{"id":3433,"date":"2017-11-13T22:51:07","date_gmt":"2017-11-13T22:51:07","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3433"},"modified":"2024-09-04T15:03:31","modified_gmt":"2024-09-04T15:03:31","slug":"resolucoes-das-questoes-de-vestibulares-recentes-por-assunto-2017-2016-2015-optica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/optica\/optica-vestibulares-recentes-por-assunto-2017-2016-2015\/resolucoes-das-questoes-de-vestibulares-recentes-por-assunto-2017-2016-2015-optica\/","title":{"rendered":"Resolu\u00e7\u00f5es das quest\u00f5es de Vestibulares recentes por assunto \u2013 2017 \u2013 2016 \u2013 2015 \u2013 \u00d3ptica"},"content":{"rendered":"

Resolu\u00e7\u00f5es das quest\u00f5es de Vestibulares recentes por assunto \u2013 <\/b><\/span><\/span><\/span>2017 \u2013 2016 \u2013 2015 \u2013 \u00d3ptica<\/b><\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span>velocidade das nanosondas corresponde a 20% da velocidade da luz<\/b><\/span><\/span>e ter\u00e1\u00a0<\/b><\/span><\/span>valor<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 20% de 3.10<\/b><\/span><\/span>8<\/b><\/span><\/sup><\/span>\u00a0m\/s\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 0,2\u00d73.10<\/b><\/span><\/span>8<\/b><\/span><\/sup><\/span>\u00a0= 0,6.10<\/b><\/span><\/span>8<\/b><\/span><\/sup><\/span>\u00a0m\/s\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 6.10<\/b><\/span><\/span>7<\/b><\/span><\/sup><\/span>\u00a0m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Um\u00a0<\/b><\/span><\/span>ano-luz \u00e9 a dist\u00e2ncia que a luz<\/b><\/span><\/span>\u00a0pode viajar em\u00a0<\/b><\/span><\/span>um ano com velocidade de 3.108\u00a0m\/s.<\/b><\/span><\/span>\u00a0Por exemplo,\u00a0em\u00a0<\/b><\/span><\/span>um ano-luz, a dist\u00e2ncia percorrida\u00a0<\/b><\/span><\/span>seria de <\/b><\/span><\/span>aproximadamente 9.460.800.000.000 (nove\u00a0trilh\u00f5es, quatrocentos e sessenta\u00a0bilh\u00f5es\u00a0e oitocentos milh\u00f5es) de\u00a0quil\u00f4metros, ou, aproximadamente 9.1015\u00a0m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

V =\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206S = 4,4\u00d79.1015\u00a0= 39,6.1015\u00a0m\u00a0\"\"\u00a06.107\u00a0=\u00a0\"\"\u00a0\"\"\u00a0\u2206t =\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t = 6,6.108\u00a0s<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ano em segundos\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>1 minuto =\u00a0<\/b><\/span><\/span>60 segundos\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>1 hora\u00a0<\/b><\/span><\/span>= 60 x 60\u00a0<\/b><\/span><\/span>= 3600 segundos\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>1 dia\u00a0<\/b><\/span><\/span>= 24h x 3600\u00a0<\/b><\/span><\/span>= 86.400 segundos\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>365 dias = 31.536.000 segundos\u00a0\"\"\u00a01 ano tem aproximadamente 3,15.107\u00a0s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u2206<\/span>t =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>21 anos<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

02- <\/span><\/span><\/span><\/span><\/span><\/p>\n

N\u00f3s conseguimos resolver esse problema por semelhan\u00e7a de tri\u00e2ngulos, observe a figura:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Os\u00a0<\/b><\/span><\/span>dois tri\u00e2ngulos formados nessa figura s\u00e3o semelhantes<\/b><\/span><\/span>, sendo assim:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span>x \u00e9 a altura da imagem, que ser\u00e1 formada no fundo da c\u00e2mera.<\/b><\/span><\/span>\u00a0Isolando ele:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Substituindo, o\u00a0<\/b><\/span><\/span>L est\u00e1 em cm (8 cm = 0,08 m)\u00a0<\/b><\/span><\/span>apenas dividir por 100<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

x = 0,04 m = 4 cm<\/b><\/span><\/span><\/span><\/span><\/p>\n

Alternativa B<\/b><\/span><\/span><\/span><\/span><\/p>\n

03- <\/span><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir:<\/b><\/span><\/span><\/p>\n

