hTP7fRx2 cMCTvG/zELabetfV+Z9n6V 6tUrzFr8N78cDlDWk/A=</lat exit> dsen(✓m) = mλ <latexit sha1 _base64="qDC1oFyzsEpGnlCWq WFtqa79a/A=">AC5HicjVHL SsNAFD3GV31XbowWATdlEQqu hGKblwqWC0pUzSqQ3Ni8lEKV Ld+7ErT/gVr9F/AP9C+MEdQi OiHJmXPvOTP3XjcJ/FRa1suYMT 4xOTVdmJmdm19YXCour5yncSY 8XvPiIBZ1l6U8CNek74MeD0Rn IVuwC/c3pGKX1xkfpxdCb7CW +G7DLyO7HJFGt4np74IjQTHk 03HJkl0vWCrfNAzN0AjJps1axZ JUtvcxRYOeghHydxMVnOGgjho cMITgiSMIBGFJ6GrBhISGuiQFx gpCv4xDzJI2oyxOGYzYHn0va dfI2Yj2yjPVao9OCegVpDSxSZq Y8gRhdZqp45l2Vuxv3gPtqe7W p7+be4XESnSJ/Uv3mflfnapFo oN9XYNPNSWaUdV5uUumu6Jubn6 pSpJDQpzCbYoLwp5WfvbZ1JpU 1656y3T8VWcqVu29PDfDm7olDd j+Oc5RcL5Ttivl3dNKqXqYj7q ANWxgi+a5hyqOcYIaeV/jAY94M jrGjXFr3H2kGmO5ZhXflnH/Dp wKm50=</latexit> Para esses pontos, a intensidade é proporcional ao valor I(✓m) ' 9I0 <latexit sha1_base64="9vel Tbm9xDwRNE4PNwopjkIBI2U=">AC3XicjVHLSsNAFD2Nr1pfV TeCm8Ei6Kakoqi7ohu7q2BbwUpI4qiDeZmZCFJ0507c+gNu9XfE P9C/8M6Yg9EJyQ5c+49Z+be6yWBkMq2XwrWwODQ8EhxtDQ2Pj E5VZ6eacs4S3e8uMgTvc9V/JARLylhAr4fpJyN/QC3vHOtnW8c 8FTKeJoT10m/DB0TyJxLHxXEeWU5xpLXKleuEy6wrRcjP2SZr OLZTrthV2yz2E9RyUEG+mnH5GV0cIYaPDCE4IijCAVxIeg5Qg42 EuEP0iEsJCRPnuEKJtBlcpwiT2j7wntDnI2or32lEbt0ykBvS kpGRZJE1NeSlifxkw8M86a/c27Zz13S7p7+VeIbEKp8T+petn/ lena1E4xoapQVBNiWF0dX7ukpmu6JuzT1UpckiI0/iI4ilh3yj7 fWZGI03tureuib+aTM3qvZ/nZnjTt6QB176P8ydor1Rrq9W13d VKfSsfdRHzWMASzXMdeygiRZ5X+MBj3iyHOvGurXuPlKtQq6Zx Zdl3b8DGz6X6A=</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL JOrOG1q195FqTRSadXxZ1v07m eKViw=</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL JOrOG1q195FqTRSadXxZ1v07m eKViw=</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL eKViw=</latexit>JOrOG1q195FqTRSadXxZ1v07m φ = 2⇡ 3 <latexi t sha1_base64 ="NwR4k5bQB4 mBA6I0aj4OqHc 8urY=">AC23 icjVHLSsNAFD 2Nr/qOCm7cBIv gqRV0Y1QdOy gn2ALZJMpzo0T cJkIkjtyp249 Qfc6v+If6B/4Z 0xglpEJyQ5c+4 9Z+be68eBSJT rvuSsfGJyan8 9Mzs3PzCor20X E+iVDJeY1EQya bvJTwQIa8poQ LejCX3+n7AG37 vUMcbl1wmIgpP 1FXM23vPBRd wTxF1Jm92ovh LPvtDpd6bFBuR WL4WBreGYX3KJ rljMKShkoIFv VyH5GCx1EYEjR B0cIRTiAh4SeU 5TgIiaujQFxk pAwcY4hZkibUh anDI/YHn3PaXe asSHtWdi1IxO CeiVpHSwQZqI 8iRhfZpj4qlx1 uxv3gPjqe92RX 8/8+oTq3B7F +6z8z/6nQtCl3 smRoE1RQbRlfH MpfUdEXf3PlSl SKHmDiNOxSXh JlRfvbZMZrE1K 5765n4q8nUrN6 zLDfFm74lDbj 0c5yjoF4ulraL O8fbhcpBNuo81 rCOTZrnLio4Qh U18r7GAx7xZL WtG+vWuvtItXK ZgXflnX/Dk+r mBA=</latexi t> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL eKViw=</latexit>JOrOG1q195FqTRSadXxZ1v07m φ = 2⇡ 3 <latexi t sha1_base64 ="NwR4k5bQB4 mBA6I0aj4OqHc 8urY=">AC23 icjVHLSsNAFD 2Nr/qOCm7cBIv gqRV0Y1QdOy gn2ALZJMpzo0T cJkIkjtyp249 Qfc6v+If6B/4Z 0xglpEJyQ5c+4 9Z+be68eBSJT rvuSsfGJyan8 9Mzs3PzCor20X E+iVDJeY1EQya bvJTwQIa8poQ LejCX3+n7AG37 vUMcbl1wmIgpP 1FXM23vPBRd wTxF1Jm92ovh LPvtDpd6bFBuR WL4WBreGYX3KJ rljMKShkoIFv VyH5GCx1EYEjR B0cIRTiAh4SeU 5TgIiaujQFxk pAwcY4hZkibUh anDI/YHn3PaXe asSHtWdi1IxO CeiVpHSwQZqI 8iRhfZpj4qlx1 uxv3gPjqe92RX 8/8+oTq3B7F +6z8z/6nQtCl3 smRoE1RQbRlfH MpfUdEXf3PlSl SKHmDiNOxSXh JlRfvbZMZrE1K 5765n4q8nUrN6 zLDfFm74lDbj 0c5yjoF4ulraL O8fbhcpBNuo81 rCOTZrnLio4Qh U18r7GAx7xZL WtG+vWuvtItXK ZgXflnX/Dk+r mBA=</latexi t> ~v0 <latexit sha1_base64="vtzB yQUyFJhYHNdgiahXtmKa8hc=">ACz3icjVHLSsNAFD2Nr1pfV ZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/ 4Z0xBbWITkhy5tx7zsy91w5dJxaG8ZrTlpZXVtfy64WNza3tne LuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHs RP412Ia8p5njXxn6DBLENXtTjhLJ7N+asz0frFklA219EVgZqCE bNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJt QFqcMi9gxfUe062SsT3vpGSs1o1NceiNS6jgiTUB5EWF5mq7iX KW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6 ljmkqiuyJvrX6oS5BASJ/GA4hFhpTzPutKE6vaZW8tFX9TmZKV e5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcY UaGuQd4hFPeNbq2q12p91/pmq5TLOPb0t7+AsR5Qq</latexit > A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL eKViw=</latexit>JOrOG1q195FqTRSadXxZ1v07m φ = 2⇡ 3 <latexi t sha1_base64 ="NwR4k5bQB4 mBA6I0aj4OqHc 8urY=">AC23 icjVHLSsNAFD 2Nr/qOCm7cBIv gqRV0Y1QdOy gn2ALZJMpzo0T cJkIkjtyp249 Qfc6v+If6B/4Z 0xglpEJyQ5c+4 9Z+be68eBSJT rvuSsfGJyan8 9Mzs3PzCor20X E+iVDJeY1EQya bvJTwQIa8poQ LejCX3+n7AG37 vUMcbl1wmIgpP 1FXM23vPBRd wTxF1Jm92ovh LPvtDpd6bFBuR WL4WBreGYX3KJ rljMKShkoIFv VyH5GCx1EYEjR B0cIRTiAh4SeU 5TgIiaujQFxk pAwcY4hZkibUh anDI/YHn3PaXe asSHtWdi1IxO CeiVpHSwQZqI 8iRhfZpj4qlx1 uxv3gPjqe92RX 8/8+oTq3B7F +6z8z/6nQtCl3 smRoE1RQbRlfH MpfUdEXf3PlSl SKHmDiNOxSXh JlRfvbZMZrE1K 5765n4q8nUrN6 zLDfFm74lDbj 0c5yjoF4ulraL O8fbhcpBNuo81 rCOTZrnLio4Qh U18r7GAx7xZL WtG+vWuvtItXK ZgXflnX/Dk+r t>mBA=</latexi ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr gkzIxy3GfbaFJTu+6tZ+KvJlOzes+K3Axv+pY0YPf7OH+Ci52qu1vdO92t1A6LUZewgU1s0z3UcMxTlAnb4kHPOLJOrOG1q195Fq TRSadXxZ1v07meKViw=</latexit> ~v0 <latexit sha1_base64="vtzB yQUyFJhYHNdgiahXtmKa8hc=">ACz3icjVHLSsNAFD2Nr1pfV ZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/ 4Z0xBbWITkhy5tx7zsy91w5dJxaG8ZrTlpZXVtfy64WNza3tne LuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHs RP412Ia8p5njXxn6DBLENXtTjhLJ7N+asz0frFklA219EVgZqCE bNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJt QFqcMi9gxfUe062SsT3vpGSs1o1NceiNS6jgiTUB5EWF5mq7iX KW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6 ljmkqiuyJvrX6oS5BASJ/GA4hFhpTzPutKE6vaZW8tFX9TmZKV e5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcY UaGuQd4hFPeNbq2q12p91/pmq5TLOPb0t7+AsR5Qq</latexit > A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL eKViw=</latexit>JOrOG1q195FqTRSadXxZ1v07m φ = 2⇡ 3 <latexi t sha1_base64 ="NwR4k5bQB4 mBA6I0aj4OqHc 8urY=">AC23 icjVHLSsNAFD 2Nr/qOCm7cBIv gqRV0Y1QdOy gn2ALZJMpzo0T cJkIkjtyp249 Qfc6v+If6B/4Z 0xglpEJyQ5c+4 9Z+be68eBSJT rvuSsfGJyan8 9Mzs3PzCor20X E+iVDJeY1EQya bvJTwQIa8poQ LejCX3+n7AG37 vUMcbl1wmIgpP 1FXM23vPBRd wTxF1Jm92ovh LPvtDpd6bFBuR WL4WBreGYX3KJ rljMKShkoIFv VyH5GCx1EYEjR B0cIRTiAh4SeU 5TgIiaujQFxk pAwcY4hZkibUh anDI/YHn3PaXe asSHtWdi1IxO CeiVpHSwQZqI 8iRhfZpj4qlx1 uxv3gPjqe92RX 8/8+oTq3B7F +6z8z/6nQtCl3 smRoE1RQbRlfH MpfUdEXf3PlSl SKHmDiNOxSXh JlRfvbZMZrE1K 5765n4q8nUrN6 zLDfFm74lDbj 0c5yjoF4ulraL O8fbhcpBNuo81 rCOTZrnLio4Qh U18r7GAx7xZL WtG+vWuvtItXK ZgXflnX/Dk+r t>mBA=</latexi ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW 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UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 