{"id":2279,"date":"2016-06-20T03:36:00","date_gmt":"2016-06-20T03:36:00","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2279"},"modified":"2024-09-02T14:05:56","modified_gmt":"2024-09-02T14:05:56","slug":"formula-equacao-dos-fabricantes-de-lentes","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/formula-equacao-dos-fabricantes-de-lentes\/","title":{"rendered":"F\u00f3rmula (equa\u00e7\u00e3o) dos Fabricantes de Lentes"},"content":{"rendered":"

F\u00f3rmula (equa\u00e7\u00e3o) dos Fabricantes de Lentes<\/b><\/span><\/span><\/span><\/p>\n

Podemos <\/b><\/span><\/span><\/span>determinar a dist\u00e2ncia focal f e a verg\u00eancia C de uma lente <\/b><\/span><\/span><\/span>conhecendo <\/b><\/span><\/span><\/span>os raios de curvatura de suas faces e os \u00edndices de refra\u00e7\u00e3o da lente e do meio que a envolve<\/b><\/span><\/span><\/span>, atrav\u00e9s da <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o dos fabricantes de lentes:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

Considera\u00e7\u00f5es importantes para a resolu\u00e7\u00e3o de exerc\u00edcios<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

Justaposi\u00e7\u00e3o de lentes<\/b><\/span><\/span><\/span><\/p>\n

Quando <\/b><\/span><\/span><\/span>justapomos duas lentes <\/b><\/span><\/span><\/span>obtemos uma <\/b><\/span><\/span><\/span>lente equivalente <\/b><\/span><\/span><\/span>cuja <\/b><\/span><\/span><\/span>verg\u00eancia ou converg\u00eancia<\/b><\/span><\/span><\/span> <\/b>C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a <\/b><\/span><\/span><\/span>soma alg\u00e9brica da verg\u00eancia de cada uma das lentes<\/b><\/span><\/span><\/span>, ou seja, <\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span>eq <\/b><\/span><\/span><\/sub><\/span>= C<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>+ C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Lembre-se que <\/b><\/span><\/span><\/span>C = 1 \/ f<\/b><\/span><\/span><\/span> e que se a <\/b><\/span><\/span><\/span>lente \u00e9 divergente f e C s\u00e3o negativos <\/b><\/span><\/span><\/span>e se a <\/b><\/span><\/span><\/span>lente \u00e9 convergente, positivos.<\/b><\/span><\/span><\/span><\/p>\n

Exemplo num\u00e9rico:<\/b><\/span><\/span><\/span><\/p>\n

Uma <\/b><\/span><\/span><\/span>lente convergente de 12di <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>justaposta a uma lente divergente de 20di.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>associa\u00e7\u00e3o funciona como uma \u00fanica lente convergente ou divergente?<\/b><\/span><\/span><\/span> Determine sua <\/b><\/span><\/span><\/span>verg\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

Lente convergente \"\"<\/b><\/span><\/span><\/span> <\/b>C<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>= +12di.<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

Lente divergente \"\" C<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= – 20di.<\/b><\/span><\/span><\/span><\/p>\n

C<\/b><\/span><\/span><\/span>eq <\/b><\/span><\/span><\/sub><\/span>= + 12di \u2013 20di<\/b><\/span><\/span><\/span>\u00a0 \"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span>eq <\/b><\/span><\/span><\/sub><\/span>= – 8di.<\/b><\/span><\/span><\/span><\/p>\n

Essa associa\u00e7\u00e3o <\/b><\/span><\/span><\/span>funciona<\/b><\/span><\/span><\/span> como uma \u00fanica lente\u00a0divergente,<\/b><\/span><\/span><\/span> pois <\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> \u00e9 negativa. <\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"\u00a0Quando <\/b><\/span><\/span><\/span>uma lente quebra<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>dist\u00e2ncia focal e consequentemente a verg\u00eancia de cada peda\u00e7o n\u00e3o se altera, pois baseado na equa\u00e7\u00e3o dos fabricantes, as superf\u00edcies externas continuam com os mesmos raios e a lente continua com mesmo \u00edndice de refra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