Cor de um corpo<\/b><\/span><\/span><\/p>\n

\u00a0<\/span>A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>luz branca do Sol ou de uma l\u00e2mpada\u00a0<\/b><\/span><\/span>qualquer \u00e9 denominada\u00a0<\/b><\/span><\/span>luz policrom\u00e1tica (v\u00e1rias cores)\u00a0<\/b><\/span><\/span>e \u00e9 composta das\u00a0<\/b><\/span><\/span>cores monocrom\u00e1ticas (uma s\u00f3 cor), vermelho, alaranjado, amarelo, verde, azul, anil e violeta.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>cor apresentada por um corpo,\u00a0<\/b><\/span><\/span>ao ser iluminado,\u00a0<\/b><\/span><\/span>depende do tipo de luz que ele reflete difusamente<\/b><\/span><\/span>\u00a0(espalha em todas as dire\u00e7\u00f5es e sentidos) e\u00a0<\/b><\/span><\/span>que chega aos olhos do observador.<\/b><\/span><\/span><\/p>\n

Um\u00a0<\/b><\/span><\/span>corpo negro absorve todas as cores<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>e um\u00a0<\/b><\/span><\/span>corpo branco reflete todas as cores.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span>Figura 1<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0O\u00a0<\/b><\/span><\/span>corpo vermelho\u00a0<\/b><\/span><\/span>reflete difusamente\u00a0<\/b><\/span><\/span>apenas a cor vermelha\u00a0<\/b><\/span><\/span>e o\u00a0<\/b><\/span><\/span>observador enxergar\u00e1 vermelho.<\/b><\/span><\/span><\/p>\n

Figura 2\u00a0\"\"<\/b><\/span><\/span>\u00a0As\u00a0<\/b><\/span><\/span>sete cores (policrom\u00e1tica branca)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>incidem<\/b><\/span><\/span>\u00a0sobre o\u00a0<\/b><\/span><\/span>corpo amarelo que reflete difusamente somente o amarelo<\/b><\/span><\/span>\u00a0e o\u00a0<\/b><\/span><\/span>observador enxergar\u00e1 essa cor.<\/b><\/span><\/span><\/p>\n

Figura 3\u00a0\"\"<\/b><\/span><\/span>\u00a0 As\u00a0<\/b><\/span><\/span>sete cores (policrom\u00e1tica branca) incidem<\/b><\/span><\/span>\u00a0sobre o\u00a0<\/b><\/span><\/span>corpo verde que reflete difusamente somente o verde\u00a0<\/b><\/span><\/span>e o observador\u00a0<\/b><\/span><\/span>enxergar\u00e1 verde.<\/b><\/span><\/span><\/p>\n

Figura 4\u00a0\"\"\u00a0O corpo negro absorve todas as cores<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>nenhuma chegar\u00e1 ao observador\u00a0<\/b><\/span><\/span>que <\/b><\/span><\/span>ver\u00e1 negro (aus\u00eancia de cores).<\/b><\/span><\/span><\/p>\n

Figura 5\u00a0\"\"<\/b><\/span><\/span>\u00a0O\u00a0<\/b><\/span><\/span>corpo violeta reflete difusamente somente o violeta\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>absorver\u00e1 o verde,\u00a0<\/b><\/span><\/span>n\u00e3o <\/b><\/span><\/span>chegando nenhuma luz ao observador, que ver\u00e1 negro.<\/b><\/span><\/span><\/p>\n

