{"id":2269,"date":"2016-06-20T03:29:01","date_gmt":"2016-06-20T03:29:01","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2269"},"modified":"2024-09-02T14:02:39","modified_gmt":"2024-09-02T14:02:39","slug":"estudo-analitico-das-lentes-esfericas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/estudo-analitico-das-lentes-esfericas\/","title":{"rendered":"Estudo Anal\u00edtico das Lentes Esf\u00e9ricas"},"content":{"rendered":"

Estudo Anal\u00edtico das Lentes Esf\u00e9ricas<\/b><\/span><\/span><\/span><\/p>\n

Introdu\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

As <\/b><\/span><\/span><\/span>lentes possuem dois focos,<\/b><\/span><\/span><\/span> sendo, um o<\/b><\/span><\/span><\/span>foco principal objeto (f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> e o outro o<\/b><\/span><\/span><\/span>foco principal imagem (f<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>). <\/b><\/span><\/span><\/span><\/p>\n

Ambos s\u00e3o sim\u00e9tricos <\/b><\/span><\/span><\/span>e est\u00e3o <\/b><\/span><\/span><\/span>localizados sobre o eixo principal.<\/b><\/span><\/span><\/span><\/p>\n

As <\/b><\/span><\/span><\/span>posi\u00e7\u00f5es e as alturas <\/b><\/span><\/span><\/span>de objetos colocados diante de uma lente esf\u00e9rica <\/b><\/span><\/span><\/span>s\u00e3o determinadas atrav\u00e9s das mesmas equa\u00e7\u00f5es estudadas nos espelhos esf\u00e9ricos.<\/b><\/span><\/span><\/span><\/p>\n

Seus sinais s\u00e3o fornecidos pelo referencial de Gauss.<\/b><\/span><\/span><\/span><\/p>\n

Referencial de Gauss \u2013 Conven\u00e7\u00e3o de sinais<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O <\/b><\/span><\/span><\/span>sistema de eixo ortogonais<\/b><\/span><\/span><\/span>tem <\/b><\/span><\/span><\/span>origem no centro \u00f3ptico da lente.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0A <\/b><\/span><\/span><\/span>origem do referencial <\/b><\/span><\/span><\/span>est\u00e1 no <\/b><\/span><\/span><\/span>centro \u00f3ptico da lente.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0O <\/b><\/span><\/span><\/span>eixo das abscissas coincide com o eixo principal<\/b><\/span><\/span><\/span> e tem <\/b><\/span><\/span><\/span>sentido contr\u00e1rio ao da luz incidente <\/b><\/span><\/span><\/span>para os <\/b><\/span><\/span><\/span>objetos<\/b><\/span><\/span><\/span> e a <\/b><\/span><\/span><\/span>favor da luz incidente para as imagens.<\/b><\/span><\/span><\/span><\/p>\n

Assim, <\/b><\/span><\/span><\/span>observe na figura acima que imagens reais <\/b><\/span><\/span><\/span>(\u00e0 direita da lente) e <\/b><\/span><\/span><\/span>objetos reais <\/b><\/span><\/span><\/span>(\u00e0 esquerda da lente) <\/b><\/span><\/span><\/span>t\u00eam abscissas positivas e que imagens virtuais (\u00e0 esquerda da lente) tem abscissas negativas<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0O <\/b><\/span><\/span><\/span>eixo das ordenadas \u00e9 orientado para cima. <\/b><\/span><\/span><\/span>Assim, a <\/b><\/span><\/span><\/span>imagem<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>e o<\/b><\/span><\/span><\/span>objeto O<\/b><\/span><\/span><\/span> tem <\/b><\/span><\/span><\/span>mesmo sin<\/b><\/span><\/span><\/span>al se a <\/b><\/span><\/span><\/span>imagem for direita em rela\u00e7\u00e3o ao objet<\/b><\/span><\/span><\/span>o e <\/b><\/span><\/span><\/span>sinais opostos se a imagem for invertida em rela\u00e7\u00e3o ao objeto.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que a dist\u00e2ncia focal\u00a0<\/b><\/span><\/span><\/span> f<\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>positiva para lentes convergentes e negativa para lentes divergentes.<\/b><\/span><\/span><\/span><\/p>\n