\"\" Observe na <\/b><\/span><\/span><\/span>express\u00e3o C = 1\/f = (n<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 1).(1\/R<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>+<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>1\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>onde <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 o \u00edndice de refra\u00e7\u00e3o da lente e n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0o \u00edndice de refra\u00e7\u00e3o do meio que a envolve, que:<\/b><\/span><\/span><\/span><\/p>\n

\"\" se a lente estiver <\/b><\/span><\/span><\/span>imersa no ar (n<\/b><\/span><\/span><\/span>m <\/b><\/span><\/span><\/sub><\/span>= 1), <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>express\u00e3o C =1\/f = (n<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 1).(1\/R<\/b><\/span><\/span><\/span>1 +\u00a0<\/b><\/span><\/span><\/sub><\/span>1\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0 <\/b><\/span><\/span><\/span>fica <\/b><\/span><\/span><\/span>C = 1\/f = (n<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 1).(1\/R<\/b><\/span><\/span><\/span>1 +\u00a0<\/b><\/span><\/span><\/sub><\/span>1\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Assim, se a <\/b><\/span><\/span><\/span>lente possuir os mesmos raios, f \u00e9 inversamente proporcional a (n<\/b><\/span><\/span><\/span>l <\/b><\/span><\/span><\/sub><\/span>– 1),<\/b><\/span><\/span><\/span> ou seja, <\/b><\/span><\/span><\/span>quanto maior for o \u00edndice de refra\u00e7\u00e3o da lente, menor ser\u00e1 sua dist\u00e2ncia focal.<\/b><\/span><\/span><\/span><\/p>\n

\"\" como <\/b><\/span><\/span><\/span>(n<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 1) <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>inversamente proporcional a f<\/b><\/span><\/span><\/span>, quanto <\/b><\/span><\/span><\/span>maior for a raz\u00e3o n<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>, menor ser\u00e1 a dist\u00e2ncia focal da lente.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\" Analise as resolu\u00e7\u00f5es dessas quest\u00f5es interessantes:<\/b><\/span><\/span><\/span><\/p>\n

01-(ITA)<\/b><\/span><\/span><\/span> <\/b>Uma <\/b><\/span><\/span><\/span>lente convergente biconvexa tem dist\u00e2ncia focal de 20cm <\/b><\/span><\/span><\/span>quando est\u00e1 <\/b><\/span><\/span><\/span>mergulhada em<\/b><\/span><\/span><\/span>\"\"<\/p>\n

ar.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

A <\/b><\/span><\/span><\/span>lente \u00e9 feita de vidro, <\/b><\/span><\/span><\/span>cujo <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o \u00e9 n<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> = 1,6.<\/b><\/span><\/span><\/span> Se a <\/b><\/span><\/span><\/span>lente \u00e9 mergulhada em um meio, menos refringente do que o material da lente<\/b><\/span><\/span><\/span>, cujo <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o \u00e9 n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span> considere as <\/b><\/span><\/span><\/span>seguintes afirma\u00e7\u00f5es e analise-as como verdadeiras ou falsas:<\/b><\/span><\/span><\/span><\/p>\n

I.<\/b><\/span><\/span><\/span> <\/b>A <\/b><\/span><\/span><\/span>dist\u00e2ncia focal n\u00e3o varia <\/b><\/span><\/span><\/span>se o <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o do meio for igual ao do material da lente.<\/b><\/span><\/span><\/span><\/p>\n

II.<\/b><\/span><\/span><\/span> <\/b>A <\/b><\/span><\/span><\/span>dist\u00e2ncia focal torna-se maior <\/b><\/span><\/span><\/span>se o <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o n<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> for maior que o do ar.<\/b><\/span><\/span><\/span><\/p>\n