YPf7OH+Ci52qu1vdO92t1A6LUZ ewgU1s0z3UcMxTlAnb4kHPOL eKViw=</latexit>JOrOG1q195FqTRSadXxZ1v07m φ = 2⇡ 3 <latexi t sha1_base64 ="NwR4k5bQB4 mBA6I0aj4OqHc 8urY=">AC23 icjVHLSsNAFD 2Nr/qOCm7cBIv gqRV0Y1QdOy gn2ALZJMpzo0T cJkIkjtyp249 Qfc6v+If6B/4Z 0xglpEJyQ5c+4 9Z+be68eBSJT rvuSsfGJyan8 9Mzs3PzCor20X E+iVDJeY1EQya bvJTwQIa8poQ LejCX3+n7AG37 vUMcbl1wmIgpP 1FXM23vPBRd wTxF1Jm92ovh LPvtDpd6bFBuR WL4WBreGYX3KJ rljMKShkoIFv VyH5GCx1EYEjR B0cIRTiAh4SeU 5TgIiaujQFxk pAwcY4hZkibUh anDI/YHn3PaXe asSHtWdi1IxO CeiVpHSwQZqI 8iRhfZpj4qlx1 uxv3gPjqe92RX 8/8+oTq3B7F +6z8z/6nQtCl3 smRoE1RQbRlfH MpfUdEXf3PlSl SKHmDiNOxSXh JlRfvbZMZrE1K 5765n4q8nUrN6 zLDfFm74lDbj 0c5yjoF4ulraL O8fbhcpBNuo81 rCOTZrnLio4Qh U18r7GAx7xZL WtG+vWuvtItXK ZgXflnX/Dk+r t>mBA=</latexi ~vφ <latexit sha1_base64="4sBYbIEmvEk6PEmdlKxp5+vahw=" >AC0XicjVHLSsNAFD3GV31XboJFsFVSUTRZdGNS0VrC1YlmU7boXkxmRKIhbf8Ct/pT4B/oX3hlT8IHohCRnzr3nzNx7/SQq XKclwlrcmp6ZrY0N7+wuLS8Ul5du0jTDJeZ3EQy6bvpTwQEa8roQLeTCT3Qj/gDb9/pONAZepiKNzNUz4Veh1I9ERzFNEXbcGnOW D0U3eSnpidFOuOFXHLPsncAtQbFO4vIzWmgjBkOGEBwRFOEAHlJ6LuHCQULcFXLiJCFh4hwjzJM2oyxOGR6xfp2aXdZsBHtWdq 1IxOCeiVpLSxRZqY8iRhfZpt4plx1uxv3rnx1Hcb0t8vEJiFXrE/qUbZ/5Xp2tR6ODA1CopsQwujpWuGSmK/rm9qeqFDkxGncpr TRSadXxZ1v07meKViw=</latexit>gkzIxy3GfbaFJTu+6tZ+KvJlOzes+K3Axv+pY0YPf7OH+Ci52qu1vdO92t1A6LUZewgU1s0z3UcMxTlAnb4kHPOLJOrOG1q195Fq ~v2φ <latexit sha1_base64="7zrF DcmPH6CrVmY4Rx81Ysb6Ub0=">AC1HicjVHLTsJAFD3UF+ID1 KWbRmLihSC0SXRjUtM5JEAIe0wITSNu2UhCAr49YfcKvfZPwD /QvjCVRidFp2p4595w7c+91AldE0rJeU8bK6tr6Rnozs7W9s5 vN7e3XIz8OGa8x3/XDpmNH3BUer0khXd4MQm6PHZc3nNGlijcmP IyE793IacA7Y3vgib5gtiSqm8u2J5zNJvPurNQOhmLezeWtgqWX uQyKCcgjWVU/94I2evDBEGMDg+SsAsbET0tFGEhIK6DGXEhIaH jHNkyBuTipPCJnZE3wHtWgnr0V7ljLSb0SkuvSE5TRyTxydSF idZup4rDMr9rfcM51T3W1KfyfJNSZWYkjsX76F8r8+VYtEH+e6B kE1BZpR1bEkS6y7om5ufqlKUoaAOIV7FA8JM+1c9NnUnkjXrnpr 6/ibVipW7VmijfGubkDLv4c5zKolwrFcuH0upyvXCSjTuMQRz iheZ6hgitUdMzf8QTno26cWvcGfefUiOVeA7wbRkPH83Alfg=< /latexit> φ = 4⇡ 3 <latexit sha1_base64="xdaL kNHa/G2Ty3SIR3gilqrj0t8=">AC23icjVHLSsNAFD2N73dVc OMmWARXJdWKbgTRjcsK9gGmSDKd6mCahMlEKLErd+LWH3Cr/yP+ gf6Fd8YIahGdkOTMufecmXuvHwciUY7zUrBGRsfGJyanpmdm5+ YXiotLjSRKJeN1FgWRbPlewgMR8roSKuCtWHKv5we86V8e6njzi stEROGJ6se83fPOQ9EVzFNEnRVX3PhC2Hu2+lKj2VNxaDbGtw Viw5ZcsexhUclBCvmpR8RkuOojAkKIHjhCKcAPCT2nqMBTFw bGXGSkDBxjgGmSZtSFqcMj9hL+p7T7jRnQ9prz8SoGZ0S0CtJaW OdNBHlScL6NvEU+Os2d+8M+Op79anv5979YhVuCD2L91n5n91u haFLnZNDYJqig2jq2O5S2q6om9uf6lKkUNMnMYdikvCzCg/+2wb TWJq1731TPzVZGpW71mem+JN35IGXPk5zmHQ2CxXquXt42p/y Af9SRWsYNmucO9nGEGurkfY0HPOLJals31q195FqFXLNMr4t6 /4dVHmYEg=</latexit> ~v0 <latexit sha1_base64="vtzB yQUyFJhYHNdgiahXtmKa8hc=">ACz3icjVHLSsNAFD2Nr1pfV ZdugkVwVRKp6LoxmUL9gFtKcl0WkPzIplUSqi49Qfc6l+Jf6B/ 4Z0xBbWITkhy5tx7zsy91w5dJxaG8ZrTlpZXVtfy64WNza3tne LuXjMOkojxBgvcIGrbVsxdx+cN4QiXt8OIW57t8pY9vpTx1oRHs RP412Ia8p5njXxn6DBLENXtTjhLJ7N+asz0frFklA219EVgZqCE bNWC4gu6GCAQwIPHD4EYRcWYno6MGEgJK6HlLiIkKPiHDMUSJt QFqcMi9gxfUe062SsT3vpGSs1o1NceiNS6jgiTUB5EWF5mq7iX KW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6 ljmkqiuyJvrX6oS5BASJ/GA4hFhpTzPutKE6vaZW8tFX9TmZKV e5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcY UaGuQd4hFPeNbq2q12p91/pmq5TLOPb0t7+AsR5Qq</latexit > A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 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KW7G/eqfKUd5vS3868PGIFboj9SzfP/K9O1iIwxLmqwaGaQsXI6 ljmkqiuyJvrX6oS5BASJ/GA4hFhpTzPutKE6vaZW8tFX9TmZKV e5blJniXt6QBmz/HuQiaJ2WzUj6tV0rVi2zUeRzgEMc0zNUcY >UaGuQd4hFPeNbq2q12p91/pmq5TLOPb0t7+AsR5Qq</latexit cos(φm,n) = −1 2 <latexit sha1_base64="RalPVEGwrBSXQVXRAkYyP8jWfyI=" >AC5HicjVHLSsRAECzj+x316MHgIijokoiF0H04lHBVWFXlmScXYfNi8lEkLBHb97Eqz/gVb9F/AP9C3vGCD4QnZCkprqrZro7S EORKd97rF6+/oHBoeGR0bHxicm7anpoyzJeM1loSJPAn8jIci5jUlVMhPUsn9KAj5cdDZ1fHjCy4zkcSH6jLlp5HfjkVLMF8R1bT nGizJFhvpuWgW0XLcXK2nJVGS/qs8LrFardpV9yqa5bzE3glqKBc+4n9hAbOkIAhRwSOGIpwCB8ZPXV4cJESd4qCOElImDhHFyOk zSmLU4ZPbIe+bdrVSzamvfbMjJrRKSG9kpQOFkiTUJ4krE9zTDw3zpr9zbswnvpul/QPSq+IWIVzYv/SfWT+V6drUWh09QgqKbUML o6Vrkpiv65s6nqhQ5pMRpfEZxSZgZ5UefHaPJTO26t76Jv5hMzeo9K3NzvOpb0oC97+P8CY5Wq95adf1grbK9U456CLOYxyLNcwPb 2M+auR9hXs84NFqWdfWjX7nmr1lJoZfFnW3RvTd5tM</latexit> A figura de intensidade para uma rede de difração com 3 fendas idênticas (quadrado da amplitude) se escreve Rede de Difração I(✓) = I0 2 4 sen ⇣ β(✓) 2 ⌘ β(✓) 2 3 5 2 [1 + 2 cos [φ(✓)]]2 <latexit sha1 _base64="4dBhd5c013V1ZN5pZ +uqi2xn+Qk=">ADYXicjVFd SxwxFL2z09aPfjaR19Cl8JKY ZkZLPVFEPtS3y4KmzWJZPN7gb niyQjyLD/xP8lPvS/ovexGyx FWkzMzJuec5CZnUt4vgu6I QvXr5aWV1bf/3m7buNaHPrVFe N4mLAq7xS5xnTIpelGBhpcnFeK 8GKLBdn2eVXWz+7EkrLqjwx17 UYFWxWyqnkzCA1jm6OetTMhWE 7ZJ8cjWOai6kZ0slUMd62VBVEi 3Lh2B51JM1QvTQt2nRBlZzNDc Jn6l4wukgfwpNPKeWV9ivVc7kU ex35rR9H3bgfu0GegsSDLvhxX EW3QGECFXBoABJRjEOTDQ+Aw hgRhq5EbQIqcQSVcXsIB19Dao EqhgyF7id4azoWdLnNtM7dwcV 8nxVegk8BE9FeoUYrsacfXGJVv 2uezWZdq9XeM/81kFsgbmyP7L t1T+r8/2YmAKe64HiT3VjrHdcZ /SuFOxOyePujKYUCNn8QTrCjF 3zuU5E+fRrnd7tszVfzilZe2ce 20DP+0u8YKTv6/zKThN+8lu/ P3e7Bob/qVdiGD9D+/wCB/A NjmEAPOgEvSAJ0s59uBZG4daDt BN4z3v4Y4TbvwDnUseQ</late xit> Os pontos de interferência destrutiva ocorrem quando a soma dos três raios é um vetor nulo. Para três fendas, é necessário construir um triângulo de lados iguais: ~v0 <latexit sha1 _base64="vtzByQUyFJhYHNdgi ahXtmKa8hc=">ACz3icjVHL SsNAFD2Nr1pfVZdugkVwVRKp6 LoxmUL9gFtKcl0WkPzIplUSqi 49Qfc6l+Jf6B/4Z0xBbWITkhy 5tx7zsy91w5dJxaG8ZrTlpZXVt fy64WNza3tneLuXjMOkojxBgv cIGrbVsxdx+cN4QiXt8OIW57t8 pY9vpTx1oRHsRP412Ia8p5njX xn6DBLENXtTjhLJ7N+asz0frF klA219EVgZqCEbNWC4gu6GCAQ wIPHD4EYRcWYno6MGEgJK6HlL iIkKPiHDMUSJtQFqcMi9gxfUe0 62SsT3vpGSs1o1NceiNS6jgiT UB5EWF5mq7iXKW7G/eqfKUd5v S3868PGIFboj9SzfP/K9O1iIw xLmqwaGaQsXI6ljmkqiuyJvrX 6oS5BASJ/GA4hFhpTzPutKE6v aZW8tFX9TmZKVe5blJniXt6QB mz/HuQiaJ2WzUj6tV0rVi2zUeR zgEMc0zNUcYUaGuQd4hFPeNb q2q12p91/pmq5TLOPb0t7+AsR 5Qq</latexit> ~v2φ <l ate xi t s ha1 _b ase 64= "7 zrF Dcm PH6 Cr VmY 4Rx 81 Ysb 6Ub 0= ">AAC 1H icj VHL TsJ AF D3U F+I D1 KWb RmL ii hSC 0SX RjU tM 5JE AIe 0wIT SNu 2U hCA r49 Yfc Kv fZP wD/ QvjC VRi dF p2p 459 5w 7c+ 91A ldE 0r JeU 8bK 6t r6R noz s7 W9s 5vN 7e3 XI z8O Ga8 x3 /XD pmN H3 BUe r0k hXd 4M Qm6 PHZ c3 nNG lij cm PIy E79 3Ia cA 7Y3 vgi b5 gti Sqm 8u 2J5 zNJ vP