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

Equa\u00e7\u00e3o de Gauss<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Verg\u00eancia C de uma lente esf\u00e9rica<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\u00a0\"\"\u00a0Observe na <\/b><\/span><\/span><\/span>express\u00e3o C=1\/f que f e C s\u00e3o inversamente proporcionais,<\/b><\/span><\/span><\/span> ou seja, <\/b><\/span><\/span><\/span>lentes de<\/b><\/span><\/span><\/span> <\/b>pequenas dist\u00e2ncias focais,<\/b><\/span><\/span><\/span> que <\/b><\/span><\/span><\/span>provocam maiores altera\u00e7\u00f5es<\/b><\/span><\/span><\/span> nas imagens dos objetos, <\/b><\/span><\/span><\/span>t\u00eam maior verg\u00eancia, na linguagem cotidiana \u201cmaior grau\u201d.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Note nas figuras que a <\/b><\/span><\/span><\/span>verg\u00eancia (grau) da lente da esquerda \u00e9 maior que a da direita.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\" Quest\u00f5es interessantes nas quais voc\u00ea deve analisar atentamente os processos de resolu\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

01-(ITA-SP)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>figura mostra uma barra LM de 10\u221a2cm de comprimento,<\/b><\/span><\/span><\/span> formando um <\/b><\/span><\/span><\/span>\u00e2ngulo de 45\u00b0 com a horizontal,<\/b><\/span><\/span><\/span> tendo o seu <\/b><\/span><\/span><\/span>centro situado a x = 30,0 cm de uma lente divergente,<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>dist\u00e2ncia focal igual a 20,0 cm<\/b><\/span><\/span><\/span>, e a <\/b><\/span><\/span><\/span>y = 10,0 cm acima do<\/b><\/span><\/span><\/span> <\/b>eixo \u00f3tico da mesma.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Determine o comprimento da imagem da barra<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>fa\u00e7a<\/b><\/span><\/span><\/span> um <\/b><\/span><\/span><\/span>desenho esquem\u00e1tico para mostrar a orienta\u00e7\u00e3o da imagem.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

Determina\u00e7\u00e3o gr\u00e1fica da imagem<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0 <\/b><\/span><\/span>utilizando os dois raios not\u00e1veis <\/b><\/span><\/span><\/span>da figura <\/b><\/span><\/span><\/span>localizamos a imagem M\u2019 do ponto M e a imagem L\u2019 do ponto L<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>unindo-os<\/b><\/span><\/span><\/span> achamos a <\/b><\/span><\/span><\/span>imagem L\u2019M\u2019 do objeto LM. <\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>figura mostra as coordenadas X e Y<\/b><\/span><\/span><\/span> dos pontos <\/b><\/span><\/span><\/span>M e L<\/b><\/span><\/span><\/span>\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>M(25;15)<\/b><\/span><\/span><\/span>\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>L(5;35):<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>coordenada P\u2019 das imagens M\u2019 e L\u2019<\/b><\/span><\/span><\/span> pela <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o dos pontos conjugados:<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Coordenada da imagem de M<\/b><\/span><\/span><\/span>\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>1\/f = 1\/P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 1\/ P\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0 \"\"\u00a0 1\/-20 = 1\/25 + 1\/P\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>P\u2019<\/b><\/span><\/span><\/span>M <\/b><\/span><\/span><\/sub><\/span>= –<\/b><\/span><\/span><\/span> <\/b>100\/9 cm = – 11,1 cm (o sinal negativo significa que a imagem \u00e9 virtual) \"\" em m\u00f3dulo P\u2019<\/b><\/span><\/span><\/span>M <\/b><\/span><\/span><\/sub><\/span>=11,1cm \"\"<\/b><\/span><\/span><\/span><\/p>\n

