{"id":2240,"date":"2016-05-23T23:52:44","date_gmt":"2016-05-23T23:52:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2240"},"modified":"2024-09-02T14:21:15","modified_gmt":"2024-09-02T14:21:15","slug":"leis-da-refracao-exercicios","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/leis-da-refracao\/leis-da-refracao-exercicios\/","title":{"rendered":"Leis da Refra\u00e7\u00e3o &#8211; Exerc\u00edcios"},"content":{"rendered":"<p align=\"CENTER\"><span style=\"color: #c00000;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: large;\">Exerc\u00edcios de vestibulares com resolu\u00e7\u00e3o comentada sobre <\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: large;\">Leis da Refra\u00e7\u00e3o<\/span><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>01-(UFSC-SC<\/b><\/span><\/span><\/span><span style=\"color: #0000cc;\"><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>) <\/b><\/span><\/span><\/span><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A figura a seguir mostra um l\u00e1pis de comprimento AB, parcialmente imerso na \u00e1gua e sendo observado por um estudante.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img fetchpriority=\"high\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_3acda298.jpg\" alt=\"\" width=\"338\" height=\"160\" name=\"Imagem 21\" align=\"BOTTOM\" border=\"0\" \/><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_d011c4e5.jpg\" alt=\"\" width=\"129\" height=\"158\" name=\"Imagem 22\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Assinale a(s) proposi\u00e7\u00e3o (\u00f5es) CORRETA(S) e indique sua soma:.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(01) O estudante v\u00ea o l\u00e1pis &#8220;quebrado&#8221; na interface ar-\u00e1gua, porque o \u00edndice de refra\u00e7\u00e3o da \u00e1gua \u00e9 maior do que o do ar.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(02) O feixe luminoso proveniente do ponto B, ao passar da \u00e1gua para o ar se afasta da normal, sofrendo desvio.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(04) O estudante v\u00ea o l\u00e1pis &#8220;quebrado&#8221; na interface ar-\u00e1gua, sendo o fen\u00f4meno explicado pelas leis da reflex\u00e3o.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(08) O observador v\u00ea o l\u00e1pis &#8220;quebrado&#8221; na interface ar-\u00e1gua porque a luz sofre dispers\u00e3o ao passar do ar para a \u00e1gua.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(16) O ponto B&#8217;, visto pelo observador, \u00e9 uma imagem virtual<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>02-(UFSC-SC)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>&#8220;A apar\u00eancia do arco-\u00edris \u00e9 causada pela dispers\u00e3o da luz do Sol, a qual sofre refra\u00e7\u00e3o pelas gotas de chuva. A luz sofre uma refra\u00e7\u00e3o inicial quando penetra na superf\u00edcie da gota de chuva; dentro da gota ela \u00e9 refletida e sofre nova refra\u00e7\u00e3o ao sair da gota. (Dispon\u00edvel em:\u00a0 &lt;http:\/\/pt.wikipedia.org\/wiki\/Arco-%C3%Adris&gt;\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Acesso em:\u00a0 25 jul. 2006.)<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Com o intuito de explicar o fen\u00f4meno, um aluno desenhou as possibilidades de caminhos \u00f3ticos de um feixe de luz monocrom\u00e1tica em uma gota d&#8217;\u00e1gua, de forma esf\u00e9rica e de centro geom\u00e9trico O, representados nas figuras A, B, C, D e E.<\/b><\/span><\/span><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_b3f03303.jpg\" alt=\"\" width=\"780\" height=\"199\" name=\"Imagem 23\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Admitindo-se que o \u00edndice de refra\u00e7\u00e3o do ar (nar) seja menor que o \u00edndice de refra\u00e7\u00e3o da \u00e1gua (n\u00e1gua), assinale a(s) proposi\u00e7\u00e3o(\u00f5es) CORRETA(S) e indique sua soma.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(01) A velocidade da luz no ar \u00e9 maior do que na \u00e1gua.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(02) A e D s\u00e3o caminhos \u00f3ticos aceit\u00e1veis.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(04) B e C s\u00e3o caminhos \u00f3ticos aceit\u00e1veis.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(08) D e E s\u00e3o caminhos \u00f3ticos aceit\u00e1veis.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(16) A e C s\u00e3o caminhos \u00f3ticos aceit\u00e1veis.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(32) B e E s\u00e3o caminhos \u00f3ticos aceit\u00e1veis.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>03-(UNIFOR)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Para responder \u00e0 quest\u00e3o que segue, utilize o esquema e as informa\u00e7\u00f5es abaixo.