Figura 6\u00a0\"\"<\/b><\/span><\/span>\u00a0O\u00a0<\/b><\/span><\/span>observador ver\u00e1 vermelho<\/b><\/span><\/span>, pois o\u00a0<\/b><\/span><\/span>corpo branco reflete difusamente todas as cores inclusive o vermelho.<\/b><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span>caso do exerc\u00edcio<\/b><\/span><\/span>, as\u00a0<\/b><\/span><\/span>palavras\u00a0<\/b><\/span><\/span>que voc\u00ea enxergar\u00e1\u00a0<\/b><\/span><\/span>em azul\u00a0<\/b><\/span><\/span>s\u00e3o\u00a0<\/b><\/span><\/span>PRETO, VERDE e VERMELHO\u00a0<\/b><\/span><\/span>e as cores ser\u00e3o\u00a0 <\/b><\/span><\/span>azul\u00a0<\/b><\/span><\/span>(as tr\u00eas palavras anteriores) e <\/b><\/span><\/span>preta<\/b><\/span><\/span>\u00a0(as\u00a0<\/b><\/span><\/span>demais radia\u00e7\u00f5es\u00a0<\/b><\/span><\/span>ser\u00e3o absorvidas fornecendo a\u00a0<\/b><\/span><\/span>sensa\u00e7\u00e3o visual de preto<\/b><\/span><\/span>).<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

04- <\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas da imagem num espelho plano<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span>ponto objeto (O)\u00a0<\/b><\/span><\/span>e o\u00a0<\/b><\/span><\/span>ponto imagem (i<\/b><\/span><\/span>)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>s\u00e3o\u00a0<\/b><\/span><\/span>sim\u00e9tricos em rela\u00e7\u00e3o ao espelho,\u00a0<\/b><\/span><\/span>ou seja,\u00a0<\/b><\/span><\/span>a dist\u00e2ncia\u00a0<\/b><\/span><\/span>do<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

dist\u00e2ncia\u00a0<\/b><\/span><\/span>do<\/b><\/span><\/span> <\/b>objeto ao espelho \u00e9 a mesma que a dist\u00e2ncia da imagem ao espelho\u00a0e contidos numa mesma reta perpendicular ao plano do espelho.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

05- <\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

06- <\/span><\/span><\/span><\/span><\/span><\/p>\n

a)\u00a0<\/b><\/span><\/span>O\u00a0<\/b><\/span><\/span>foco F\u00a0<\/b><\/span><\/span>est\u00e1 localizado na\u00a0<\/b><\/span><\/span>dist\u00e2ncia m\u00e9dia entre o centro de curvatura C e o espelho.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f\u00a0<\/b><\/span><\/span>= R\/2 = 10\/2\u00a0<\/b><\/span><\/span>= 5 cm\u00a0<\/b><\/span><\/span>(positivo, espelho c\u00f4ncavo).<\/b><\/span><\/span><\/span><\/span><\/p>\n

P = 3 cm<\/b><\/span><\/span><\/span><\/span><\/p>\n

P\u2019 = ?<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o dos pontos conjugados\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0=\u00a0\"\"\u00a0<\/b><\/span><\/span>+<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>=<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>+<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"–<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>=<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>=<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>P\u2019 = \u2013 7,5 cm\u00a0<\/b><\/span><\/span>(P\u2019 < 0, imagem virtual)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

i=<\/b><\/span><\/span>\u00a0?<\/b><\/span><\/span><\/span><\/span><\/p>\n

o =\u00a0<\/b><\/span><\/span>– 2cm (invertida)<\/b><\/span><\/span><\/span><\/span><\/p>\n

P =<\/b><\/span><\/span>\u00a03 cm<\/b><\/span><\/span><\/span><\/span><\/p>\n

P\u2019 =<\/b><\/span><\/span>\u00a0\u2013 7,5 cm<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>=<\/b><\/span><\/span>\u00a0\u2013\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0= \u2013\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>i =<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>= 5 cm\u00a0<\/b><\/span><\/span>( a imagem \u00e9 direita em rela\u00e7\u00e3o ao objeto e tem 5cm de altura).<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

