{"id":2208,"date":"2016-05-14T23:44:28","date_gmt":"2016-05-14T23:44:28","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2208"},"modified":"2024-09-02T13:42:41","modified_gmt":"2024-09-02T13:42:41","slug":"dioptro-plano","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/dioptro-plano\/","title":{"rendered":"Dioptro Plano"},"content":{"rendered":"

Dioptro Plano <\/b><\/span><\/span><\/span><\/p>\n

Um <\/b><\/span><\/span><\/span>dioptro plano<\/b><\/span><\/span><\/span> \u00e9 um <\/b><\/span><\/span><\/span>sistema refrator<\/b><\/span><\/span><\/span> que apresenta <\/b><\/span><\/span><\/span>dois meios transparentes e homog\u00eaneos<\/b><\/span><\/span><\/span> <\/b>separados<\/b><\/span><\/span><\/span> por uma <\/b><\/span><\/span><\/span>superf\u00edcie<\/b><\/span><\/span><\/span> ou, mais especificamente <\/b><\/span><\/span><\/span>consiste num conjunto de dois meios opticamente homog\u00eaneos e transparentes separados por uma superf\u00edcie plana como, por exemplo,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

o ar em contato com a \u00e1gua parada e transparente de uma piscina (dioptro plano \u201car-\u00e1gua\u201d)<\/b><\/span><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o das imagens num dioptro plano<\/b><\/span><\/span><\/span><\/p>\n

Quando <\/b><\/span><\/span><\/span>observamos uma pessoa mergulhada <\/b><\/span><\/span><\/span>numa <\/b><\/span><\/span><\/span>piscina cheia de \u00e1gua<\/b><\/span><\/span><\/span>, estamos <\/b><\/span><\/span><\/span>observando<\/b><\/span><\/span><\/span> sua <\/b><\/span><\/span><\/span>imagem virtual, que est\u00e1 acima do objeto.<\/b><\/span><\/span><\/span><\/p>\n

Da mesma maneira, <\/b><\/span><\/span><\/span>se estivermos debaixo da \u00e1gua, e olharmos <\/b><\/span><\/span><\/span>para um corpo<\/b><\/span><\/span><\/span> fora da piscina, estaremos observando sua <\/b><\/span><\/span><\/span>imagem virtual, que est\u00e1 acima do objeto.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

\"\"<\/p>\n

Essa posi\u00e7\u00e3o, diferente da real<\/b><\/span><\/span><\/span>, em que <\/b><\/span><\/span><\/span>enxergamos a imagem virtual,<\/b><\/span><\/span><\/span> chamamos de\u00a0<\/b><\/span><\/span><\/span>posi\u00e7\u00e3o (ou dist\u00e2ncia) aparente.<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>figura 1 <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>ponto objeto real P<\/b><\/span><\/span><\/span> est\u00e1 dentro <\/b><\/span><\/span><\/span>da \u00e1gua (mais refringente, maior \u00edndice de refra\u00e7\u00e3o) e emitindo dois raios de luz<\/b><\/span><\/span><\/span>, um <\/b><\/span><\/span><\/span>vertical que<\/b><\/span><\/span><\/span> <\/b>n\u00e3o sofre desvio<\/b><\/span><\/span><\/span> e o outro <\/b><\/span><\/span><\/span>obl\u00edquo que, ao se refratar para o ar, afasta-se da normal atingindo<\/b><\/span><\/span><\/span> os <\/b><\/span><\/span><\/span>olhos do observador<\/b><\/span><\/span><\/span> e determina a <\/b><\/span><\/span><\/span>imagem virtual (P\u2019),<\/b><\/span><\/span><\/span> <\/b>obtida <\/b><\/span><\/span><\/span>no <\/b><\/span><\/span><\/span>prolongamento desse raio<\/b><\/span><\/span><\/span> e onde ele <\/b><\/span><\/span><\/span>intercepta o raio de luz vertical, fornecendo essa imagem P\u2019,<\/b><\/span><\/span><\/span> <\/b>acima do objeto real P.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0Na <\/b><\/span><\/span><\/span>figura II<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>ponto objeto real P est\u00e1 no ar (menos refringente, menor \u00edndice de refra\u00e7\u00e3o)<\/b><\/span><\/span><\/span> e emitindo <\/b><\/span><\/span><\/span>dois raios de luz, um vertical que n\u00e3o sofre<\/b><\/span><\/span><\/span> desvio e o <\/b><\/span><\/span><\/span>outro obl\u00edquo que, ao refratar-se para\u00a0 a \u00e1gua aproxima-se da normal<\/b><\/span><\/span><\/span> atingindo os <\/b><\/span><\/span><\/span>olhos do observador e determina a imagem virtual P\u2019, acima do objeto real P.<\/b><\/span><\/span><\/span><\/p>\n

