{"id":11036,"date":"2022-05-20T09:00:00","date_gmt":"2022-05-20T12:00:00","guid":{"rendered":"https:\/\/professorpinguim.com.br\/blog\/?p=11036"},"modified":"2023-08-06T21:00:31","modified_gmt":"2023-08-07T00:00:31","slug":"questoes-sobre-aplicacao-da-lei-de-snell-na-refracao-da-luz","status":"publish","type":"post","link":"https:\/\/professorpinguim.com.br\/blog\/questoes-sobre-aplicacao-da-lei-de-snell-na-refracao-da-luz\/","title":{"rendered":"Quest\u00f5es sobre aplica\u00e7\u00e3o da Lei de Snell na Refra\u00e7\u00e3o da Luz"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Na \u00f3ptica Geom\u00e9trica a Lei de Snell \u00e9 a segunda Lei da Refra\u00e7\u00e3o da Luz. Ela permite relacionar o \u00edndice de refra\u00e7\u00e3o do meio com o seno do \u00e2ngulo que o raio de luz faz com a reta normal \u00e0 superf\u00edcie de separa\u00e7\u00e3o entre os meios.<\/p>\n\n\n\n<!--more-->\n\n\n\n<p class=\"wp-block-paragraph\"><strong>n<sub>1<\/sub>.sen i = n<sub>2<\/sub>.sen r<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img fetchpriority=\"high\" decoding=\"async\" src=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2022\/05\/Captura-de-Tela-2022-05-19-a\u0300s-19.42.25.png\" alt=\"\" class=\"wp-image-11037\" width=\"296\" height=\"292\" srcset=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2022\/05\/Captura-de-Tela-2022-05-19-a\u0300s-19.42.25.png 666w, https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2022\/05\/Captura-de-Tela-2022-05-19-a\u0300s-19.42.25-300x296.png 300w\" sizes=\"(max-width: 296px) 100vw, 296px\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">No desenho acima, a luz passou do meio menos para o mais refringente. O raio se aproxima da normal<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Exerc\u00edcios<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>1. (Unesp 2022)&nbsp;<\/strong> A figura representa um raio de luz monocrom\u00e1tica propagando- se pelo ar (n = 1), incidindo na superf\u00edcie de um bloco feito de material homog\u00eaneo e transparente com um \u00e2ngulo de incid\u00eancia de 60\u00b0 e refratando-se com um \u00e2ngulo de refra\u00e7\u00e3o r.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/dFgub_09x0COP_fn4bb4xT174hKdOqwBew8Gq9KR-j21pOyv6fJrVKzOhRW-hqTJVq5mAHDm-J2sHna90bd6Q3LtTrdVhoyBqZeE47SgYRMTfwb4b7UYSSgUlqeHa8Sc07o05VZ58Y3sWcuIAQ\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/fktFBTjqMnZh9E6pGyRv1Iu47ZEEAtab6XFJ98bsnFcT3r5UUF4gbhSVu7pR34PN6jfL1sdVHr2O8wgx-SoDb-gXPscGk51dwWaJloRC3BOZu5eKwqnXPYwbIXPrv03rDcranAOmJTL5CfsnBQ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Sabendo que o \u00e2ngulo limite de incid\u00eancia para refra\u00e7\u00e3o da luz desse bloco para o ar \u00e9 de 30\u00b0 e considerando os valores indicados na tabela, o valor de r, quando o \u00e2ngulo de incid\u00eancia no ar for 60\u00b0, \u00e9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 25,6\u00b0&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 29,3\u00b0&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 30,0\u00b0&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 35,3\u00b0&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 45,0\u00b0&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Resposta <strong>1: <\/strong>A<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/l8nhV9qrs7AIRwwgEKjAZgSlzh7xT1BSJnQfUNavUmRYDH4CcH9_dFBFa7pEVvnpmHQUX3LEJ8P8sx-mDY781ERuWVvD-aLlKpK08zohGXJosRV6Du3jCq1YlKmDvXdmBuYA3W0GKsFlA6Q3BA\" alt=\"\" width=\"631\" height=\"438\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. (Fmj 2022)<\/strong>&nbsp; Um raio de luz monocrom\u00e1tica se propaga no ar, cujo \u00edndice de refra\u00e7\u00e3o absoluto \u00e9 igual a 1,0, e incide na superf\u00edcie de uma l\u00e2mina de vidro formando um \u00e2ngulo com a reta normal \u00e0 superf\u00edcie. Ao penetrar no vidro, o raio passa a formar um \u00e2ngulo com a reta normal.