urN QOh mLe ze Wtg qWX uQ yKC cgj WV U/9 4I2 evD BE GMM Dg+ Ss Asb ET0 tF GEh IK6 DGX Eh IaH jHH Nk yBu Tip PC JnZ E3w Ht Wgn r0V 7lj LS b0S kuv SE 5TR yTx ydSF idZ up4 rD Mr9 rfc M5 1T3 W1K fy fJN SZW Ykj sX 76F 8r8 +V YtE H+e 6B kE1 BZp R1b Ek S6y 7om 5u fql KUo aA OIV 7FA 8J M+1 c9N nUn kj Xrn pr6 /i bVi pW7 Vm ijf Gub kkD Lv 4c5 zKo lw rFc uH0 up yvX CSj TuM QR zih eZ6 hg itUdM zf 8QT no2 6c Wvc Gfe fUi OV eA7 wbR kP H83 Alf g= </l ate >xit ~vφ <latexit sha1 _base64="4sBYbIEmvEk6PEmd lKxp5+vahw=">AC0XicjVHL SsNAFD3GV31XboJFsFVSUTRZ dGNS0VrC1YlmU7boXkxmRKIh bf8Ct/pT4B/oX3hlT8IHohCRn zr3nzNx7/SQqXKclwlrcmp6Zr Y0N7+wuLS8Ul5du0jTDJeZ3E Qy6bvpTwQEa8roQLeTCT3Qj/gD b9/pONAZepiKNzNUz4Veh1I9 ERzFNEXbcGnOWD0U3eSnpidFO uOFXHLPsncAtQbFO4vIzWmgjB kOGEBwRFOEAHlJ6LuHCQULcFX LiJCFh4hwjzJM2oyxOGR6xfp2 aXdZsBHtWdq1IxOCeiVpLSxR ZqY8iRhfZpt4plx1uxv3rnx1Hc b0t8vEJiFXrE/qUbZ/5Xp2tR 6ODA1CopsQwujpWuGSmK/rm9 qeqFDkxGncprgkzIxy3GfbaFJ Tu+6tZ+KvJlOzes+K3Axv+pY0 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>AC7nicjVHLTtAFD24BVKeoSzZWI2QKDI4SHYVIraDctUagISRpE9TJR/BiNx0iRlW9gx65iyw90Sz8D8QflL7gzdaQCQjCW7 TPn3nNm7r2hjESmPe9+yvnwcXpmtvJpbn5hcWm5uvK5k6W5YrzN0ihVJ2GQ8UgkvK2FjviJVDyIw4gfh8PvJn58wVUm0uSnHkl+Fgf 9RPQECzR3eqmLweiW8Tbyfjrji+FH/Ge3vB7KmBFMi52x1uxr0R/oDe71ZpX9+xyX4JGCWoVyut3sHOVIw5IjBkUATjhAgo+cU DXiQxJ2hIE4REjbOMcYcaXPK4pQREDukb592pyWb0N54ZlbN6JSIXkVKF+ukSlPETanuTaeW2fDvuZdWE9ztxH9w9IrJlZjQOxbuk nme3WmFo0eDm0NgmqSljHVsdIlt10xN3f/q0qTgyTO4HOK8LMKid9dq0ms7Wb3gY2/tdmGtbsWZmb48HckgbceD7Ol6CzU2/s1fd/ 7NWa38pRV7CGL9igeR6giSO0CbvS/zGLf40rlyfjnX/1KdqVKzifLuXkEB+ygYA=</latexit> 1  n < 3 <latexit sha1_base64="548t3mW+pUSJCrX5vaJIypyRoPI=" >ACzXicjVHLSsNAFD2Nr1pfVZdugkVwVRKt6MJF0Y07K9gHtkWS6bQO5kUyEUrVrT/gVn9L/AP9C+MKahFdEKSM+fec2buvW7ki URa1mvOmJqemZ3LzxcWFpeWV4qra40kTGPG6yz0wrjlOgn3RMDrUkiPt6KYO7r8aZ7fazizRseJyIMzuUw4l3fGQSiL5gjibqwOx4 3A/PQ3L0slqypZc5CewMlJCtWlh8Qc9hGBI4YMjgCTswUFCTxs2LETEdTEiLiYkdJzjDgXSpTFKcMh9pq+A9q1MzagvfJMtJrR KR69MSlNbJEmpLyYsDrN1PFUOyv2N+R9lR3G9Lfzbx8YiWuiP1LN878r07VItHga5BUE2RZlR1LHNJdVfUzc0vVUlyiIhTuEfxmD DTynGfTa1JdO2qt46Ov+lMxao9y3JTvKtb0oDtn+OcBI2dsl0p751VStWjbNR5bGAT2zTPfVRxghrq5B3gEU94Nk6N1Lg17j9TjVym Wce3ZTx8AOJxkgI=</latexit> Rede de Difracdo A figura de intensidade para uma rede de difragdo com 3 fendas idénticas (quadrado da amplitude) se escreve 30) \} ° sen (20) > 2 Um esboco do grafico da intensidade (normalizada com I0=1 w/ m*2) da parte da interferéncia como fungdo do angulo phi é dado a seguir. : ol L) \ A A / p • Difração por três fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Difração por N (>3) fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Resolução de um espectrógrafo: poder de resolução. Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. Rede de Difração Rede de Difracdo Considere agora uma rede de difracdo composta por N fendas idénticas e igualmente espacadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicagado em espectroscopia. A figura ao lado mostra um conjunto | [\ de 8 fendas. O padrdo de | oA intensidade, pelo mesmo argumento | 4S f anterior, vale: | eZ 1 = tal [SE lag Lo a LY’ Yee FT dsen Oe... O maximo ocorre quando a diferenca de caminho entre duas fendas adjacentes é um multiplo inteiro de comprimentos de onda: dsen@ = ma. Rede de Difracdo Considere agora uma rede de difracdo composta por N fendas idénticas e igualmente espacadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicagdo em espectroscopia. A figura ao lado mostra um conjunto | [\ de 8 fendas. O padrdo de | oA intensidade, pelo mesmo argumento | P| — SS P anterior, vale: | +A 10) =) [OY s < Vamos determinar a contribuicdo do % i’ | padrdo de intensidade devido a YAR cen Bee | interferéncia das N fendas. (©) maximo ocorre quando a diferenca de caminho entre duas fendas adjacentes é um multiplo inteiro de comprimentos de onda: dsen@ = mia. Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. A interferência construtiva ocorre quando a diferença de fase da luz advinda de cada fenda é um múltiplo de 2pi. Todas as ondas das fendas vão interferir construtivamente. Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. A interferência construtiva ocorre quando a diferença de fase da luz advinda de cada fenda é um múltiplo de 2pi. Todas as ondas das fendas vão interferir construtivamente. dsen(✓) = mλ (m = 0, ±1, ±2, ±3, ...) Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. A interferência construtiva ocorre quando a diferença de fase da luz advinda de cada fenda é um múltiplo de 2pi. Todas as ondas das fendas vão interferir construtivamente. dsen(✓) = mλ (m = 0, ±1, ±2, ±3, ...) Entre dois máximos existem pontos de mínimo e pontos de máximos locais, quando apenas algumas fendas interferem construtivamente Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. Para duas fendas, existe apenas um mínimo entre dois máximos. Isso não é verdade com N fendas entre os “máximos principais” deduzidos anteriormente. As posições dos mínimos podem ser obtidas geometricamente vai método de fasores. Considere o caso com 8 fendas: Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. Para duas fendas, existe apenas um mínimo entre dois máximos. Isso não é verdade com N fendas entre os “máximos principais” deduzidos anteriormente. As posições dos mínimos podem ser obtidas geometricamente vai método de fasores. Considere o caso com 8 fendas: Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. Para duas fendas, existe apenas um mínimo entre dois máximos. Isso não é verdade com N fendas entre os “máximos principais” deduzidos anteriormente. As posições dos mínimos podem ser obtidas geometricamente vai método de fasores. Considere o caso com 8 fendas: d λsen(✓) = m N (m = ±1, ±2, ..., ±N − 1) Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. A intensidade devido à interferência para uma rede de difração é: Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. A intensidade devido à interferência para uma rede de difração é: Rede de Difração Considere agora uma rede de difração composta por N fendas idênticas e igualmente espaçadas. O limite de Fraunhofer é obtido com R>>Nd. Essa rede possui aplicação em