07- <\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

08- <\/span><\/span><\/span>Pelas caracter\u00edsticas dos espelhos, podemos excluir algumas alternativas.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Por mais que um\u00a0<\/b><\/span><\/span>espelho plano\u00a0<\/b><\/span><\/span>ajudasse, ele\u00a0<\/b><\/span><\/span>n\u00e3o seria capaz de modificar a imagem<\/b><\/span><\/span>, sendo assim\u00a0<\/b><\/span><\/span>n\u00e3o aumentaria seu campo de vis\u00e3o<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Um\u00a0<\/b><\/span><\/span>espelho c\u00f4ncavo converge\u00a0<\/b><\/span><\/span>os raios para um ponto, com isso\u00a0<\/b><\/span><\/span>n\u00e3o poderia ampliar o campo de vis\u00e3o<\/b><\/span><\/span>\u00a0do maquinista. Para um espelho ser considerado\u00a0<\/b><\/span><\/span>divergente\u00a0<\/b><\/span><\/span>ele precisa ter sua\u00a0<\/b><\/span><\/span>verg\u00eancia negativa<\/b><\/span><\/span>, quanto <\/b><\/span><\/span>maior esse valor, em m\u00f3dulo, maior o campo de vis\u00e3o gerado<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sendo assim<\/b><\/span><\/span>, a alternativa B \u00e9 a correta<\/b><\/span><\/span>. Um espelho\u00a0<\/b><\/span><\/span>convexo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>aumenta o campo de vis\u00e3o<\/b><\/span><\/span>, e se a\u00a0<\/b><\/span><\/span>verg\u00eancia for grande mais expandido \u00e9 esse campo visual<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

09- <\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span>comportamento do bal\u00e3o de Natal<\/b><\/span><\/span>\u00a0como espelho\u00a0<\/b><\/span><\/span>\u00e9 id\u00eantico ao fornecido por um espelho esf\u00e9rico<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>convexo,<\/b><\/span><\/span>\u00a0com as\u00a0<\/b><\/span><\/span>seguintes caracter\u00edsticas<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

Constru\u00e7\u00e3o geom\u00e9trica de imagens nos espelhos esf\u00e9ricos convexos<\/b><\/span><\/span><\/p>\n

Espelho convexo<\/b><\/span><\/span>\u00a0\"\"\u00a0Para\u00a0<\/b><\/span><\/span>qualquer localiza\u00e7\u00e3o do objeto Independentemente\u00a0<\/b><\/span><\/span>de sua posi\u00e7\u00e3o, a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

imagem ter\u00e1 sempre:<\/b><\/span><\/span><\/p>\n

Natureza\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>virtual<\/b><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>atr\u00e1s do espelho e entre V e F<\/b><\/span><\/span>\u00a0e observe nas\u00a0figuras acima\u00a0que\u00a0\u00e0\u00a0<\/b><\/span><\/span>medida que o objeto se aproxima do espelho, a imagem tamb\u00e9m se aproxima e aumenta de tamanho, mas est\u00e1 sempre entre V e F.<\/b><\/span><\/span><\/p>\n

Tamanho<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>menor que o do objeto<\/b><\/span><\/span><\/p>\n

Orienta\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span>\u00a0direita em rela\u00e7\u00e3o ao objeto<\/b><\/span><\/span><\/p>\n

Utilidades:<\/b><\/span><\/span>\u00a0Os\u00a0<\/b><\/span><\/span>espelhos convexos<\/b><\/span><\/span>\u00a0s\u00e3o empregados como\u00a0<\/b><\/span><\/span>retrovisores em ve\u00edculos, cabines de seguran\u00e7a, elevadores, etc.<\/b><\/span><\/span><\/p>\n

Sua\u00a0<\/b><\/span><\/span>vantagem\u00a0<\/b><\/span><\/span>sobre o espelho plano, nesse particular, \u00e9 ter\u00a0maior campo visual.\u00a0T\u00eam, entretanto, o\u00a0<\/b><\/span><\/span>inconveniente<\/b><\/span><\/span>\u00a0de n\u00e3o darem no\u00e7\u00e3o da dist\u00e2ncia.<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

10- <\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

11- <\/span><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Na figura 1 a calota esf\u00e9rica est\u00e1 se comportando como um espelho esf\u00e9rico c\u00f4ncavo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