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_bc04d723.jpg\" alt=\"\" width=\"378\" height=\"256\" name=\"graphics6\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>S &#8211; representa a superf\u00edcie de separa\u00e7\u00e3o entre os meios transparentes e homog\u00eaneos I e II. r<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>, r<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e r<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&#8211; representam raios luminosos<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Assinale a alternativa que preenche corretamente as lacunas da seguinte frase:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Se r<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>, r<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e r<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0forem, respectivamente, raios ____________, ____________ e ____________, o meio\u00a0I \u00e9 mais\u00a0 _______________ que o meio II.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) incidente &#8211; refletido &#8211; refratado \u2013 refletor\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) refratado &#8211; incidente &#8211; refletido \u2013 refringente<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c) incidente &#8211; refletido &#8211; refratado \u2013 refringente\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d) refletido &#8211; refratado &#8211; incidente &#8211; refringente<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) refletido &#8211; refratado &#8211; incidente \u2013 refletor<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>04-(UFRJ-RJ)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A figura mostra uma estrela localizada no ponto O, emitindo um raio de luz que se propaga at\u00e9 a Terra.\u00a0Ao atingir a atmosfera, o raio desvia-se da trajet\u00f3ria retil\u00ednea original, fazendo com que um\u00a0observador na Terra veja a imagem da estrela na posi\u00e7\u00e3o I. O desvio do raio de luz deve-se ao fato de o \u00edndice de refra\u00e7\u00e3o absoluto\u00a0da atmosfera variar com a altitude, do modo indicado na figura, respondendo se o \u00edndice de refra\u00e7\u00e3o absoluto\u00a0cresce ou diminui \u00e0 medida que a altitude aumenta. (Na figura a espessura da atmosfera e o\u00a0desvio do raio\u00a0foram grandemente exagerados para mostrar com clareza o fen\u00f4meno.)<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_81fb67f5.jpg\" alt=\"\" width=\"297\" height=\"180\" name=\"Imagem 25\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>05-(Cefet)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Durante o dia, quando um raio luminoso solar, atravessando a camada atmosf\u00e9rica, atinge a polu\u00edda cidade de S\u00e3o Paulo, sua trajet\u00f3ria prov\u00e1vel, devido ao fen\u00f4meno da refra\u00e7\u00e3o, \u00e9 descrita em uma das figuras a seguir. Assinale a alternativa que representa essa prov\u00e1vel trajet\u00f3ria.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_e5f61d21.png\" alt=\"\" width=\"765\" height=\"223\" name=\"Imagem 75\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>06-(UEPB)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Ao viajar num dia quente por uma estrada asfaltada, \u00e9 comum enxergarmos ao longe uma &#8220;po\u00e7a d\u2019 \u00e1gua&#8221;.\u00a0Sabemos que em dias de alta temperatura as camadas de ar, nas proximidades do solo, s\u00e3o\u00a0mais quentes que\u00a0as camadas superiores.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_2ecee88d.jpg\" alt=\"\" width=\"448\" height=\"160\" name=\"Imagem 27\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Como explicamos essa miragem?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) Devido ao aumento de temperatura a luz sofre dispers\u00e3o.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) A densidade e o \u00edndice de refra\u00e7\u00e3o absoluto diminuem com o aumento da temperatura. Os raios\u00a0rasantes\u00a0incidentes do Sol alcan\u00e7am o \u00e2ngulo limite e h\u00e1 reflex\u00e3o total.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c) Devido ao aumento de temperatura, ocorre refra\u00e7\u00e3o com desvio.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d) Ocorre reflex\u00e3o simples devido ao aumento da temperatura.