espectroscopia. A intensidade devido à interferência para uma rede de difração é: Rede de Difração A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] Rede de Difração A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] E(✓) = N−1 X j=0 E0 cos (!t + jφ) Rede de Difração A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] E(✓) = N−1 X j=0 E0 cos (!t + jφ) Uma forma de obtermos uma expressão fechada para a amplitude é com os fasores. Veja que podemos fazer: Rede de Difração A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] E(✓) = N−1 X j=0 E0 cos (!t + jφ) Uma forma de obtermos uma expressão fechada para a amplitude é com os fasores. Veja que podemos fazer: E(✓) = Re 8 < : N−1 X j=0 E0ei(!t+jφ) 9 = ; Rede de Difração A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] E(✓) = N−1 X j=0 E0 cos (!t + jφ) Uma forma de obtermos uma expressão fechada para a amplitude é com os fasores. Veja que podemos fazer: E(✓) = Re 8 < : N−1 X j=0 E0ei(!t+jφ) 9 = ; = Re 8 < :E0ei!t N X j=0 [eiφ]j 9 = ; Rede de Difração A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] Note que a soma no slide anterior pode ser realizada analiticamente: Rede de Difração 1 + x + x2 + · · · + xN−1 = 1 − xN 1 − x <latexit sha1_base64="COaW ybmuTcf1S23CuwN9T8dM7b8=">AC8nicjVHLSsRAECzj+x316 CW4CMKySyKXgTRiydRcHVhVyWZndVgXkwmoR8hTdv4tUf8Kof If6B/oU9YwQfiE5IUlPdVTPd7SWBn0rbfu4xev6BwaHhkdGx8 YnJs2p6f0zgTjDRYHsWh6bsoDP+IN6cuANxPB3dAL+IF3tqniB +dcpH4c7cnLhB+G7knkd3mSqKOzZpTvaheHC1W26wTy5Rgvl1z CmvNane6wmW5U1NUShQHJsVu27rZf0ETgkqKNdObD6hjQ5iMGQ IwRFBEg7gIqWnBQc2EuIOkRMnCPk6zlFghLQZXHKcIk9o+8J7V olG9FeaZazeiUgF5BSgvzpIkpTxBWp1k6nmlnxf7mnWtPdbdL+ nulV0isxCmxf+k+Mv+rU7VIdLGqa/CpkQzqjpWumS6K+rm1qeq JDkxCncobgzLTyo8+W1qS6dtVbV8dfdKZi1Z6VuRle1S1pwM 73cf4E+4t1Z6m+vLtUWd8oRz2EWcxhgea5gnVsYQcN8r7CPR7wa Ej2rgxbt9TjZ5SM4Mvy7h7A0q2oF8=</latexit> A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] Note que a soma no slide anterior pode ser realizada analiticamente: Rede de Difração 1 + x + x2 + · · · + xN−1 = 1 − xN 1 − x <latexit sha1_base64="COaW ybmuTcf1S23CuwN9T8dM7b8=">AC8nicjVHLSsRAECzj+x316 CW4CMKySyKXgTRiydRcHVhVyWZndVgXkwmoR8hTdv4tUf8Kof If6B/oU9YwQfiE5IUlPdVTPd7SWBn0rbfu4xev6BwaHhkdGx8 YnJs2p6f0zgTjDRYHsWh6bsoDP+IN6cuANxPB3dAL+IF3tqniB +dcpH4c7cnLhB+G7knkd3mSqKOzZpTvaheHC1W26wTy5Rgvl1z CmvNane6wmW5U1NUShQHJsVu27rZf0ETgkqKNdObD6hjQ5iMGQ IwRFBEg7gIqWnBQc2EuIOkRMnCPk6zlFghLQZXHKcIk9o+8J7V olG9FeaZazeiUgF5BSgvzpIkpTxBWp1k6nmlnxf7mnWtPdbdL+ nulV0isxCmxf+k+Mv+rU7VIdLGqa/CpkQzqjpWumS6K+rm1qeq JDkxCncobgzLTyo8+W1qS6dtVbV8dfdKZi1Z6VuRle1S1pwM 73cf4E+4t1Z6m+vLtUWd8oRz2EWcxhgea5gnVsYQcN8r7CPR7wa Ej2rgxbt9TjZ5SM4Mvy7h7A0q2oF8=</latexit> N−1 X j=0 ⇥ eiφ⇤j = 1 − eiNφ 1 − eiφ <latexit sha1_base64="0ctY Z8VNW97cprSwzyQkY/K1XCs=">ADEnicjVFNT9tAEH24H1CgJ bTHXiwiJC5EdgWC1JEL70UYkAUpxE9maTLPhL6zVSZPlf8E96 6w1x5Y64tpfyL5hdFqktqmAt2/ezHu7sxPlsSiU593MOC9evn o9O/dmfmHx7bulxvL7wyIrJeMdlsWZPI7Cgsci5R0lVMyPc8nDJ Ir5UXT6WePzrgsRJYeqGnOe0k4TsVIsFARNWh8DYoyGVQnO17d r/bW/TqI+Uh1eb8SQT4RtRtIMZ6oXv/E3XGD4UiGrPLXdXrP5Gs b3QeDRtNreWa5j4FvQRN27WeNawQYIgNDiQcKRThGCEKerw4S EnroeKOElImDxHjXnSlTFqSIk9pS+Y4q6lk0p1p6FUTPaJaZXk tLFKmkyqpOE9W6uyZfGWbP/86Mpz7blP6R9UqIVZgQ+5TuofK5 Ot2LwgjbpgdBPeWG0d0x61KaW9End/oSpFDTpzGQ8pLwswoH+ 7ZNZrC9K7vNjT536ZSszpmtrbErT4lDdj/d5yPweGnlr/R2vy20 Wzv2lHP4SNWsEbz3EIbX7CPDnl/xw1+4pdz7vxwLpzL+1Jnxmo+ 4K/lXN0BnH+vTg=</latexit> A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas é: E(✓) = E0 cos (!t) + E0 cos (!t + φ) + E0 cos (!t + 2φ) + ... + +E0 cos [!t + (N − 1)φ] Note que a soma no slide anterior pode ser realizada analiticamente: N X j=0 [eiφ]j = ei(N−1)φ/2 sen(Nφ/2) sen(φ/2) Rede de Difração 1 + x + x2 + · · · + xN−1 = 1 − xN 1 − x <latexit sha1_base64="COaW ybmuTcf1S23CuwN9T8dM7b8=">AC8nicjVHLSsRAECzj+x316 CW4CMKySyKXgTRiydRcHVhVyWZndVgXkwmoR8hTdv4tUf8Kof If6B/oU9YwQfiE5IUlPdVTPd7SWBn0rbfu4xev6BwaHhkdGx8 YnJs2p6f0zgTjDRYHsWh6bsoDP+IN6cuANxPB3dAL+IF3tqniB +dcpH4c7cnLhB+G7knkd3mSqKOzZpTvaheHC1W26wTy5Rgvl1z CmvNane6wmW5U1NUShQHJsVu27rZf0ETgkqKNdObD6hjQ5iMGQ IwRFBEg7gIqWnBQc2EuIOkRMnCPk6zlFghLQZXHKcIk9o+8J7V olG9FeaZazeiUgF5BSgvzpIkpTxBWp1k6nmlnxf7mnWtPdbdL+ nulV0isxCmxf+k+Mv+rU7VIdLGqa/CpkQzqjpWumS6K+rm1qeq JDkxCncobgzLTyo8+W1qS6dtVbV8dfdKZi1Z6VuRle1S1pwM 73cf4E+4t1Z6m+vLtUWd8oRz2EWcxhgea5gnVsYQcN8r7CPR7wa Ej2rgxbt9TjZ5SM4Mvy7h7A0q2oF8=</latexit> N−1 X j=0 ⇥ eiφ⇤j = 1 − eiNφ 1 − eiφ <latexit sha1_base64="0ctY Z8VNW97cprSwzyQkY/K1XCs=">ADEnicjVFNT9tAEH24H1CgJ bTHXiwiJC5EdgWC1JEL70UYkAUpxE9maTLPhL6zVSZPlf8E96 6w1x5Y64tpfyL5hdFqktqmAt2/ezHu7sxPlsSiU593MOC9evn o9O/dmfmHx7bulxvL7wyIrJeMdlsWZPI7Cgsci5R0lVMyPc8nDJ Ir5UXT6WePzrgsRJYeqGnOe0k4TsVIsFARNWh8DYoyGVQnO17d r/bW/TqI+Uh1eb8SQT4RtRtIMZ6oXv/E3XGD4UiGrPLXdXrP5Gs b3QeDRtNreWa5j4FvQRN27WeNawQYIgNDiQcKRThGCEKerw4S EnroeKOElImDxHjXnSlTFqSIk9pS+Y4q6lk0p1p6FUTPaJaZXk tLFKmkyqpOE9W6uyZfGWbP/86Mpz7blP6R9UqIVZgQ+5TuofK5 Ot2LwgjbpgdBPeWG0d0x61KaW9End/oSpFDTpzGQ8pLwswoH+ 7ZNZrC9K7vNjT536ZSszpmtrbErT4lDdj/d5yPweGnlr/R2vy20 Wzv2lHP4SNWsEbz3EIbX7CPDnl/xw1+4pdz7vxwLpzL+1Jnxmo+ 4K/lXN0BnH+vTg=</latexit> A amplitude do campo elétrico, considerando apenas a interferência entre as N fendas, é: E(✓) = E0 sen(Nφ/2) sen(φ/2) cos [!t + (N − 1)φ/2] Rede de Difração A amplitude do campo elétrico, considerando apenas a interferéncia entre as N fendas, é: A intensidade de uma rede de difracdo é dada - por ch tonslioora' 1. Ei til tyb til til Rede de Difracdo A amplitude do campo elétrico, considerando apenas a interferéncia entre as N fendas, é: A intensidade de uma rede de difracdo é dada - por hit d | — 27d TT tp MM sen) _1,|_ 5 Tit Rede de Difracdo A amplitude do campo elétrico, considerando apenas a interferéncia entre as N fendas, é: A intensidade de uma rede de difracdo é dada - _ por hit d | — 27d 27a ———-T— * | (60) —= TZ sen(9) B(@) —= sent?) —, | 5 Tit Rede de Difracdo A amplitude do campo elétrico, considerando apenas a interferéncia entre as N fendas, é: sen(N@/2) E(¢) = Ey ———_— N —1)@/2 0) = Be iy) coslwt + (N — 1)0/2 A intensidade de uma rede de difracdo é dada - por i seal /aly eenteteh 2) , | | — sen[9(9)/2] Jin L B()/2 J ais d 27d 277A TT * — (60) = TZ sen(9) B(@) = sent?) TT * Note que os pontos de interferéncia | P OE construtiva (destrutiva) com todas as fendas joy Li podem ser recuperados. Rede de Difracdo A amplitude do campo elétrico, considerando apenas a interferéncia entre as N fendas, é: E(0) = Bo cos |wt + (N — 1)¢/2| A intensidade de uma rede de difracdo é dada - por ub senlvol@e | enti) ‘| sen[9(9)/2] Jin L B()/2 J ais Ty * | — 27d 27a +- * Ca o(@) = TZ sen(9) B(@) = TZ sen(9) rT * Til @ = 2mn I « NI Exemplo. Os comprimentos de onda das extremidades do espectro visível são aproximadamente 380nm (violeta) e 750nm (vermelho). a) calcule a largura angular do espectro visível de primeira ordem produzido por uma rede plana com 600 fendas por milímetro quando a luz branca incide perpendicularmente sobre a rede. b) os espectros de primeira e segunda ordem se superpõe? E os de segunda e terceira ordem? Rede de Difração Exemplo. Os comprimentos de onda das extremidades do espectro visível são aproximadamente 380nm (violeta) e 750nm (vermelho). a) calcule a largura angular do espectro visível de primeira ordem produzido por uma rede plana com 600 fendas por milímetro quando a luz branca incide perpendicularmente sobre a rede. b) os espectros de primeira e segunda ordem se superpõe? E os de segunda e terceira ordem? Rede de Difração Exemplo. Os comprimentos de onda das extremidades do espectro visível são aproximadamente 380nm (violeta) e 750nm (vermelho). a) calcule a largura angular do espectro visível de primeira ordem produzido por uma rede plana com 600 fendas por milímetro quando a luz branca incide perpendicularmente sobre a rede. b) os espectros de primeira e segunda ordem se superpõe? E os de segunda e terceira ordem? Rede de Difração Exemplo. Os comprimentos de onda das extremidades do espectro visível são aproximadamente 380nm (violeta) e 750nm (vermelho). a) calcule a largura angular do espectro visível de primeira ordem produzido por uma rede plana com 600 fendas por milímetro quando a luz branca incide perpendicularmente sobre a rede. b) os espectros de primeira e segunda ordem se superpõe? E os de segunda e terceira ordem? Rede de Difração Exemplo. Os comprimentos de onda das extremidades do espectro visível são aproximadamente 380nm (violeta) e 750nm (vermelho). a) calcule a largura angular do espectro visível de primeira ordem produzido por uma rede plana com 600 fendas por milímetro quando a luz branca incide perpendicularmente sobre a rede. b) os espectros de primeira e segunda ordem se superpõe? E os de segunda e terceira ordem? Rede de Difração Exemplo. Os comprimentos de onda das extremidades do espectro visível são aproximadamente 380nm (violeta) e 750nm (vermelho). a) calcule a largura angular do espectro visível de primeira ordem produzido por uma rede plana com 600 fendas por milímetro quando a luz branca incide perpendicularmente sobre a rede. b) os espectros de primeira e segunda ordem se superpõe? E os de segunda e terceira ordem? Rede de Difração • Difração por três fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Difração por N (>3) fendas igualmente espaçadas: padrão de intensidade no regime de Fraunhofer; • Resolução de um espectrógrafo: poder de resolução. Rede de Difração As redes de difração são amplamente empregadas para medir o espectro da luz emitida por uma fonte, uma técnica chamada de espectroscopia ou espectrometria. Rede de Difração Um parâmetro importante de uma rede de difração é: Poder de resolução cromático R, definido por: Tal que o espectrômetro consegue resolver (diferenciar) dois comprimentos de onda próximos. Rede de Difração Um parâmetro importante de uma rede de difração é: Poder de resolução cromático R, definido por: Tal que o espectrômetro consegue resolver (diferenciar) dois comprimentos de onda próximos. Exemplo. Átomos de sódio aquecidos emitem intensamente radiação com os comprimentos de onda 589,0nm e 589,59nm. Um espectrômetro conseguirá distinguir esses comprimentos de onda se: R ≥ 589, 0nm 0, 59nm = 998 Rede de Difração Um parâmetro importante de uma rede de difração é: Poder de resolução cromático R, definido por: Tal que o espectrômetro consegue resolver (diferenciar) dois comprimentos de onda próximos. Exemplo. Átomos de sódio aquecidos emitem intensamente radiação com os comprimentos de onda 589,0nm e 589,59nm. Um espectrômetro conseguirá distinguir esses comprimentos de onda se: R ≥ 589, 0nm 0, 59nm = 998 = λmenor λmaior − λmenor <latexit sha1_base64="fJad1gq3mnyVRHvpqADXPqbHkvc=" >ADCnicjVHLSgMxFD0d3+qSzeDRXBjmYqiG0EUwaWCtYVWSiZNTgvMhlBSv/AP3HnTtz6A65E/QH9C2/iCD7RDNzcu45J7mJn wQy1Z73UHD6+gcGh4ZHRsfGJyanitMzh2mcKS6qPA5iVfdZKgIZiaqWOhD1RAkW+oGo+afbpl47EyqVcXSgzxNxFLjSHYkZ5qoVnF no9nuKMa7zYBMbdbqNlXohiKVa/3hWSyKWfhK1iySt7drjfQSUHJeRjLy7eo4k2YnBkCEQRMOwJDS0AFHhLijtAlThGSti7Q wyh5M1IJUjBiT+l7TLNGzkY0N5mpdXNaJaBXkdPFAnli0inCZjX1jObNjfsrs20+ztnP5+nhUSq3FC7F+d+V/faYXjQ7WbQ+Sek osY7rjeUpmT8Xs3P3QlaEhDiD21RXhLl1vp+zaz2p7d2cLbP1Z6s0rJnzXJvhxeySLrjy9Tq/g8PlcmWlvLq/Utrcyq96GHOYxyLd 5xo2sYs9VCn7End4xJNz4Vw5187Nm9Qp5J5ZfBrO7SufvK1D</latexit> Rede de Difração Rede de Difracdo Para ilustrar a possibilidade de se diferenciar dois comprimentos de onda através da figura de intensidade. Considere o padrdo de difracdo com N=5 fendas idénticas e dois comprimentos de onda que se relacionam pela formula: Amaior — I, O5Amenor O grafico a seguir mostra o padrdo de intensidade para os dois comprimentos de onda, cujos maximos principais sdo m=O, 1, 2, 3 4e5. : | | | 0 ei JOLT p1,g2 Rede de Difracdo Para m = O, as posigdes dos maximos para ambos os comprimentos de onda sdo idénticas. Para m = 1, 2 e 3, as posigdes dos maximos sdo diferentes porém “muito proximas". A possibilidade para diferenciar os comprimentos de onda é com m maior ou igual a 4. 30 ° neohome As figuras de ery intensidade mostram que Sook | os comprimentos de onda ionts : podem ser “resolvidos" (diferenciad a 10 : os) de uma melhor TT Ey | maneira para maiores > ob wh af rofl! Bouin! NAAN WAATATIA valores de m. 5 0 : 5 10 15 20 25 30 35 sevneneed soe m=O Amaior — 1, O5Amenor Rede de Difracdo Para m = O, as posigdes dos maximos para ambos os comprimentos de onda sdo idénticas. Para m = 1, 2 e 3, as posigdes dos maximos sdo diferentes porém “muito proximas". A possibilidade para diferenciar os comprimentos de onda é com m maior ou igual a 4. 