M\u00f3dulo do espelho c\u00f4ncavo = m\u00f3dulo do espelho convexo = f = 30 cm.<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) Na figura 2<\/b><\/span><\/span>\u00a0a calota esf\u00e9rica se comporta como um\u00a0<\/b><\/span><\/span>espelho esf\u00e9rico convexo\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>dist\u00e2ncia focal<\/b><\/span><\/span><\/span><\/span><\/p>\n

f = \u2013 30 cm\u00a0<\/b><\/span><\/span>(espelhos convexos possuem f < 0)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

12- <\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Podemos\u00a0<\/b><\/span><\/span>excluir<\/b><\/span><\/span>\u00a0de imediato\u00a0<\/b><\/span><\/span>o espelho plano<\/b><\/span><\/span>, pois nele\u00a0<\/b><\/span><\/span>n\u00e3o h\u00e1 como se modificar a imagem<\/b><\/span><\/span>. Uma imagem formada por um\u00a0<\/b><\/span><\/span>espelho convexo <\/b><\/span><\/span>deve ser,\u00a0<\/b><\/span><\/span>obrigatoriamente,\u00a0<\/b><\/span><\/span>virtual e menor que o objeto<\/b><\/span><\/span>. Por a imagem ser\u00a0<\/b><\/span><\/span>expandida dessa forma<\/b><\/span><\/span>, s\u00f3 podemos trabalhar com um\u00a0<\/b><\/span><\/span>espelho c\u00f4ncavo\u00a0<\/b><\/span><\/span>de modo que a\u00a0<\/b><\/span><\/span>imagem seja virtual<\/b><\/span><\/span>, conforme esquema abaixo.<\/b><\/span><\/span><\/p>\n

Objeto entre o foco F e o v\u00e9rtice V ou entre o foco F e o espelho<\/b><\/span><\/span>\"\"<\/p>\n

A imagem ter\u00e1:<\/b><\/span><\/span><\/p>\n

Natureza\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>virtual (obtida na interse\u00e7\u00e3o do prolongamento dos raios de luz )\u00a0\"\"\u00a0n\u00e3o pode ser projetada, fotografada, etc., atr\u00e1s do espelho.<\/b><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>atr\u00e1s do espelho<\/b><\/span><\/span><\/p>\n

Tamanho\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>maior que o do objeto<\/b><\/span><\/span><\/p>\n

Orienta\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>direita em rela\u00e7\u00e3o ao objeto<\/b><\/span><\/span><\/p>\n

Utilidades:<\/b><\/span><\/span>\u00a0s\u00e3o\u00a0<\/b><\/span><\/span>empregados com frequ\u00eancia\u00a0<\/b><\/span><\/span>quando se deseja obter uma\u00a0<\/b><\/span><\/span>imagem virtual e ampliada de um objeto, como \u00e9 o caso dos espelhos de barbear, toalete, de dentista, espelho de otorrinolaringologia, etc.<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

13- \"\"<\/span><\/span><\/span><\/p>\n

Como o\u00a0<\/b><\/span><\/span>enunciado afirma\u00a0<\/b><\/span><\/span>que a\u00a0<\/b><\/span><\/span>imagem conjugada\u00a0<\/b><\/span><\/span>est\u00e1 situada<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>no\u00a0<\/b><\/span><\/span>plano focal (f) do espelho\u00a0\"\"<\/b><\/span><\/span><\/p>\n

a\u00a0<\/b><\/span><\/span>dist\u00e2ncia da imagem ao espelho\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>\u2019<\/b><\/span><\/span>\u00a0\u00e9 a\u00a0<\/b><\/span><\/span>pr\u00f3pria dist\u00e2ncia focal\u00a0<\/b><\/span><\/span>f<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>do espelho\u00a0\"\"\u00a0<\/b><\/span><\/span>P\u2019 = f = 0,2 m.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a0i = \u2013\u00a0\"\"\u00a0\"\"\u00a0i = \u2013 0,04m\u00a0\"\"\u00a0<\/b><\/span><\/span>i = \u2013 4 cm (a imagem tem\u00a0<\/b><\/span><\/span>altura de 4 cm<\/b><\/span><\/span>\u00a0e \u00e9 invertida em rela\u00e7\u00e3o ao objeto)<\/b><\/span><\/span><\/p>\n