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>07- (UFPR)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um raio de luz atravessa tr\u00eas meios \u00f3pticos de \u00edndices de refra\u00e7\u00e3o absolutos n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>, n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>,\u00a0 conforme a figura:<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_93b17403.jpg\" alt=\"\" width=\"395\" height=\"165\" name=\"Imagem 28\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Sendo paralelas as superf\u00edcies de separa\u00e7\u00e3o do meio 2 com os outros dois meios, \u00e9 correto afirmar que:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c) n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d) n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0&gt; n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>08-(Ufpe)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>O \u00edndice de refra\u00e7\u00e3o, n, de um vidro de quartzo depende do comprimento de onda da luz, \u03bb, conforme indica o gr\u00e1fico a seguir. Calcule o \u00e2ngulo de refra\u00e7\u00e3o \u04e8 para luz com \u03bb = 400 nm incidindo sobre uma pe\u00e7a de quartzo, conforme a figura. Considere o \u00edndice de refra\u00e7\u00e3o do ar igual a 1,00.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_8558f8f4.jpg\" alt=\"\" width=\"515\" height=\"194\" name=\"Imagem 29\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>a) arc sen 0,07\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>b) arc sen 0,13\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>c) arc sen 0,34\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>d) arc sen 0,59\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) arc sen 0,73<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>09-(FATEC-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um estreito feixe de luz monocrom\u00e1tica, proveniente do ar, incide na superf\u00edcie de um vidro formando \u00e2ngulo de 49\u00b0 com a normal \u00e0 superf\u00edcie no ponto de incid\u00eancia.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>DADOS<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>n(ar) = 1,00<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>n(vidro) = 1,50<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>sen 49\u00b0 = 0,75<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>cos 49\u00b0 = 0,66<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Nessas condi\u00e7\u00f5es, o feixe luminoso refratado forma com a dire\u00e7\u00e3o do feixe incidente \u00e2ngulo de<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_99e94842.png\" alt=\"\" width=\"766\" height=\"21\" name=\"Imagem 76\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>10-(PUCCAMP-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Uma onda eletromagn\u00e9tica vis\u00edvel possui, no ar ou no v\u00e1cuo , velocidade de 3,00.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0m\/s e no vidro de 1,73.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0m\/s. Essa onda, propagando no ar, incide sobre uma superf\u00edcie plana de vidro com \u00e2ngulo de incid\u00eancia de 60\u00b0.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_a7706221.jpg\" alt=\"\" width=\"594\" height=\"181\" name=\"Imagem 30\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>O \u00e2ngulo de refra\u00e7\u00e3o da onda, no vidro, vale:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>Dados:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>sen 30\u00b0 = cos 60\u00b0 = 0,50; sen 60\u00b0 = cos 30\u00b0 = 0,87<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_358d0547.png\" alt=\"\" width=\"765\" height=\"21\" name=\"Imagem 78\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>11-(UFRJ-RJ) <\/b><\/span><\/span><\/span><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um raio luminoso que se propaga no ar &#8220;n(ar) =1&#8221; incide obliquamente sobre um meio transparente de \u00edndice de refra\u00e7\u00e3o n, fazendo um \u00e2ngulo de 60\u00b0 com a normal. Nessa situa\u00e7\u00e3o,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Verifica-se que o raio refletido \u00e9 perpendicular ao raio refratado, como ilustra a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_8b663c84.jpg\" alt=\"\" width=\"411\" height=\"173\" name=\"Imagem 79\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Calcule o \u00edndice de refra\u00e7\u00e3o do meio e o desvio sofrido pelo raio luminoso.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>12-(Ufg)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Como ilustrado na figura, a luz colimada de uma fonte F incide no espelho E, no ar, e \u00e9 refletida para a face maior do prisma reto P. A luz emerge da face horizontal do prisma, formando com ela um \u00e2ngulo reto.