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/0q8AlrndTm3Vxt4Mpm-04sPkHL9MWT6bXpe2PUnfsC6C8OlRf6czPha504cmVhRVXZMIAxY_YGtXinyzswxNLMraS0I2pA_OAJXph8ucMKnw6pP5L1OKRGSIwevUWpNItJWREuQhNF4M_QNvow\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Sabendo que a luz se propaga no ar com velocidade de 3.10<sup>8<\/sup> m\/s que sen = 0,8&nbsp; e sen = 0,6 a velocidade de propaga\u00e7\u00e3o da luz no vidro que constitui a l\u00e2mina \u00e9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 1,15.10<sup>8<\/sup> m\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 1,80.10<sup>8<\/sup> m\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 4,00.10<sup>8<\/sup> m\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 1,40.10<sup>8<\/sup> m\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 2,25.10<sup>8<\/sup> m\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Resposta <strong>2: <\/strong>E<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/wNCyb-eWB2L2iLKR641mpE1Sn_OdNU1_05gah_7fmk56aAajxfuHOp3MUpch-heklqTF6c66kElhCz3f1BdZxN1jzBk7RzjurIuTntKuud00gyZnHoJ0aUZUQI4k38vZeDakfDaW96_BNyNkAQ\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3. (Albert Einstein &#8211; Medicina 2022)<\/strong>&nbsp; Em p\u00e9, sobre a borda de uma piscina de 5 m de profundidade, uma pessoa aponta um laser para uma boia dentro da \u00e1gua, que est\u00e1 presa por um fio de comprimento L no fundo dessa piscina. Por\u00e9m, devido ao desvio sofrido pelo laser, ele atinge o fundo da piscina exatamente no ponto em que o fio que prende a boia est\u00e1 fixo.<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/_bsY_jBf7J_TFyBqtyKXpDAExEAGFwphsi3jr667-jslZJ2M7h3kw-djVeVcSlKq7iX6iH_RaDrayH_aKPfdD9haprnfcR_CPAfJo7ut8GPP7sEgtfRz9CqQeMAxrdK2AV4xtxZfvCedHNKExw\" alt=\"\" width=\"397\" height=\"271\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Sabendo que o \u00edndice de refra\u00e7\u00e3o do ar e da \u00e1gua s\u00e3o, respectivamente, n<sub>Ar<\/sub> = 1,0 e n<sub>\u00c1gua<\/sub> = 1,4, que sen 30\u00b0 = 0,5 e adotando sen 45\u00b0 = 0,7 e tg 30\u00b0 = 0,6, o comprimento do fio que prende a boia ao fundo da piscina \u00e9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 3,0 m.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 4,0 m.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 2,0 m.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 3,5 m.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 2,5 m.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Resposta <strong>3: <\/strong>C<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/zfgF44fj2iYLIokoNihyESSGWs4M9NiVZi-YCkPuCYhboOLWFHydDGVcqJIwFL0WoSh0SfEWt-GyxkT5fEi_NxC3SETo3Pdseq4Z06yhxTm7mUW9PMzw9Zx9MzBrir5R0K-Nl9tOSTK3Kw2klA\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/YxXXEyINfqiO6zz0OnQ83sWO4ReNNKxcrIq2vr9PNNiCeIM4-AJHywO6DqjYWZRrW-91S2WQwZzAA1je_XUhtegygxiEiLWNxx2WF5ZE5ccj5Tt6TRb1cq23d-9rG_74rgsuGTXAWKqRdzvqCA\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/YpCQsfpygvHtaJcdi6akZrbC1H5sI3-8h-dv4TVpYO9Q-VSu-2tHViCEPhD9QAcBQ4Xf5nmXabQevupzSm1o5JjJNbveSw0jImcpNTKnCyg8DckGoTzIANMrNCm2z7QDsOMLs2GBMJel47426w\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>4. (Unicamp 