14- <\/span><\/span><\/span>Veja abaixo\u00a0<\/b><\/span><\/span>a representa\u00e7\u00e3o do raio de luz no plano cartesiano:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

15- <\/span><\/span><\/span>Vamos pensar em todas as alternativas:<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso de A\u00a0<\/b><\/span><\/span>h\u00e1 uma refra\u00e7\u00e3o<\/b><\/span><\/span>, pois\u00a0<\/b><\/span><\/span>a luz passou de um ambiente para outro<\/b><\/span><\/span>, o que\u00a0<\/b><\/span><\/span>n\u00e3o muda \u00e9 o \u00e2ngulo de refra\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Em B o \u00e2ngulo \u00e9 de 0\u00b0,\u00a0<\/b><\/span><\/span>pois est\u00e1\u00a0<\/b><\/span><\/span>localizado em cima do eixo normal.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Na alternativa C, a frequ\u00eancia n\u00e3o muda<\/b><\/span><\/span>\u00a0quando se troca de ambiente,\u00a0<\/b><\/span><\/span>a velocidade se altera<\/b><\/span><\/span>. Sendo assim,\u00a0<\/b><\/span><\/span>a alternativa D \u00e9 a correta<\/b><\/span><\/span>, pois q<\/b><\/span><\/span>uando se modifica a velocidade de propaga\u00e7\u00e3o o comprimento de onda tamb\u00e9m se altera,<\/b><\/span><\/span>\u00a0basta analisar a f\u00f3rmula abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

V \u00e9 a velocidade de propaga\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u00e9 o comprimento de onda<\/b><\/span><\/span><\/span><\/span><\/p>\n

f \u00e9 a frequ\u00eancia<\/b><\/span><\/span><\/span><\/span><\/p>\n

16-<\/span><\/span><\/span>
\nLei de Snell-Descartes (refra\u00e7\u00e3o da luz)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

nar.sen45o\u00a0= n\u00e1gua.sen<\/b><\/span><\/span>r\u00a0\"\"\u00a01.\"\"\/2 = 1,5.senr\u00a0\"\"\u00a0<\/b><\/span><\/span>senr\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"0,4<\/b><\/span><\/span>7\u00a0\"\"\u00a0pelos\u00a0<\/b><\/span><\/span>valores <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecidos<\/b><\/span><\/span> no enunciado voc\u00ea observa que\u00a0\"\"<\/b><\/span><\/span>\u00a0<\u00a0\"\"\u00a0, ou seja, menor que 30o.<\/b><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

17- <\/b><\/span><\/span><\/span><\/p>\n

Vamos usar a imagem a seguir como apoio:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Nela podemos observar que\u00a0<\/b><\/span><\/span><\/span>quanto mais refringente<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>(quanto maior o \u00edndice de refra\u00e7\u00e3o),<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>mais o raio refratado se aproxima da reta normal\u00a0<\/b><\/span><\/span><\/span>(reta que usamos de refer\u00eancia), no caso do exerc\u00edcio, a linha pontilhada. Ao observarmos o exerc\u00edcio, podemos concluir que\u00a0<\/b><\/span><\/span>o meio A tem que ser o lant\u00e2nio, pois s\u00f3 assim h\u00e1 uma aproxima\u00e7\u00e3o em um caso e um distanciamento no outro<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span>j\u00e1 que ele \u00e9 o valor intermedi\u00e1rio. Ainda sobra o problema de B e C, que podemos concluir observando a imagem de apoio, que\u00a0<\/b><\/span><\/span><\/span>em B se trata de diamante (h\u00e1 uma aproxima\u00e7\u00e3o \u00e0 reta normal)\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span>C se trata de vidro crown (h\u00e1 um afastamento \u00e0 reta normal).<\/b><\/span><\/span><\/span><\/p>\n