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>O espelho E \u00e9 perpendicular \u00e0 face maior do prisma. Sabendo que a luz incide na dire\u00e7\u00e3o horizontal e que \u03b1 = 30\u00b0, calcule o \u00edndice de refra\u00e7\u00e3o do prisma. Dado: n(ar) =1,0.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_4a0888cb.jpg\" alt=\"\" width=\"593\" height=\"193\" name=\"Imagem 32\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>13-(Fuvest-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>O espa\u00e7o percorrido pela luz que incide perpendicularmente a uma face de um cubo s\u00f3lido feito de material<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_a756fb2f.jpg\" alt=\"\" width=\"492\" height=\"223\" name=\"Imagem 33\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>transparente, antes, durante e ap\u00f3s a incid\u00eancia, \u00e9 dado, em fun\u00e7\u00e3o do tempo, pelo gr\u00e1fico s x t (dist\u00e2ncia x tempo) ao lado. Determine:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) o \u00edndice de refra\u00e7\u00e3o da luz do meio mais refringente em rela\u00e7\u00e3o ao menos refringente.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0b) o comprimento da aresta do cubo.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>14-(UFRGS)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A figura representa um raio de luz monocrom\u00e1tica que se refrata na superf\u00edcie plana de separa\u00e7\u00e3o de dois meios transparentes, cujos \u00edndices de refra\u00e7\u00e3o s\u00e3o n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>. Com base nas medidas expressas na figura, onde C \u00e9 uma circunfer\u00eancia, pode-se calcular a raz\u00e3o n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\/n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0dos \u00edndices de refra\u00e7\u00e3o desses meios. Qual das alternativas apresenta corretamente o valor dessa raz\u00e3o?<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_b702d28d.jpg\" alt=\"\" width=\"481\" height=\"273\" name=\"Imagem 34\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a)\u00a02\/3.\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b)\u00a03\/4\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c)\u00a01.\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d)\u00a04\/3.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e)\u00a03\/2.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>15-(PUCCAMP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b> De uma lanterna colocada no ar (n = 1,0) sai um estreito feixe de luz que incide na superf\u00edcie de separa\u00e7\u00e3o entre o ar e um l\u00edquido transparente, refratando-se conforme mostra a figura .O \u00edndice de refra\u00e7\u00e3o do l\u00edquido \u00e9:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_bb8f1894.jpg\" alt=\"\" width=\"457\" height=\"231\" name=\"Imagem 35\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_b269f20f.png\" alt=\"\" width=\"765\" height=\"23\" name=\"Imagem 80\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>16-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A figura representa o corte transversal de um tanque. Quando o mesmo est\u00e1 vazio, o observador O, no ar (n<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>ar<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>=1), visa o ponto A. Suponha agora o tanque completamente cheio de um l\u00edquido de \u00edndice de refra\u00e7\u00e3o n.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_fef2d2f4.jpg\" alt=\"\" width=\"425\" height=\"278\" name=\"Imagem 36\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Determine o valor m\u00ednimo de n que faz o observador ver o ponto B sob o mesmo raio visual.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>17-(FUVEST-SP<\/b><\/span><\/span><\/span><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>) As figuras a e b indicam os raios de luz incidente i e refratado r na interface entre o meio 1 e os meios 2 e 3, respectivamente.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_2ad814a2.jpg\" alt=\"\" width=\"770\" height=\"242\" name=\"Imagem 37\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) Represente graficamente a refra\u00e7\u00e3o de um raio de luz que passa do meio 2 para o meio 3.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) Um desses tr\u00eas meios \u00e9 o v\u00e1cuo. Qual deles? Justifique.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>18-(PUCCAMP-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Uma onda eletromagn\u00e9tica vis\u00edvel possui, no ar ou no v\u00e1cuo, velocidade de 3,00.