M\u2019<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> = 11,1 cm.<\/b><\/span><\/span><\/span><\/p>\n

Coordenada da imagem de L \"\"<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>1\/f = 1\/P<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 1\/ P\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span>\u00a0 1\/-20 = 1\/35 + 1\/P\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>P\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>= – 140\/11 cm = – 12,7 cm \u00a0\"\" em m\u00f3dulo P\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span> = L\u2019<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> = 12,7 cm. <\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0 <\/span>c\u00e1lculo da coordenada Y\u2019 das imagens M\u2019 e L\u2019<\/b><\/span><\/span><\/span> pela <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o do aumento linear transversal<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0 <\/span>Y\u2019<\/b><\/span><\/span><\/span>L <\/b><\/span><\/span><\/sub><\/span>\/ Y<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>= – P\u2019<\/b><\/span><\/span><\/span>L <\/b><\/span><\/span><\/sub><\/span>\/ P<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0 \"\"\u00a0 Y\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\/5= – ( – 140\/11)\/35\u00a0 \"\"\u00a0 Y\u2019<\/b><\/span><\/span><\/span>L <\/b><\/span><\/span><\/sub><\/span>= 20\/11 cm \"\" <\/b><\/span><\/span><\/span>Y\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span> = 1,8 cm.<\/b><\/span><\/span><\/span><\/p>\n

\"\" Y\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/Y<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>= – P\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0 \"\"\u00a0 Y\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/15 = – ( – 100\/9)\/25\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>Y\u2019<\/b><\/span><\/span><\/span>M <\/b><\/span><\/span><\/sub><\/span>= 20\/3 cm = 6,7 cm.<\/b><\/span><\/span><\/span><\/p>\n

\"\" coordenadas M\u2019<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> e M\u2019<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> do ponto <\/b><\/span><\/span><\/span>M\u2019(11,1;6,7)<\/b><\/span><\/span><\/span><\/p>\n

\"\" coordenadas L\u2019<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> e L\u2019<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> do ponto <\/b><\/span><\/span><\/span>L\u2019(12,7;1,8)<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0 <\/span>Comprimento c da imagem da barra<\/b><\/span><\/span><\/span>\u00a0 \"\"\u00a0 c<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span>= (P\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 P\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (Y\u2019<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 Y\u2019<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 \"\"\u00a0 c<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span>= { – 100\/9 \u2013 ( – 140\/11)}<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (20\/3 \u2013 20\/11)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 \"\"\u00a0 c<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span>\u2248 256.000\/9.800\u00a0 \"\"\u00a0 c<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span>\u2248 26\u00a0 \"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>c \u2248 5,1cm.<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>Um <\/b><\/span><\/span><\/span>objeto luminoso e uma tela de proje\u00e7\u00e3o est\u00e3o separados pela dist\u00e2ncia D = 80 cm.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Existem duas posi\u00e7\u00f5es<\/b><\/span><\/span><\/span> em que uma <\/b><\/span><\/span><\/span>lente convergente de dist\u00e2ncia focal f = 15 cm,<\/b><\/span><\/span><\/span> colocada entre o <\/b><\/span><\/span><\/span>objeto e a tela, produz uma imagem real na tela.<\/b><\/span><\/span><\/span> <\/b>\"\"<\/p>\n

Calcule a <\/b><\/span><\/span><\/span>dist\u00e2ncia, em cm, entre estas duas posi\u00e7\u00f5es.<\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

Observe a <\/b><\/span><\/span><\/span>figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