30 hen Leonean As figuras’ de ery intensidade mostram que Sook | : os comprimentos de onda ionts : 3 podem ser “resolvidos" (diferenciad = 10 : : : os) de uma melhor ee | maneira para maiores : tthe Al Wisshi post hAAN WAAIAIA valores de m. 5 0 4 5 0) 158 20 25 30 35 BEEBE SBE etait AY) m=O m=2 Amaior — 1, O5Amenor Rede de Difracdo Para m = O, as posigdes dos maximos para ambos os comprimentos de onda sdo idénticas. Para m = 1, 2 e 3, as posigdes dos maximos sdo diferentes porém “muito proximas". A possibilidade para diferenciar os comprimentos de onda é com m maior ou igual a 4. te sesee poossoeda poosecoon As figuras de p25 3 3 intensidade mostram que $20 : : 3 os comprimentos de onda ohte : 3 : 3 podem ser “resolvidos" (diferenciad - os) de uma melhor : FL: iy 3 3 | maneira para maiores : ° 0 F a Pp ob 0 : 2 . ; 30 . - 35 valores de m. $1,62 m = 0 m= 2 m=A4 Vamos determinar uma condição para a resolução de dois comprimentos de onda. O critério de Rayleigh estabelece que se a posição do m-ésimo máximo principal para um determinado comprimento de onda for maior ou igual que a posição que a intensidade se anula para o outro comprimento de onda, ambos podem ser resolvidos. Rede de Difração A condição com N fendas para a igualdade é obtida com as formulas: dsen(✓) = mλmaior <lat exit s ha1_ba se64="H 9GjVK/ /lzWpx Um6ZW07 rTNGa6 0=">A C8Hicj VHLSgM xFD2Or/ qunQT LIilK lUdCMU3 bisYLX QSslMo wbnRSYj lNKPcO dO3PoD bvUrxD/ Qv/Amj uAD0Qwz c3LuPS e593pJ IFPtus9 DzvDI6 Nh4YWJ yanpmdq 4v3CU xpnyRc OPg1g1P Z6KQEa ioaUOR DNRgode I69iz 0TP74UK pVxdKh 7iTgJ+ VkT6XP NVGd4n qX9dsq ZKmIBqt tfS40X 2M7LGT tgEy6vG OjIZex GnSKJb fs2sV+g koOSsh XPS4+oY 0uYvjI EIgi YcgCOlp 4UKXCT EnaBPn CIkbVxg gEnSZp QlKIMT e0HfM9q 1cjaiv fFMrdq nUwJ6FS kZVkgT U54ibE5 jNp5Z 8P+5t2 3nuZuPf p7uVdI rMY5sX /pPjL/q zO1aJx i29Ygq abEMqY6 P3fJbF fMzdmn qjQ5JMQ Z3KW4I uxb5Uef mdWktn bTW27j LzbTsGb v57kZX s0tacC V7+P8CY 42ypVq efOgWq rt5qMuY AnLWKV 5bqGf dTRIO8r 3OMBj4 5yrp0b5 /Y91Rn KNYv4s py7Nwi0 oFM=</ latexi t> dsen(✓) = ✓ m + 1 N ◆ λmenor <lat exit s ha1_ba se64="m J3fp0 Foxg0 hdAg1LH QtGQ4S 8=">A DC3icj VHLShx BFD124j M+xrjM psgQGB GbjHEj SAGJKt gIKOCL UN1Tc1M Yb+orh ak6U/w T9xlF7L 1B9wE0 Q9I/iK3 Ki0kim g13X3q 3HtO1b0 3ymNVG N+/nvB evJycmp 6ZnXs1 v7C41F p+vV9kp RayJ7I 404cRL 2SsUtkz ysTyMN eSJ1EsD 6KTjzZ +cCp1o bL0qznL 5XHCR6 kaKsEN Uf3W7oB VoU5YI dO6E5q xNHyVb EwlkPT SdgaC4 eaiyqoq 891qNV obFYpSP 4D3nfC RKaZrv ut/13 WIPQdC ANpq1l 7V+IsQA GQRKJ BIYQjH 4CjoOUI AHzlx 6iI04S Ui0vUmC NtSVmS MjixJ/Q d0e6oY VPaW8/ CqQWdEt OrScnw jQZ5W nC9jTm4 qVztux j3pXzt Hc7o3/U eCXEGo yJfUp3 l/lcna3 FYIhNV 4OimnLH 2OpE41 K6rtib s3+qMuS QE2fxg OKasHD Kuz4zpy lc7ba3 3MV/uU zL2r1oc kv8tre kAQf3x /kQ7K93 g43u+y 8b7e2dZ tQzeIO 36NA8P 2Abn7CH Hnlf4A o3uPXO vW/ed+/ H31Rvo tGs4L/ lXf4BGu irDg= </late xit> ! <latexit sha1_base64= "MPaxa9zmzYUAzcGEz2n69FsgxC8=">ACznicjV HLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUF+4C2SJ O26F5MZlUSilu/QG3+lniH+hfeGdMQS2iE5KcOfe cO3PvdWOfJ9KyXnPG0vLK6lp+vbCxubW9U9zdayR KjxW9yI/Ei3XSZjPQ1aXPqsFQvmBK7Pmu7oUsWbY yYSHoU3chKzbuAMQt7niOJancEHwylI0R0d1sW VL3MR2BkoIVu1qPiCDnqI4CFAIYQkrAPBwk9bdi wEBPXxZQ4QYjrOMBfKmpGKkcIgd0XdAu3bGhrRX ORPt9ugUn15BThNH5IlIJwir0wdT3Vmxf6We6pzq rtN6O9muQJiJYbE/uWbK/rU7VI9HGua+BU6wZV Z2XZUl1V9TNzS9VScoQE6dwj+KCsKed8z6b2pPo2l VvHR1/0rFqr2XaVO8q1vSgO2f41wEjZOyXSmfXld K1Yts1Hkc4BDHNM8zVHGFGuq6494wrNRM8bGzLj/ lBq5zLOPb8t4+ADp5QW</latexit> Vamos determinar uma condição para a resolução de dois comprimentos de onda. O critério de Rayleigh estabelece que se a posição do m-ésimo máximo principal para um determinado comprimento de onda for maior ou igual que a posição que a intensidade se anula para o outro comprimento de onda, ambos podem ser resolvidos. Rede de Difração A condição com N fendas para a igualdade é obtida com as formulas: dsen(✓) = mλmaior <lat exit s ha1_ba se64="H 9GjVK/ /lzWpx Um6ZW07 rTNGa6 0=">A C8Hicj VHLSgM xFD2Or/ qunQT LIilK lUdCMU3 bisYLX QSslMo wbnRSYj lNKPcO dO3PoD bvUrxD/ Qv/Amj uAD0Qwz c3LuPS e593pJ IFPtus9 DzvDI6 Nh4YWJ yanpmdq 4v3CU xpnyRc OPg1g1P Z6KQEa ioaUOR DNRgode I69iz 0TP74UK pVxdKh 7iTgJ+ VkT6XP NVGd4n qX9dsq ZKmIBqt tfS40X 2M7LGT tgEy6vG OjIZex GnSKJb fs2sV+g koOSsh XPS4+oY 0uYvjI EIgi YcgCOlp 4UKXCT EnaBPn CIkbVxg gEnSZp QlKIMT e0HfM9q 1cjaiv fFMrdq nUwJ6FS kZVkgT U54ibE5 jNp5Z 8P+5t2 3nuZuPf p7uVdI rMY5sX /pPjL/q zO1aJx i29Ygq abEMqY6 P3fJbF fMzdmn qjQ5JMQ Z3KW4I uxb5Uef mdWktn bTW27j LzbTsGb v57kZX s0tacC V7+P8CY 42ypVq efOgWq rt5qMuY AnLWKV 5bqGf dTRIO8r 3OMBj4 5yrp0b5 /Y91Rn KNYv4s py7Nwi0 oFM=</ latexi t> dsen(✓) = ✓ m + 1 N ◆ λmenor <lat exit s ha1_ba se64="m J3fp0 Foxg0 hdAg1LH QtGQ4S 8=">A DC3icj VHLShx BFD124j M+xrjM psgQGB GbjHEj SAGJKt gIKOCL UN1Tc1M Yb+orh ak6U/w T9xlF7L 1B9wE0 Q9I/iK3 Ki0kim g13X3q 3HtO1b0 3ymNVG N+/nvB evJycmp 6ZnXs1 v7C41F p+vV9kp RayJ7I 404cRL 2SsUtkz ysTyMN eSJ1EsD 6KTjzZ +cCp1o bL0qznL 5XHCR6 kaKsEN Uf3W7oB VoU5YI dO6E5q xNHyVb EwlkPT SdgaC4 eaiyqoq 891qNV obFYpSP 4D3nfC RKaZrv ut/13 WIPQdC ANpq1l 7V+IsQA GQRKJ BIYQjH 4CjoOUI AHzlx 6iI04S Ui0vUmC NtSVmS MjixJ/Q d0e6oY VPaW8/ CqQWdEt OrScnw jQZ5W nC9jTm4 qVztux j3pXzt Hc7o3/U eCXEGo yJfUp3 l/lcna3 FYIhNV 4OimnLH 2OpE41 K6rtib s3+qMuS QE2fxg OKasHD Kuz4zpy lc7ba3 3MV/uU zL2r1oc kv8tre kAQf3x /kQ7K93 g43u+y 8b7e2dZ tQzeIO 36NA8P 2Abn7CH Hnlf4A o3uPXO vW/ed+/ H31Rvo tGs4L/ lXf4BGu irDg= </late xit> ! <latexit sha1_base64= "MPaxa9zmzYUAzcGEz2n69FsgxC8=">ACznicjV HLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUF+4C2SJ O26F5MZlUSilu/QG3+lniH+hfeGdMQS2iE5KcOfe cO3PvdWOfJ9KyXnPG0vLK6lp+vbCxubW9U9zdayR KjxW9yI/Ei3XSZjPQ1aXPqsFQvmBK7Pmu7oUsWbY yYSHoU3chKzbuAMQt7niOJancEHwylI0R0d1sW VL3MR2BkoIVu1qPiCDnqI4CFAIYQkrAPBwk9bdi wEBPXxZQ4QYjrOMBfKmpGKkcIgd0XdAu3bGhrRX ORPt9ugUn15BThNH5IlIJwir0wdT3Vmxf6We6pzq rtN6O9muQJiJYbE/uWbK/rU7VI9HGua+BU6wZV Z2XZUl1V9TNzS9VScoQE6dwj+KCsKed8z6b2pPo2l VvHR1/0rFqr2XaVO8q1vSgO2f41wEjZOyXSmfXld K1Yts1Hkc4BDHNM8zVHGFGuq6494wrNRM8bGzLj/ lBq5zLOPb8t4+ADp5QW</latexit> m (λmaior − λmenor) = λmenor N <latexit