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0m\/s e no vidro 1,73.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0m\/s. Essa onda, propagando no ar, incide sobre uma superf\u00edcie plana de vidro com \u00e2ngulo de incid\u00eancia de 60\u00b0. O \u00e2ngulo de refra\u00e7\u00e3o da onda, no vidro, vale:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Dados:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>sen 30\u00b0 = cos 60\u00b0 = 0,50\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 sen 60\u00b0 = cos 30\u00b0 = 0,87<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_db534fff.png\" alt=\"\" width=\"764\" height=\"22\" name=\"Imagem 20\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>19-(UNESP-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um raio de luz monocrom\u00e1tica incide sobre a superf\u00edcie plana de um bloco de vidro de tal modo que o raio refletido R forma um \u00e2ngulo de 90<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0com o raio refratado r. O \u00e2ngulo entre o raio incidente I e a superf\u00edcie de separa\u00e7\u00e3o dos dois meios mede 32<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>, como mostra a figura.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_81c66b1a.jpg\" alt=\"\" width=\"306\" height=\"200\" name=\"Imagem 38\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Os \u00e2ngulos de incid\u00eancia e de refra\u00e7\u00e3o medem respectivamente:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) 62<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 32<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) 58<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 32<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c) 90<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 38<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d) 32<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 90<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) 58<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 45<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>20-(PUC-RS)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A figura abaixo representa a trajet\u00f3ria seguida por um raio luminoso de uma lanterna, at\u00e9 o olho de um observador que se encontra dentro da \u00e1gua.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_97319934.png\" alt=\"\" width=\"379\" height=\"172\" name=\"Imagem 24\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A luz segue essa trajet\u00f3ria por ser o caminho mais:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) r\u00e1pido\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) lento\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c) longo\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d) curto\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) desviado<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>21-(UNESP-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um raio de luz monocrom\u00e1tica, propagando-se no ar, incide sobre a superf\u00edcie de um hemisf\u00e9rio maci\u00e7o de raio R, e emerge perpendicularmente \u00e0 face plana, a uma dist\u00e2ncia R\/2 do eixo \u00f3ptico, conforme figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_c36b7ed2.jpg\" alt=\"\" width=\"462\" height=\"223\" name=\"Imagem 39\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>O \u00edndice de refra\u00e7\u00e3o do material do hemisf\u00e9rio para esse raio de luz, \u00e9 n=\u221a2. Calcule o desvio angular sofrido pelo raio ao atravessar o hemisf\u00e9rio.<\/b><\/span><\/span><\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>22-(FUVEST-SP)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Dois sistemas \u00f3ticos, D<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0e D<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>2,<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0s\u00e3o utilizados para analisar uma l\u00e2mina de tecido biol\u00f3gico a partir de dire\u00e7\u00f5es diferentes. Em uma an\u00e1lise, a luz fluorescente, emitida por um indicador incorporado a uma pequena estrutura, presente no tecido, \u00e9 captada, simultaneamente, pelos dois sistemas, ao longo das dire\u00e7\u00f5es tracejadas. Levando-se em conta o desvio da luz pela refra\u00e7\u00e3o, dentre as posi\u00e7\u00f5es indicadas, aquela que poderia corresponder \u00e0 localiza\u00e7\u00e3o real dessa estrutura no tecido \u00e9:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_5d4af95a.jpg\" alt=\"\" width=\"496\" height=\"194\" name=\"Imagem 40\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Suponha que o tecido biol\u00f3gico seja transparente \u00e0 luz e tenha \u00edndice de refra\u00e7\u00e3o uniforme, semelhante ao da \u00e1gua.