1\/15 = 1\/S + 1\/ 80 \u2013 S)<\/b><\/span><\/span><\/span>\u00a0 \"\" 1\/15 = (80 \u2013 S) + S \/ S.(80 \u2013 S)\u00a0 \"\"\u00a0 1200 = 80S \u2013 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 80S + 1200 = 0\u00a0 \"\"<\/b><\/span><\/span><\/span>\u00a0 \u0394 = B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>– 4.A.C\u00a0 \"\"\u00a0 <\/b><\/span><\/span><\/span>\u0394 = 40\u00a0 \"\"<\/b><\/span><\/span><\/span> \u00a0S = – B \u00b1\u221a(\u0394)\/2.A\u00a0 \"\"\u00a0 S<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>= (80 + 40)\/2\u00a0 \"\" S<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>= 60 cm e <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>= 80 \u2013 60 <\/b><\/span><\/span><\/span>= 20cm<\/b><\/span><\/span><\/span>\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 S<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= (80 \u2013 40)\/2\u00a0 \"\"\u00a0\u00a0S<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= 20cm\u00a0e\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span> <\/b>= 80 \u2013 20 <\/b><\/span><\/span><\/span>= 60cm.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Portanto, para que a <\/b><\/span><\/span><\/span>imagem seja real e n\u00edtida sobre a tela <\/b><\/span><\/span><\/span>existe duas\u00a0 posi\u00e7\u00f5es, <\/b><\/span><\/span><\/span>uma<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>20cm <\/b><\/span><\/span><\/span>da lente e a <\/b><\/span><\/span><\/span>outra a 40 cm da lente.<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> <\/b>O <\/b><\/span><\/span><\/span>sistema de imagens\u00a0street view<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>dispon\u00edvel na internet permite <\/b><\/span><\/span><\/span>a visualiza\u00e7\u00e3o de v\u00e1rios lugares do mundo atrav\u00e9s de fotografias de alta defini\u00e7\u00e3o, tomadas\u00a0em 360 graus, no n\u00edvel da rua.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\"\"<\/p>\n

Em uma\u00a0<\/b><\/span><\/span><\/span>c\u00e2mera fotogr\u00e1fica tradicional,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>imagem \u00e9 gravada em um\u00a0filme fotogr\u00e1fico\u00a0<\/b><\/span><\/span><\/span>para posterior <\/b><\/span><\/span><\/span>revela\u00e7\u00e3o. <\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o da lente <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>ajustada<\/b><\/span><\/span><\/span> de modo a <\/b><\/span><\/span><\/span>produzir a imagem no filme <\/b><\/span><\/span><\/span>colocado na <\/b><\/span><\/span><\/span>parte posterior da c\u00e2mera.<\/b><\/span><\/span><\/span><\/p>\n

Considere uma <\/b><\/span><\/span><\/span>c\u00e2mera para a qual um\u00a0objeto muito distante <\/b><\/span><\/span><\/span>fornece <\/b><\/span><\/span><\/span><\/p>\n

uma <\/b><\/span><\/span><\/span>imagem pontual <\/b><\/span><\/span><\/span>no filme em uma\u00a0<\/b><\/span><\/span><\/span>posi\u00e7\u00e3o P\u2019 = 5 cm. <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

<\/a> \"\"<\/p>\n

A <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o da posi\u00e7\u00e3o da imagem pedida <\/b><\/span><\/span><\/span>devido \u00e0 <\/b><\/span><\/span><\/span>troca da posi\u00e7\u00e3o do objeto vale \u0394P\u2019 <\/b><\/span><\/span><\/span>= 5,26 \u2013 5,0 \"\" <\/b><\/span><\/span><\/span>\u0394P\u2019= 0,26cm.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Estudo Anal\u00edtico das Lentes Esf\u00e9ricas Introdu\u00e7\u00e3o As lentes possuem dois focos, sendo, um ofoco principal objeto (fo) e o outro ofoco principal imagem (fi). Ambos s\u00e3o sim\u00e9tricos e est\u00e3o localizados sobre o eixo principal. As posi\u00e7\u00f5es e as alturas de objetos colocados diante de uma lente esf\u00e9rica s\u00e3o determinadas atrav\u00e9s das mesmas equa\u00e7\u00f5es estudadas nos espelhos esf\u00e9ricos. Seus sinais s\u00e3o<\/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-2269","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2269","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=2269"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2269\/revisions"}],"predecessor-version":[{"id":2278,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2269\/revisions\/2278"}],"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=2269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}