sha1_base64="l7Pr0W9VfEWpjF4wGRC3Y6iH7/c=" >ADG3icjVFPSxtBH3ZqrW2amyPvSyGgj0YNmJpL4VQLz2JheYPuBJmJ5NkyOwfZmcLYclH8Zv01lvxKuLVk6X9EP3NuIKplXaW3 X3zfu+9md9MlCmZmyC4rHmPlpZXHq8+WXv6bH1js71vJunheaiw1OV6n7EcqFkIjpGiX6mRYsjpToRdMDW+9ETqXafLZzDJxErN xIkeSM0PUoN6NQyVGZidU5BmyQRnq2I+ZTPV8d5ETCXGhluOJe2/98PhSDNe/kUzLw/ng3ojaAZu+PdBqwINVOMorV8gxBApOArE EhgCsw5PQco4UAGXEnKInThKSrC8yxRt6CVIUjNgpfc0O67YhOY2M3duTqsoejU5fbwiT0o6Tdiu5rt64ZIt+1B26TLt3mb0j6 qsmFiDCbH/8t0q/9dnezEY4Z3rQVJPmWNsd7xKdyp2J37d7oylJARZ/GQ6powd87bc/adJ3e927Nlrn7tlJa1c15pC/ywu6QLbv15 nfdBd6/Z2m+bTfaH+ornoVL7GNHbrPt2jI47QoeyvuMJP/PJOvW/ed+/sRurVKs8LAzv/DfOL7O6</latexit> Vamos determinar uma condição para a resolução de dois comprimentos de onda. O critério de Rayleigh estabelece que se a posição do m-ésimo máximo principal para um determinado comprimento de onda for maior ou igual que a posição que a intensidade se anula para o outro comprimento de onda, ambos podem ser resolvidos. Rede de Difração A condição com N fendas para a igualdade é obtida com as formulas: dsen(✓) = mλmaior <lat exit s ha1_ba se64="H 9GjVK/ /lzWpx Um6ZW07 rTNGa6 0=">A C8Hicj VHLSgM xFD2Or/ qunQT LIilK lUdCMU3 bisYLX QSslMo wbnRSYj lNKPcO dO3PoD bvUrxD/ Qv/Amj uAD0Qwz c3LuPS e593pJ IFPtus9 DzvDI6 Nh4YWJ yanpmdq 4v3CU xpnyRc OPg1g1P Z6KQEa ioaUOR DNRgode I69iz 0TP74UK pVxdKh 7iTgJ+ VkT6XP NVGd4n qX9dsq ZKmIBqt tfS40X 2M7LGT tgEy6vG OjIZex GnSKJb fs2sV+g koOSsh XPS4+oY 0uYvjI EIgi YcgCOlp 4UKXCT EnaBPn CIkbVxg gEnSZp QlKIMT e0HfM9q 1cjaiv fFMrdq nUwJ6FS kZVkgT U54ibE5 jNp5Z 8P+5t2 3nuZuPf p7uVdI rMY5sX /pPjL/q zO1aJx i29Ygq abEMqY6 P3fJbF fMzdmn qjQ5JMQ Z3KW4I uxb5Uef mdWktn bTW27j LzbTsGb v57kZX s0tacC V7+P8CY 42ypVq efOgWq rt5qMuY AnLWKV 5bqGf dTRIO8r 3OMBj4 5yrp0b5 /Y91Rn KNYv4s py7Nwi0 oFM=</ latexi t> dsen(✓) = ✓ m + 1 N ◆ λmenor <lat exit s ha1_ba se64="m J3fp0 Foxg0 hdAg1LH QtGQ4S 8=">A DC3icj VHLShx BFD124j M+xrjM psgQGB GbjHEj SAGJKt gIKOCL UN1Tc1M Yb+orh ak6U/w T9xlF7L 1B9wE0 Q9I/iK3 Ki0kim g13X3q 3HtO1b0 3ymNVG N+/nvB evJycmp 6ZnXs1 v7C41F p+vV9kp RayJ7I 404cRL 2SsUtkz ysTyMN eSJ1EsD 6KTjzZ +cCp1o bL0qznL 5XHCR6 kaKsEN Uf3W7oB VoU5YI dO6E5q xNHyVb EwlkPT SdgaC4 eaiyqoq 891qNV obFYpSP 4D3nfC RKaZrv ut/13 WIPQdC ANpq1l 7V+IsQA GQRKJ BIYQjH 4CjoOUI AHzlx 6iI04S Ui0vUmC NtSVmS MjixJ/Q d0e6oY VPaW8/ CqQWdEt OrScnw jQZ5W nC9jTm4 qVztux j3pXzt Hc7o3/U eCXEGo yJfUp3 l/lcna3 FYIhNV 4OimnLH 2OpE41 K6rtib s3+qMuS QE2fxg OKasHD Kuz4zpy lc7ba3 3MV/uU zL2r1oc kv8tre kAQf3x /kQ7K93 g43u+y 8b7e2dZ tQzeIO 36NA8P 2Abn7CH Hnlf4A o3uPXO vW/ed+/ H31Rvo tGs4L/ lXf4BGu irDg= </late xit> ! <latexit sha1_base64= "MPaxa9zmzYUAzcGEz2n69FsgxC8=">ACznicjV HLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUF+4C2SJ O26F5MZlUSilu/QG3+lniH+hfeGdMQS2iE5KcOfe cO3PvdWOfJ9KyXnPG0vLK6lp+vbCxubW9U9zdayR KjxW9yI/Ei3XSZjPQ1aXPqsFQvmBK7Pmu7oUsWbY yYSHoU3chKzbuAMQt7niOJancEHwylI0R0d1sW VL3MR2BkoIVu1qPiCDnqI4CFAIYQkrAPBwk9bdi wEBPXxZQ4QYjrOMBfKmpGKkcIgd0XdAu3bGhrRX ORPt9ugUn15BThNH5IlIJwir0wdT3Vmxf6We6pzq rtN6O9muQJiJYbE/uWbK/rU7VI9HGua+BU6wZV Z2XZUl1V9TNzS9VScoQE6dwj+KCsKed8z6b2pPo2l VvHR1/0rFqr2XaVO8q1vSgO2f41wEjZOyXSmfXld K1Yts1Hkc4BDHNM8zVHGFGuq6494wrNRM8bGzLj/ lBq5zLOPb8t4+ADp5QW</latexit> m (λmaior − λmenor) = λmenor N <latexit sha1_base64="l7Pr0W9VfEWpjF4wGRC3Y6iH7/c=" >ADG3icjVFPSxtBH3ZqrW2amyPvSyGgj0YNmJpL4VQLz2JheYPuBJmJ5NkyOwfZmcLYclH8Zv01lvxKuLVk6X9EP3NuIKplXaW3 X3zfu+9md9MlCmZmyC4rHmPlpZXHq8+WXv6bH1js71vJunheaiw1OV6n7EcqFkIjpGiX6mRYsjpToRdMDW+9ETqXafLZzDJxErN xIkeSM0PUoN6NQyVGZidU5BmyQRnq2I+ZTPV8d5ETCXGhluOJe2/98PhSDNe/kUzLw/ng3ojaAZu+PdBqwINVOMorV8gxBApOArE EhgCsw5PQco4UAGXEnKInThKSrC8yxRt6CVIUjNgpfc0O67YhOY2M3duTqsoejU5fbwiT0o6Tdiu5rt64ZIt+1B26TLt3mb0j6 qsmFiDCbH/8t0q/9dnezEY4Z3rQVJPmWNsd7xKdyp2J37d7oylJARZ/GQ6powd87bc/adJ3e927Nlrn7tlJa1c15pC/ywu6QLbv15 nfdBd6/Z2m+bTfaH+ornoVL7GNHbrPt2jI47QoeyvuMJP/PJOvW/ed+/sRurVKs8LAzv/DfOL7O6</latexit> mN = λmenor ∆λ <latexit sha1_base64="0S6Mo1HX1202jqy4KvoDSFY42lQ=" >AC9XicjVHLShxBFD12TNTJw06ydFM4CFkNPWHEbAISXbgKE8ioMD0M1TU1prH6QXW1QZr5jezcSb5AbfJL4T8QfwLb1VKUIeg1 XT3qXPvOVX3qRUaWi6M9C8Gjx8ZOl5ZXW02fPX6yGL1/tV0WthRyIQhX6MOGVGkuByY1Sh6WvIsUfIgOd6x8YMTqau0yD+b01K OMn6Up9NUcEPUOIyj+w9iydTzUTKxJO+LiJdcYymRd6NmviXakMZz42G4ftqBO5xeZB14M2/OoX4W/EmKCAQI0MEjkMYQWOip4h uohQEjdCQ5wmlLq4xAwt0taUJSmDE3tM3yPaDT2b0956Vk4t6BRFryYlwZpCsrThO1pzMVr52zZ/3k3ztPe7ZT+ifKiDX4Qux9u vMh+psLQZTvHM1pFRT6RhbnfAuteuKvTm7UZUh5I4iycU14SFU173mTlN5Wq3veUu/tdlWtbuhc+tcWlvSQPu3h3nPNh/2+n2Opuf eu3tD37Uy1jDOt7QPLewjT30MSDvb7jAT/wKvgZnwXnw/V9qsOA1r3FrBT+uAMFro0s=</latexit> Vamos determinar uma condição para a resolução de dois comprimentos de onda. O critério de Rayleigh estabelece que se a posição do m-ésimo máximo principal para um determinado comprimento de onda for maior ou igual que a posição que a intensidade se anula para o outro comprimento de onda, ambos podem ser resolvidos. Rede de Difração A condição com N fendas para a igualdade é obtida com as formulas: dsen(✓) = mλmaior <lat exit s ha1_ba se64="H 9GjVK/ /lzWpx Um6ZW07 rTNGa6 0=">A C8Hicj VHLSgM xFD2Or/ qunQT LIilK lUdCMU3 bisYLX QSslMo wbnRSYj lNKPcO dO3PoD bvUrxD/ Qv/Amj uAD0Qwz c3LuPS e593pJ IFPtus9 DzvDI6 Nh4YWJ yanpmdq 4v3CU xpnyRc OPg1g1P Z6KQEa ioaUOR DNRgode I69iz 0TP74UK pVxdKh 7iTgJ+ VkT6XP NVGd4n qX9dsq ZKmIBqt tfS40X 2M7LGT tgEy6vG OjIZex GnSKJb fs2sV+g koOSsh XPS4+oY 0uYvjI EIgi YcgCOlp 4UKXCT EnaBPn CIkbVxg gEnSZp QlKIMT e0HfM9q 1cjaiv fFMrdq nUwJ6FS kZVkgT U54ibE5 jNp5Z 8P+5t2 3nuZuPf p7uVdI rMY5sX /pPjL/q zO1aJx i29Ygq abEMqY6 P3fJbF fMzdmn qjQ5JMQ Z3KW4I uxb5Uef mdWktn bTW27j LzbTsGb v57kZX s0tacC V7+P8CY 42ypVq efOgWq rt5qMuY AnLWKV 5bqGf dTRIO8r 3OMBj4 5yrp0b5 /Y91Rn KNYv4s py7Nwi0 oFM=</ latexi t> dsen(✓) = ✓ m + 1 N ◆ λmenor <lat exit s ha1_ba se64="m J3fp0 Foxg0 hdAg1LH QtGQ4S 8=">A DC3icj VHLShx BFD124j M+xrjM psgQGB GbjHEj SAGJKt gIKOCL UN1Tc1M Yb+orh ak6U/w T9xlF7L 1B9wE0 Q9I/iK3 Ki0kim g13X3q 3HtO1b0 3ymNVG N+/nvB evJycmp 6ZnXs1 v7C41F p+vV9kp RayJ7I 404cRL 2SsUtkz ysTyMN eSJ1EsD 6KTjzZ +cCp1o bL0qznL 5XHCR6 kaKsEN Uf3W7oB VoU5YI dO6E5q xNHyVb EwlkPT SdgaC4 eaiyqoq 891qNV obFYpSP 4D3nfC RKaZrv ut/13 WIPQdC ANpq1l 7V+IsQA GQRKJ BIYQjH 4CjoOUI AHzlx 6iI04S Ui0vUmC NtSVmS MjixJ/Q d0e6oY VPaW8/ CqQWdEt OrScnw jQZ5W nC9jTm4 qVztux j3pXzt Hc7o3/U eCXEGo yJfUp3 l/lcna3 