Observe que,<\/b><\/span><\/span><\/span> <\/b>em ambos os casos a imagem est\u00e1 sempre acima do objeto.<\/b><\/span><\/span><\/span><\/p>\n

Dedu\u00e7\u00e3o da equa\u00e7\u00e3o do dioptro plano<\/b><\/span><\/span><\/span><\/p>\n

Aplicando<\/b><\/span><\/span><\/span> a lei de Snell-Descartes, por exemplo, no primeiro caso acima, temos:<\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>1.<\/b><\/span><\/span><\/sub><\/span>seni = n<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>senr (I)<\/b><\/span><\/span><\/span> \"\"\u00a0 considerando os <\/b><\/span><\/span><\/span>\u00e2ngulos muito pequenos de modo que o observador<\/b><\/span><\/span><\/span> <\/b>esteja bem pr\u00f3ximo da normal<\/b><\/span><\/span><\/span>\u00a0de maneira que o <\/b><\/span><\/span><\/span>seno e a tangente de cada \u00e2ngulo sejam aproximadamente iguais ou seja,\u00a0seni<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2248\u00a0tgi\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2248\u00a0a\/d<\/b><\/span><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>(II)\u00a0\u00a0e\u00a0\u00a0senr<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2248\u00a0tgr\u00a0\u2248 a\/d<\/b><\/span><\/span><\/span>i\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0\u00a0(III).<\/b><\/span><\/span><\/span><\/p>\n

Substituindo (II) e (III) em (I)\u00a0\"\"\u00a0n<\/b><\/span><\/span><\/span>1.<\/b><\/span><\/span><\/sub><\/span>d<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= n<\/b><\/span><\/span><\/span>2.<\/b><\/span><\/span><\/sub><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Como utilizar a equa\u00e7\u00e3o do dioptro-plano<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo:<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>dist\u00e2ncia vertical da imagem \u00e0 superf\u00edcie de separa\u00e7\u00e3o<\/b><\/span><\/span><\/span> dos meios A e B <\/b><\/span><\/span><\/span>(dist\u00e2ncia aparente, em que o observador enxerga a imagem)<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>dist\u00e2ncia vertical do objeto, at\u00e9 a superf\u00edcie de separa\u00e7\u00e3o <\/b><\/span><\/span><\/span>dos meios A e B. <\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o do meio onde est\u00e1 o observador <\/b><\/span><\/span><\/span>(no caso, meio A)<\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o do meio onde est\u00e1 o objeto e a imagem<\/b><\/span><\/span><\/span> (no caso, meio B)<\/b><\/span><\/span><\/span><\/p>\n

Um exemplo de aplica\u00e7\u00e3o:\u00a0<\/b><\/span><\/span><\/span><\/p>\n

De acordo com o <\/b><\/span><\/span><\/span>desenho, onde voc\u00ea tem um peixe e uma andorinha se visualizando mutuamente.<\/b><\/span><\/span><\/span> Considerando o <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o absoluto da \u00e1gua como n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = 1,3<\/b><\/span><\/span><\/span> e o <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o absoluto do ar como n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span> = 1,0, pede-se:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>\"\"<\/p>\n