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_fd162fb2.png\" alt=\"\" width=\"766\" height=\"18\" name=\"Imagem 12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>23-(UFRJ-RJ)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um raio luminoso proveniente do ar atravessa uma placa de vidro de 4,0 cm de espessura e \u00edndice de refra\u00e7\u00e3o 1,5<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_24900191.jpg\" alt=\"\" width=\"414\" height=\"227\" name=\"Imagem 41\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Sabendo que o \u00e2ngulo de incid\u00eancia \u0161 do raio luminoso \u00e9 tal que sen \u03b8 = 0,90 e que o \u00edndice de refra\u00e7\u00e3o do ar \u00e9 1, calcule a dist\u00e2ncia que a luz percorre ao atravessar a placa.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>24- (UNESP-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_5ddf0de7.jpg\" alt=\"\" width=\"153\" height=\"79\" name=\"Imagem 42\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Considere um raio de luz monocrom\u00e1tico de comprimento de onda \u03bb, que incide com \u00e2ngulo \u03b8<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>i<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0em uma das faces de um prisma de vidro que est\u00e1 imerso no ar, atravessando-o como indica a figura.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_855a769.jpg\" alt=\"\" width=\"421\" height=\"224\" name=\"Imagem 43\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Sabendo que o \u00edndice de refra\u00e7\u00e3o do vidro em rela\u00e7\u00e3o ao ar diminui com o aumento do comprimento de onda<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>do raio de luz que atravessa o prisma, assinale a alternativa que melhor representa a trajet\u00f3ria de outro raio<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>de luz de comprimento 1,5 \u03bb, que incide <\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>sobre esse mesmo prisma de vidro.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_79ae944a.jpg\" alt=\"\" width=\"782\" height=\"111\" name=\"Imagem 44\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>25-(UNICAMP-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_b195f902.jpg\" alt=\"\" width=\"174\" height=\"113\" name=\"Imagem 45\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A radia\u00e7\u00e3o Cerenkov ocorre quando uma part\u00edcula carregada atravessa um meio isolante com uma velocidade maior do que a velocidade da luz nesse meio. O estudo desse efeito rendeu a Pavel A. Cerenkov e colaboradores o pr\u00eamio Nobel de F\u00edsica de 1958. Um exemplo desse fen\u00f4meno pode ser observado na \u00e1gua usada para refrigerar reatores<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_e9e84b12.jpg\" alt=\"\" width=\"777\" height=\"184\" name=\"Imagem 46\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>nucleares, em que ocorre a emiss\u00e3o de luz azul devido \u00e0s part\u00edculas de alta energia que atravessam a \u00e1gua.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) Sabendo-se que o \u00edndice de refra\u00e7\u00e3o da \u00e1gua \u00e9 n = 1,3, calcule a velocidade m\u00e1xima das part\u00edculas na \u00e1gua para que n\u00e3o ocorra a radia\u00e7\u00e3o Cerenkov. A velocidade da luz no v\u00e1cuo \u00e9 c = 3,0.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m\/s.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) A radia\u00e7\u00e3o Cerenkov emitida por uma part\u00edcula tem a forma de um cone, como ilustrado na figura abaixo, pois a sua velocidade, v<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>p<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0, \u00e9 maior do que a velocidade da luz no meio, v<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>l<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0. Sabendo que o cone formado tem um \u00e2ngulo \u03b8<\/b><\/span><\/span><span style=\"font-family: Symbol,serif;\"><span style=\"font-size: medium;\"><b>\uf020<\/b><\/span><\/span><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>= 50\u00ba e que a radia\u00e7\u00e3o emitida percorreu uma dist\u00e2ncia d = 1,6m em t = 12ns, calcule v<\/b><\/span><\/span><sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>p<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_86a3842a.jpg\" alt=\"\" width=\"528\" height=\"200\" name=\"Imagem 47\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Dados: cos50\u00ba = 0,64 e sen50\u00ba = 0,76.