2022)<\/strong>&nbsp;&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/9eX7_UTn99hP4Ixj6gpgyyx7TBFsLSk8j1CTXBGZnNlH8T2prDP5sCX0ovOE2BcLSxsSZg5EQkXYwuEMGMjG6lLHhGMjSa_-Z7uyvG9usjXtuMDHgI4ufjEyCrRe6oPTI5kGGhlCoM5aTDdTvA\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Uma imagem capturada recentemente pela sonda <em>Perseverance<\/em> na superf\u00edcie de Marte mostrou o que parece ser um arco-\u00edris no c\u00e9u daquele planeta. Na Terra, um arco-\u00edris surge como resultado da decomposi\u00e7\u00e3o da luz branca do Sol por refra\u00e7\u00e3o nas got\u00edculas quase esf\u00e9ricas de \u00e1gua, suspensas na atmosfera. Em Marte, contudo, n\u00e3o h\u00e1 chuva. Portanto, a origem do arco-\u00edris ainda \u00e9 controversa. Em rela\u00e7\u00e3o ao fen\u00f4meno de forma\u00e7\u00e3o do arco-\u00edris na Terra, quando a luz solar incide em uma got\u00edcula de \u00e1gua, \u00e9 correto afirmar que<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) o \u00edndice de refra\u00e7\u00e3o da \u00e1gua para as diferentes cores da luz branca \u00e9 menor do que o do ar; assim, no interior das got\u00edculas, os raios de luz das diferentes cores se aproximam mais da reta normal \u00e0 interface entre os meios de refra\u00e7\u00e3o, quanto maior for o \u00edndice de refra\u00e7\u00e3o corresponde \u00e0quela cor, na \u00e1gua.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) o \u00edndice de refra\u00e7\u00e3o da \u00e1gua para as diferentes cores da luz branca \u00e9 menor do que o do ar; assim, os raios de luz das diferentes cores, no interior das got\u00edculas, se afastam mais da reta normal \u00e0 interface entre os meios de refra\u00e7\u00e3o, quanto maior for o \u00edndice de refra\u00e7\u00e3o corresponde \u00e0quela cor, na \u00e1gua.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) o \u00edndice de refra\u00e7\u00e3o da \u00e1gua para as diferentes cores da luz branca \u00e9 maior do que o do ar; assim, os raios de luz das diferentes cores, no interior das got\u00edculas, se aproximam mais da reta normal \u00e0 interface entre os meios de refra\u00e7\u00e3o, quanto maior for o \u00edndice de refra\u00e7\u00e3o corresponde \u00e0quela cor, na \u00e1gua.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) o \u00edndice de refra\u00e7\u00e3o da \u00e1gua para as diferentes cores da luz branca \u00e9 maior do que o do ar; assim, os raios de luz das diferentes cores, no interior das got\u00edculas, se afastam mais da reta normal \u00e0 interface entre os meios de refra\u00e7\u00e3o, quanto maior for o \u00edndice de refra\u00e7\u00e3o corresponde \u00e0quela cor, na \u00e1gua.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Resposta <strong>4. <\/strong>C<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para uma raio de luz branca que se propaga no ar e&nbsp; atinge uma gota, o \u00e2ngulo de incid\u00eancia \u00e9 o mesmo para as cores componentes.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ent\u00e3o, de acordo com a lei de Snell:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/cSTmjDifa05cRcUw9k9lNMakP5FK9LBKxI7wsL4gX0Hh8_lLLPGl_L_13fm4clLcq1KOOPU5vm74h5Wth_zWzHjnmb3KzE5EvmeNU8jHlTWBwCEChvgJAcX0k_qG4eFQcuqfkVVYcTzcjXbBrQ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Como r &lt; 90\u00b0 e o \u00edndice de refra\u00e7\u00e3o da \u00e1gua \u00e9 maior que o do ar para todas as cores, essa express\u00e3o mostra que o que o \u00e2ngulo de refra\u00e7\u00e3o \u00e9 menor (o raio aproxima-se mais da normal) para a cor de maior \u00edndice de refra\u00e7\u00e3o na \u00e1gua.