a)<\/b><\/span><\/span><\/span> Determine a <\/b><\/span><\/span><\/span>altura aparente com que o peixe ver\u00e1 a andorinha<\/b><\/span><\/span><\/span> sabendo que <\/b><\/span><\/span><\/span>ela se encontra a 10m da superf\u00edcie livre da \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o<\/b><\/span><\/span><\/span> do meio onde est\u00e1 o <\/b><\/span><\/span><\/span>observador, no caso, o peixe<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>= n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1,3<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>i <\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/sub><\/span>\u00edndice de refra\u00e7\u00e3o<\/b><\/span><\/span><\/span> do meio onde est\u00e1 o<\/b><\/span><\/span><\/span> <\/b> <\/b>objeto e a imagem,<\/b><\/span><\/span><\/span> <\/b>no caso, o ar<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> <\/b>=n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1,0.<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>dist\u00e2ncia vertical do objeto (andorinha), at\u00e9 a superf\u00edcie de separa\u00e7\u00e3o \"\" d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 10 m.<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> \"\" dist\u00e2ncia vertical da imagem da andorinha at\u00e9 a superf\u00edcie de separa\u00e7\u00e3o \"\" d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = ?<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.1,0 = 10.1,3<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 13 m<\/b><\/span><\/span><\/span> (altura aparente da andorinha) altura de sua imagem vista pelo peixe em rela\u00e7\u00e3o \u00e0 superf\u00edcie da \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span> Determine a <\/b><\/span><\/span><\/span>profundidade aparente com que a andorinha ver\u00e1 o peixe <\/b><\/span><\/span><\/span>sabendo que <\/b><\/span><\/span><\/span>ele se encontra a uma profundidade de 2,6 m da superf\u00edcie livre da \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o<\/b><\/span><\/span><\/span> do meio onde est\u00e1 o <\/b><\/span><\/span><\/span>observador, no caso, a andorinha<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>= n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1,0<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>i <\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/sub><\/span>\u00edndice de refra\u00e7\u00e3o<\/b><\/span><\/span><\/span> do meio onde est\u00e1 o<\/b><\/span><\/span><\/span> <\/b> <\/b>objeto e a imagem,<\/b><\/span><\/span><\/span> <\/b>no caso, a \u00e1gua<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> <\/b>= n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1,3.<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>dist\u00e2ncia vertical do objeto (peixe), at\u00e9 a superf\u00edcie de separa\u00e7\u00e3o \"\" d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 2.6 m.<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> \"\" dist\u00e2ncia vertical da imagem do peixe at\u00e9 a superf\u00edcie de separa\u00e7\u00e3o \"\" d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = ?<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.1,3 = 2,6.1,0<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 2,6\/1,3<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 2,0 m<\/b><\/span><\/span><\/span> (profundidade aparente do peixe <\/b><\/span><\/span><\/span>em rela\u00e7\u00e3o \u00e0 superf\u00edcie<\/b><\/span><\/span><\/span>) altura aparente do peixe visto pela andorinha.<\/b><\/span><\/span><\/span><\/p>\n

O que voc\u00ea deve saber, informa\u00e7\u00f5es e dicas<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Em qualquer <\/b><\/span><\/span><\/span>situa\u00e7\u00e3o a imagem est\u00e1 sempre acima do objeto<\/b><\/span><\/span><\/span> e ambos na <\/b><\/span><\/span><\/span>mesma reta vertical.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o da imagem depende do ponto de onde o observador estiver olhando.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Se ele <\/b><\/span><\/span><\/span>estiver numa dire\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span>aproximadamente perpendicular \u00e0 superf\u00edcie da \u00e1gua, a express\u00e3o d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \u00e9 sempre v\u00e1lida.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

\u00c0 medida que ele se afasta, a imagem vai parecendo cada vez mais elevada.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Um <\/b><\/span><\/span><\/span>gato que observa um peixe dentro de um aqu\u00e1rio enxerga-o mais pr\u00f3ximo da parede do aqu\u00e1rio (observador no ar).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ao <\/b><\/span><\/span><\/span>mesmo tempo, o peixe observa o gato mais distante do aqu\u00e1rio (observador na \u00e1gua).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0<\/span>Se voc\u00ea quiser pode resolver o exerc\u00edcio de n\u00famero 10 com a respectiva resolu\u00e7\u00e3o, que \u00e9 sobre esse assunto.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

\"\"<\/p>\n

Exerc\u00edcio interessante:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Confira os exerc\u00edcios com resolu\u00e7\u00f5es comentadas<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Dioptro Plano Um dioptro plano \u00e9 um sistema refrator que apresenta dois meios transparentes e homog\u00eaneos separados por uma superf\u00edcie ou, mais especificamente consiste num conjunto de dois meios opticamente homog\u00eaneos e transparentes separados por uma superf\u00edcie plana como, por exemplo, o ar em contato com a \u00e1gua parada e transparente de uma piscina (dioptro plano \u201car-\u00e1gua\u201d) Localiza\u00e7\u00e3o das imagens<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2137,"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-2208","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2208","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=2208"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2208\/revisions"}],"predecessor-version":[{"id":2216,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2208\/revisions\/2216"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2137"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2208"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}