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>26-(CEFET-MG)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_b2983751.jpg\" alt=\"\" width=\"180\" height=\"103\" name=\"Imagem 48\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>A dura\u00e7\u00e3o do dia terrestre \u00e9 ligeiramente ____________ devido a___________ da luz solar na atmosfera.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Os termos que completam, corretamente, as lacunas s\u00e3o<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) menor, reflex\u00e3o.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b) maior, difra\u00e7\u00e3o.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0c) maior, refra\u00e7\u00e3o.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d) menor, absor\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) maior, interfer\u00eancia.<\/b><\/span><\/span><\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>27-(UFPR-PR)\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_46858938.jpg\" alt=\"\" width=\"144\" height=\"68\" name=\"Imagem 49\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Ao incidir sobre um prisma de vidro, um feixe de luz branca \u00e9 decomposto em v\u00e1rias cores. Esse fen\u00f4meno acontece porque as ondas eletromagn\u00e9ticas de diferentes comprimentos de onda se propagam no vidro com diferentes velocidades, de modo que o \u00edndice de refra\u00e7\u00e3o n tem valor diferente para cada comprimento de onda. O estudo das propriedades \u00f3ticas de um peda\u00e7o de vidro forneceu o gr\u00e1fico ao lado para o \u00edndice de refra\u00e7\u00e3o n em fun\u00e7\u00e3o do comprimento de onda \u0443 da luz. Suponha a velocidade da luz no v\u00e1cuo igual a 3,0.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0m\/s.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_5f7d465d.jpg\" alt=\"\" width=\"776\" height=\"192\" name=\"Imagem 50\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Com base nos conceitos de \u00f3tica e nas informa\u00e7\u00f5es do gr\u00e1fico, assinale a alternativa correta.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a) Luz com comprimento de onda entre 450 nm e 550 nm se propaga no vidro com velocidades de mesmo m\u00f3dulo.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>b) A frequ\u00eancia da luz com comprimento de onda 600 nm \u00e9 de 3,6.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>\u00a0Hz.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>c) O maior \u00edndice de refra\u00e7\u00e3o corresponde \u00e0 luz com menor frequ\u00eancia.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>d) No vidro, a luz com comprimento de onda 700 nm tem uma velocidade, em m\u00f3dulo, de 2,5.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8\u00a0<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m\/s.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>e) O menor \u00edndice de refra\u00e7\u00e3o corresponde \u00e0 luz com menor velocidade de propaga\u00e7\u00e3o no vidro.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>28-(FMABC)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_589cb0b.jpg\" alt=\"\" width=\"473\" height=\"121\" name=\"Imagem 51\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um aluno, utilizando uma fonte luminosa cujo comprimento de onda vale 6.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>-7<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m, incide perpendicularmente um feixe de luz sobre a \u00e1gua, cujo \u00edndice de refra\u00e7\u00e3o vale 4\/3, de um aqu\u00e1rio totalmente preenchido, com o objetivo de iluminar um peixe que se encontra a 20cm de profundidade. Considerando que a dist\u00e2ncia entre a fonte luminosa e a superf\u00edcie da \u00e1gua \u00e9 de 10cm, o aluno lembrou-se das aulas de F\u00edsica em que o professor havia dito que o \u00edndice de refra\u00e7\u00e3o do ar vale 1 e que a velocidade da luz no v\u00e1cuo vale 3.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>ms. Fez, ent\u00e3o, algumas observa\u00e7\u00f5es sobre a luz no interior da \u00e1gua:<\/b><\/span><\/span><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_df194b92.jpg\" alt=\"\" width=\"395\" height=\"264\" name=\"Imagem 52\" align=\"LEFT\" border=\"0\" hspace=\"12\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>I. A frequ\u00eancia, a velocidade e o comprimento de onda da luz incidente devem ter sofrido altera\u00e7\u00f5es uma vez que a \u00e1gua tem \u00edndice de refra\u00e7\u00e3o bem maior que o ar e a incid\u00eancia foi perpendicular.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>II. Como a incid\u00eancia da luz foi perpendicular, apenas a frequ\u00eancia da luz variou e n\u00e3o houve altera\u00e7\u00f5es na velocidade e no comprimento de onda da luz no interior da \u00e1gua.