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para as diferentes cores do espectro vis\u00edvel, a ordem <strong>crescente<\/strong> de \u00edndice de refra\u00e7\u00e3o \u00e9:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>&nbsp;Vermelho<\/strong><strong>&nbsp; &nbsp; Alaranjado &nbsp; Amarelo <\/strong><strong><\/strong><strong>Verde&nbsp; &nbsp; Azul<\/strong><strong> <\/strong><strong>Anil&nbsp; <\/strong><strong><\/strong><strong>Violeta<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;Portanto, essa \u00e9 a mesma ordem de aproxima\u00e7\u00e3o da normal, conforme a figura.<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/wBLciol_Wr52E41R9JTPj46HR2hYTYFtn5wbrtuglyXVn7FAv29Il1wzsiVsNBWnwFyT1V3NNRjLx6mHOenu6BujpWOfAnEu8VwtAhLSWI_d_ziTEtotmHhBsehpAvlxz53nQIm0s-3iFh1VSg\" alt=\"\" width=\"432\" height=\"285\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>5. (Unesp 2021)<\/strong>&nbsp; Um semicilindro circular reto de raio R est\u00e1 imerso no ar e \u00e9 atingido por um raio de luz monocrom\u00e1tica que incide perpendicularmente no ponto A de uma de suas faces planas. Ap\u00f3s atravess\u00e1-lo, esse raio emerge pelo ponto B contido na superf\u00edcie circular do semicilindro. As figuras indicam as duas situa\u00e7\u00f5es.<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/1yB2YZLqgsINyDQaIfCzD984GIjGL7rZITgxVF0Eb4ke7D11K-4aywo9GeO9EyLv1hQRlx1sNWvdya0ivlC5vitowAo9jUpJOaGCnAnDeZPt4oZAJWltMv7gfUkWa2I4kZUS4BOQ34-_dfBxKw\" alt=\"\" width=\"280\" height=\"423\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/1yB2YZLqgsINyDQaIfCzD984GIjGL7rZITgxVF0Eb4ke7D11K-4aywo9GeO9EyLv1hQRlx1sNWvdya0ivlC5vitowAo9jUpJOaGCnAnDeZPt4oZAJWltMv7gfUkWa2I4kZUS4BOQ34-_dfBxKw\" alt=\"\" width=\"273\" height=\"412\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Considerando sen37\u00b0=0,6 e que o \u00edndice de refra\u00e7\u00e3o absoluto do ar \u00e9 n<sub>ar<\/sub> = 1 o \u00edndice de refra\u00e7\u00e3o absoluto do material de que o semicilindro \u00e9 feito \u00e9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 1,2.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 1,4.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 1,6.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 1,8.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 2,0.&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>5. <\/strong>A<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/H1o3Z4Hvv16OQiwL-AfiOwS92xw4s9VrfIJuSpzuNtk_ar1_3fUtvKoghga2eboFuAEzfiND8YiLcEy9WvfVmpap0uU-jnssQb_d1uDLg2UVnETCFHVgoU1frZVdqOIyiI-B7oE9cAYAqNKwTA\" alt=\"\" width=\"277\" height=\"255\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/rg_67_bHyFghNqtBawnHZFPTRAlyzuOvIDhnn40ndj1d78GaPgzCmW8HhQItPGTZIBoAISWthmLsT8BrjrE-S-viEKQDKfv7Y7Glq4aXndgVHiYYzcDcgZcC5XOUgoCsmFMaOzo7hh4jqfOoTw\" alt=\"\" width=\"320\" height=\"207\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Para aprender mais:<\/h3>\n\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe title=\"Refra\u00e7\u00e3o da Luz - Exerc\u00edcios - Prof Pinguim\" width=\"800\" height=\"450\" src=\"https:\/\/www.youtube.com\/embed\/aBQISDFAV4s?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">Espero que voc\u00ea tenha entendido um pouco melhor a <strong>Lei de Snell<\/strong>. E se quiser ajuda para melhorar seu n\u00edvel de F\u00edsica em outras mat\u00e9rias,<a href=\"https:\/\/professorpinguim.com.br\/\">\u00a0<strong>entre em contato comigo e escolha o curso de F\u00edsica mais adequado para voc\u00ea<\/strong><\/a><strong>!