III.<\/b><\/span><\/span><\/span> Neste exemplo, <\/b><\/span><\/span><\/span>uma <\/b><\/span><\/span><\/span>maior diferen\u00e7a entre os \u00edndices de refra\u00e7\u00e3o do material da lente e do meio implica numa menor dist\u00e2ncia focal.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

III-<\/b><\/span><\/span><\/span>\u00a0Verdadeira, veja justificativa anterior.<\/b><\/span><\/span><\/span><\/p>\n

02-(UFC-CE)<\/b><\/span><\/span><\/span> <\/b>Uma <\/b><\/span><\/span><\/span>lente esf\u00e9rica delgada,<\/b><\/span><\/span><\/span> constitu\u00edda de <\/b><\/span><\/span><\/span>material<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o n,<\/b><\/span><\/span><\/span> est\u00e1 <\/b><\/span><\/span><\/span>imersa no ar (n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=1,00). <\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>lente<\/b><\/span><\/span><\/span> tem <\/b><\/span><\/span><\/span>dist\u00e2ncia focal f <\/b><\/span><\/span><\/span>e suas <\/b><\/span><\/span><\/span>superf\u00edcies esf\u00e9ricas tem raios de curvatura R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Esses par\u00e2metros <\/b><\/span><\/span><\/span>obedecem a uma rela\u00e7\u00e3o<\/b><\/span><\/span><\/span>, conhecida como <\/b><\/span><\/span><\/span>\u201cequa\u00e7\u00e3o dos fabricantes\u201d, mostrada abaixo.<\/b><\/span><\/span><\/span><\/p>\n

\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\u00a01\/f=(n-1).(1\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>+1\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

Suponha uma <\/b><\/span><\/span><\/span>lente biconvexa de raios de curvatura iguais (R<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>= R<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= R), dist\u00e2ncia focal f <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o n = 1,8 (figura 1).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Essa <\/b><\/span><\/span><\/span>lente \u00e9 partida <\/b><\/span><\/span><\/span>dando origem a <\/b><\/span><\/span><\/span>duas lentes plano-convexas iguais (figura 2).<\/b><\/span><\/span><\/span><\/p>\n

Calcule o <\/b><\/span><\/span><\/span>valor da dist\u00e2ncia focal <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>cada uma das novas lentes. <\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes de a <\/b><\/span><\/span><\/span>lente ser partida, <\/b><\/span><\/span><\/span>ela <\/b><\/span><\/span><\/span>possui duas faces convexas <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>raio R <\/b><\/span><\/span><\/span>tal que, <\/b><\/span><\/span><\/span>pela equa\u00e7\u00e3o dos fabricantes fornece o valor do raio R em fun\u00e7\u00e3o da dist\u00e2ncia focal f \"\" 1\/f = (n \u2013 1).(1\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + 1\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>) \"\"<\/b><\/span><\/span><\/span>\u00a0 1\/f=(1,8-1).(1\/R+1\/R)\u00a0 \"\"\u00a0 1\/f=(0,8).2\/R\u00a0 \"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>R=1,6f\u00a0(I)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Confira os exerc\u00edcios com resolu\u00e7\u00e3o comentada<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

F\u00f3rmula (equa\u00e7\u00e3o) dos Fabricantes de Lentes Podemos determinar a dist\u00e2ncia focal f e a verg\u00eancia C de uma lente conhecendo os raios de curvatura de suas faces e os \u00edndices de refra\u00e7\u00e3o da lente e do meio que a envolve, atrav\u00e9s da equa\u00e7\u00e3o dos fabricantes de lentes: \u00a0\u00a0\u00a0\u00a0 Considera\u00e7\u00f5es importantes para a resolu\u00e7\u00e3o de exerc\u00edcios Recordando: \u00a0 Justaposi\u00e7\u00e3o de lentes<\/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-2279","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2279","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=2279"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2279\/revisions"}],"predecessor-version":[{"id":2288,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2279\/revisions\/2288"}],"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=2279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}