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>III. O comprimento de onda e a velocidade variam no interior da \u00e1gua e valem respectivamente 4,5.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>-7<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m e 2,25.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m\/s, mas a frequ\u00eancia permanece inalterada.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>IV. Como a lanterna estava pr\u00f3xima da superf\u00edcie da \u00e1gua do aqu\u00e1rio, apenas a velocidade da luz no interior da \u00e1gua sofreu varia\u00e7\u00e3o e seu valor passou a ser de 2,25.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m\/s.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>V. Como a lanterna estava pr\u00f3xima da superf\u00edcie da \u00e1gua do aqu\u00e1rio, a incid\u00eancia foi perpendicular e o \u00edndice de refra\u00e7\u00e3o da \u00e1gua \u00e9 maior que o do ar, a frequ\u00eancia e o comprimento de onda da luz no interior da \u00e1gua sofreram varia\u00e7\u00f5es e seus valores passaram a ser 2,25.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>8<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Hz e 4,5.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>-7<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>m.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Com rela\u00e7\u00e3o \u00e0s observa\u00e7\u00f5es feitas pelo aluno, est\u00e1 correta apenas<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>(A) I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (B) II\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (C) III\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (D) IV\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (E) V<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><a name=\"_GoBack\"><\/a> <span style=\"color: #0000cc;\"><span style=\"font-family: Arial Black,serif;\"><span style=\"font-size: medium;\"><b>29-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_1c2ae720.jpg\" alt=\"\" width=\"420\" height=\"124\" name=\"Imagem 53\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>Um raio de luz monocrom\u00e1tico que se propaga no ar (\u00edndice de refra\u00e7\u00e3o = 1) atinge a superf\u00edcie de separa\u00e7\u00e3o com um meio homog\u00eaneo e transparente, sob determinado \u00e2ngulo de incid\u00eancia, diferente de 0\u00ba. Considerando os meios da tabela abaixo, aquele para o qual o raio luminoso tem o menor desvio \u00e9<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/optica\/lei-refracao\/i_bc04d0bee499cfe4_html_c0a0a365.jpg\" alt=\"\" width=\"515\" height=\"125\" name=\"Imagem 54\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial,serif;\"><span style=\"font-size: medium;\"><b>a)\u00a0 \u00c1gua\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b)\u00a0 \u00c1lcool et\u00edlico\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c) Diamante\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d) Glicerina\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0e)\u00a0 Vidro comum<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<h3><a title=\"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Leis da Refra\u00e7\u00e3o\" href=\"http:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/leis-da-refracao\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-leis-da-refracao\/\"><span style=\"color: #000080;\">Confira o gabarito e resolu\u00e7\u00e3o comentada<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Exerc\u00edcios de vestibulares com resolu\u00e7\u00e3o comentada sobre Leis da Refra\u00e7\u00e3o &nbsp; 01-(UFSC-SC) A figura a seguir mostra um l\u00e1pis de comprimento AB, parcialmente imerso na \u00e1gua e sendo observado por um estudante. Assinale a(s) proposi\u00e7\u00e3o (\u00f5es) CORRETA(S) e indique sua soma:. (01) O estudante v\u00ea o l\u00e1pis &#8220;quebrado&#8221; na interface ar-\u00e1gua, porque o \u00edndice de refra\u00e7\u00e3o da \u00e1gua \u00e9 maior<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2237,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-2240","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2240","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=2240"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2240\/revisions"}],"predecessor-version":[{"id":10946,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2240\/revisions\/10946"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2237"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}