<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/professorpinguim.com.br\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"488\" src=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1-1024x488.png\" alt=\"Extensivo de Fisica para Vestibulares Paulistas\" class=\"wp-image-19816\" srcset=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1-1024x488.png 1024w, https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1-300x143.png 300w, https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1-768x366.png 768w, https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1-670x319.png 670w, https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1-1060x505.png 1060w, https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2023\/02\/EXTENSIVO-BLOG-1.png 1280w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>SAIBA MAIS<\/strong><br>\ud83d\udc27&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/blog\/questoes-sobre-equacoes-em-espelhos-esfericos\/\" target=\"_blank\" rel=\"noreferrer noopener\">Exerc\u00edcios sobre equa\u00e7\u00f5es em espelhos esf\u00e9ricos<\/a><a href=\"https:\/\/professorpinguim.com.br\/blog\/primeira-lei-de-newton-inercia\/\"><br><\/a>\ud83d\udc27&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/blog\/leis-de-newton-exercicios-de-blocos-e-atrito\/\">Leis de Newton \u2013 Exerc\u00edcios de Blocos e Atrito<br><\/a>\ud83d\udc27&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/blog\/introducao-a-optica-com-exercicios-resolvidos\/\" target=\"_blank\" rel=\"noreferrer noopener\">Introdu\u00e7\u00e3o sobre \u00e0 \u00f3ptica<\/a><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Me acompanhe nas redes sociais: curta a minha&nbsp;<a href=\"https:\/\/www.facebook.com\/professorpinguim\/\"><strong>p\u00e1gina no Facebook<\/strong><\/a><strong>,<\/strong>&nbsp;me siga no&nbsp;<a href=\"https:\/\/www.instagram.com\/professorpinguim\/\"><strong>Instagram<\/strong><\/a><strong>,<\/strong>&nbsp;se inscreva no&nbsp;<a href=\"https:\/\/www.youtube.com\/c\/PinguimV%C3%ADdeoF%C3%ADsica\/?sub_confirmation=1\"><strong>Youtube<\/strong><\/a>&nbsp;e participe do meu canal oficial no&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/telegram\"><strong>Telegram<\/strong><\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Na \u00f3ptica Geom\u00e9trica a Lei de Snell \u00e9 a segunda Lei da Refra\u00e7\u00e3o da Luz. Ela permite relacionar o \u00edndice de refra\u00e7\u00e3o do meio com o seno do \u00e2ngulo que o raio de luz faz com a reta normal \u00e0 superf\u00edcie de separa\u00e7\u00e3o entre os meios.<\/p>\n","protected":false},"author":1,"featured_media":11040,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[744,764,228,656],"tags":[904,903,525,906,905],"class_list":["post-11036","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-exercicios-resolvidos","category-ondas","category-optica","category-termologia","tag-lei-da-refracao-da-luz","tag-lei-de-snell","tag-optica-geometrica","tag-raio-de-luz","tag-refracao"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Quest\u00f5es sobre aplica\u00e7\u00e3o da Lei de Snell na Refra\u00e7\u00e3o da Luz - Professor Pinguim | F\u00edsica<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/professorpinguim.com.br\/blog\/questoes-sobre-aplicacao-da-lei-de-snell-na-refracao-da-luz\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Quest\u00f5es sobre aplica\u00e7\u00e3o da Lei de Snell na Refra\u00e7\u00e3o da Luz - Professor Pinguim | F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Na \u00f3ptica Geom\u00e9trica a Lei de Snell \u00e9 a segunda Lei da Refra\u00e7\u00e3o da Luz. 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