FYIhNV 4OimnLH 2OpE41 K6rtib s3+qMuS QE2fxg OKasHD Kuz4zpy lc7ba3 3MV/uU zL2r1oc kv8tre kAQf3x /kQ7K93 g43u+y 8b7e2dZ tQzeIO 36NA8P 2Abn7CH Hnlf4A o3uPXO vW/ed+/ H31Rvo tGs4L/ lXf4BGu irDg= </late xit> ! <latexit sha1_base64= "MPaxa9zmzYUAzcGEz2n69FsgxC8=">ACznicjV HLSsNAFD2Nr1pfVZdugkVwVRKp6LoxmUF+4C2SJ O26F5MZlUSilu/QG3+lniH+hfeGdMQS2iE5KcOfe cO3PvdWOfJ9KyXnPG0vLK6lp+vbCxubW9U9zdayR KjxW9yI/Ei3XSZjPQ1aXPqsFQvmBK7Pmu7oUsWbY yYSHoU3chKzbuAMQt7niOJancEHwylI0R0d1sW VL3MR2BkoIVu1qPiCDnqI4CFAIYQkrAPBwk9bdi wEBPXxZQ4QYjrOMBfKmpGKkcIgd0XdAu3bGhrRX ORPt9ugUn15BThNH5IlIJwir0wdT3Vmxf6We6pzq rtN6O9muQJiJYbE/uWbK/rU7VI9HGua+BU6wZV Z2XZUl1V9TNzS9VScoQE6dwj+KCsKed8z6b2pPo2l VvHR1/0rFqr2XaVO8q1vSgO2f41wEjZOyXSmfXld K1Yts1Hkc4BDHNM8zVHGFGuq6494wrNRM8bGzLj/ lBq5zLOPb8t4+ADp5QW</latexit> m (λmaior − λmenor) = λmenor N <latexit sha1_base64="l7Pr0W9VfEWpjF4wGRC3Y6iH7/c=" >ADG3icjVFPSxtBH3ZqrW2amyPvSyGgj0YNmJpL4VQLz2JheYPuBJmJ5NkyOwfZmcLYclH8Zv01lvxKuLVk6X9EP3NuIKplXaW3 X3zfu+9md9MlCmZmyC4rHmPlpZXHq8+WXv6bH1js71vJunheaiw1OV6n7EcqFkIjpGiX6mRYsjpToRdMDW+9ETqXafLZzDJxErN xIkeSM0PUoN6NQyVGZidU5BmyQRnq2I+ZTPV8d5ETCXGhluOJe2/98PhSDNe/kUzLw/ng3ojaAZu+PdBqwINVOMorV8gxBApOArE EhgCsw5PQco4UAGXEnKInThKSrC8yxRt6CVIUjNgpfc0O67YhOY2M3duTqsoejU5fbwiT0o6Tdiu5rt64ZIt+1B26TLt3mb0j6 qsmFiDCbH/8t0q/9dnezEY4Z3rQVJPmWNsd7xKdyp2J37d7oylJARZ/GQ6powd87bc/adJ3e927Nlrn7tlJa1c15pC/ywu6QLbv15 nfdBd6/Z2m+bTfaH+ornoVL7GNHbrPt2jI47QoeyvuMJP/PJOvW/ed+/sRurVKs8LAzv/DfOL7O6</latexit> mN = λmenor ∆λ <latexit sha1_base64="0S6Mo1HX1202jqy4KvoDSFY42lQ=" >AC9XicjVHLShxBFD12TNTJw06ydFM4CFkNPWHEbAISXbgKE8ioMD0M1TU1prH6QXW1QZr5jezcSb5AbfJL4T8QfwLb1VKUIeg1 XT3qXPvOVX3qRUaWi6M9C8Gjx8ZOl5ZXW02fPX6yGL1/tV0WthRyIQhX6MOGVGkuByY1Sh6WvIsUfIgOd6x8YMTqau0yD+b01K OMn6Up9NUcEPUOIyj+w9iydTzUTKxJO+LiJdcYymRd6NmviXakMZz42G4ftqBO5xeZB14M2/OoX4W/EmKCAQI0MEjkMYQWOip4h uohQEjdCQ5wmlLq4xAwt0taUJSmDE3tM3yPaDT2b0956Vk4t6BRFryYlwZpCsrThO1pzMVr52zZ/3k3ztPe7ZT+ifKiDX4Qux9u vMh+psLQZTvHM1pFRT6RhbnfAuteuKvTm7UZUh5I4iycU14SFU173mTlN5Wq3veUu/tdlWtbuhc+tcWlvSQPu3h3nPNh/2+n2Opuf eu3tD37Uy1jDOt7QPLewjT30MSDvb7jAT/wKvgZnwXnw/V9qsOA1r3FrBT+uAMFro0s=</latexit> Dessa forma, concluímos que o poder de resolução de uma rede de difração com N fendas é R = mN, ao se monitorar o m-ésimo máximo. A posição de um máximo central na figura de intensidade de uma rede de difração depende do comprimento de onda: dsen(✓m) = mλ Rede de Difração A posição de um máximo central na figura de intensidade de uma rede de difração depende do comprimento de onda: dsen(✓m) = mλ se o comprimento de onda muda para λ + dλ o ângulo do m-ésimo máximo central muda para ✓m + d✓m Rede de Difração A posição de um máximo central na figura de intensidade de uma rede de difração depende do comprimento de onda: dsen(✓m) = mλ se o comprimento de onda muda para λ + dλ o ângulo do m-ésimo máximo central muda para ✓m + d✓m Tal que: d cos (✓m)d✓m = mdλ Rede de Difração A posição de um máximo central na figura de intensidade de uma rede de difração depende do comprimento de onda: dsen(✓m) = mλ se o comprimento de onda muda para λ + dλ o ângulo do m-ésimo máximo central muda para ✓m + d✓m Tal que: Um critério para podermos resolver (diferenciar) esses dois comprimentos de onda é: d cos (✓m)d✓m = mdλ ✓max m (λ + dλ) ≥ ✓min m (λ) Rede de Difração A posição de um máximo central na figura de intensidade de uma rede de difração depende do comprimento de onda: dsen(✓m) = mλ se o comprimento de onda muda para λ + dλ o ângulo do m-ésimo máximo central muda para ✓m + d✓m Tal que: Um critério para podermos resolver (diferenciar) esses dois comprimentos de onda é: Ou seja: o máximo de um comprimento de onda se encontra no mínimo do outro comprimento de onda. d cos (✓m)d✓m = mdλ ✓max m (λ + dλ) ≥ ✓min m (λ) Rede de Difração Note que ✓max m (λ + dλ) = ✓max m (λ) + d✓m Rede de Difração Note que ✓max m (λ + dλ) = ✓max m (λ) + d✓m = ✓max m (λ) + m d cos ✓m dλ Rede de Difração Note que ✓max m (λ + dλ) = ✓max m (λ) + d✓m = ✓max m (λ) + m d cos ✓m dλ O ângulo para a interferência destrutiva é encontrado com 2⇡d λ sen(✓min m ) = 2m⇡ + 2⇡ N Rede de Difração Note que ✓max m (λ + dλ) = ✓max m (λ) + d✓m = ✓max m (λ) + m d cos ✓m dλ O ângulo para a interferência destrutiva é encontrado com Definindo uma variação angular como: ✓min m = ✓max m + ∆✓m 2⇡d λ sen(✓min m ) = 2m⇡ + 2⇡ N Rede de Difração Note que ✓max m (λ + dλ) = ✓max m (λ) + d✓m = ✓max m (λ) + m d cos ✓m dλ O ângulo para a interferência destrutiva é encontrado com Definindo uma variação angular como: ✓min m = ✓max m + ∆✓m Encontramos 2⇡d λ sen(✓min m ) = 2m⇡ + 2⇡ N 2⇡d λ [sen(✓max m ) + cos (✓max m )∆✓m] = 2m⇡ + 2⇡ N Rede de Difração Note que ✓max m (λ + dλ) = ✓max m (λ) + d✓m = ✓max m (λ) + m d cos ✓m dλ O ângulo para a interferência destrutiva é encontrado com Definindo uma variação angular como: ✓min m = ✓max m + ∆✓m Encontramos 2⇡d λ sen(✓min m ) = 2m⇡ + 2⇡ N 2⇡d λ [sen(✓max m ) + cos (✓max m )∆✓m] = 2m⇡ + 2⇡ N ∆✓m = λ Nd cos (✓m) Rede de Difração Portanto, dado um comprimento de onda de referecia e uma rede de difração, devemos ter m d cos (✓m)dλ ≥ λ Nd cos (✓m) Rede de Difração Portanto, dado um comprimento de onda de referecia e uma rede de difração, devemos ter m d cos (✓m)dλ ≥ λ Nd cos (✓m) Simplificando os termos, a desigualdade fica reescrita como: dλ λ ≥ 1 mN Rede de Difração Portanto, dado um comprimento de onda de referecia e uma rede de difração, devemos ter m d cos (✓m)dλ ≥ λ Nd cos (✓m) Simplificando os termos, a desigualdade fica reescrita como: dλ λ ≥ 1 mN O lado direito é justamente o inverso do poder de resolução da dada rede de difração, já que esse valor determina a variação mínima para diferenciar dois comprimentos de onda. R = mN Rede de Difração Exemplo. Qual o número mínimo de fendas seria necessário em uma rede para resolver o dupleto de sódio na quarta ordem? Rede de Difração Exemplo. Qual o número mínimo de fendas seria necessário em uma rede para resolver o dupleto de sódio na quarta ordem? 998 = R = mN = 4N Rede de Difração Exemplo. Qual o número mínimo de fendas seria necessário em uma rede para resolver o dupleto de sódio na quarta ordem? 998 = R = mN = 4N Nmin = 998 4 ' 250 Rede de Difração