R \u2013 E<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

18- <\/b><\/span><\/span><\/span><\/p>\n

a)\u00a0<\/b><\/span><\/span>O\u00a0<\/b><\/span><\/span>\u00edndice de refra\u00e7\u00e3o absoluto<\/b><\/span><\/span>\u00a0aplicado ao\u00a0<\/b><\/span><\/span>material\u00a0<\/b><\/span><\/span>do\u00a0<\/b><\/span><\/span>cilindro <\/b><\/span><\/span>constitu\u00eddo de um\u00a0<\/b><\/span><\/span>meio homog\u00eaneo e transparente \u00e9\u00a0<\/b><\/span><\/span>fornecido por:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

n = 1,6 e c = 3.10<\/b><\/span><\/span>8<\/b><\/span>\u00a0m\/s\u00a0\"\"\u00a0n =<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>1,6 =<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 1,875 m\/s<\/b><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Aplicando a\u00a0<\/b><\/span><\/span>lei de Snell-Descartes<\/b><\/span><\/span>\u00a0na refra\u00e7\u00e3o do feixe de luz em A,\u00a0<\/b><\/span><\/span>do ar para o cilindro\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>n<\/b><\/span><\/span>ar<\/b><\/span>.seni = n<\/b><\/span><\/span>.<\/b><\/span>senr<\/b><\/span><\/span>\u00a0\"\"\u00a01.sen53<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>\u00a0= 1,6.senr\u00a0\"\"\u00a01.0,8 = 1,6.senr\u00a0\"\"\u00a0senr =\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 0,5\u00a0\"\"\u00a0<\/b><\/span><\/span>r = 30<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Se o<\/b><\/span><\/span>\u00a0\u00e2ngulo de incid\u00eancia\u00a0<\/b><\/span><\/span>(com a normal N)<\/b><\/span><\/span>\u00a0em A \u00e9 de 53<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>\u00e2ngulo de emerg\u00eancia em B\u00a0<\/b><\/span><\/span>(com a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>normal N)<\/b><\/span><\/span>\u00a0tamb\u00e9m ser\u00e1 de 53<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>,\u00a0<\/b><\/span><\/span>pois os meios <\/b><\/span><\/span>s\u00e3o os mesmos.<\/b><\/span><\/span><\/p>\n

Como o tri\u00e2ngulo AOB \u00e9 is\u00f3sceles (dois lados iguais R) ent\u00e3o seus \u00e2ngulos opostos s\u00e3o iguais (30<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>).<\/b><\/span><\/span><\/p>\n

No<\/b><\/span><\/span>\u00a0tri\u00e2ngulo ABC o \u00e2ngulo\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span>\u00a0\u00e9 externo\u00a0<\/b><\/span><\/span>o que implica que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u03b1\u00a0<\/b><\/span><\/span>= 2\u00d723\u00a0<\/b><\/span><\/span>= 46<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span><\/p>\n

 <\/p>\n

19- <\/b><\/span><\/span><\/span>Breve teoria<\/b><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>figura 1\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>ponto objeto real P<\/b><\/span><\/span>\u00a0est\u00e1 dentro\u00a0<\/b><\/span><\/span>da \u00e1gua (mais refringente, maior \u00edndice de refra\u00e7\u00e3o) e emitindo dois raios de luz<\/b><\/span><\/span>, um\u00a0<\/b><\/span><\/span>vertical que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>n\u00e3o sofre desvio<\/b><\/span><\/span>\u00a0e o outro\u00a0<\/b><\/span><\/span>obl\u00edquo que, ao se refratar para o ar, afasta-se da normal atingindo<\/b><\/span><\/span>\u00a0os\u00a0<\/b><\/span><\/span>olhos do observador<\/b><\/span><\/span>\u00a0e determina a\u00a0<\/b><\/span><\/span>imagem virtual (P\u2019),<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>obtida\u00a0<\/b><\/span><\/span>no\u00a0<\/b><\/span><\/span>prolongamento desse raio<\/b><\/span><\/span>\u00a0e onde ele\u00a0<\/b><\/span><\/span>intercepta o raio de luz vertical, fornecendo essa imagem P\u2019,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>acima do objeto real P.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0<\/span>Na\u00a0<\/b><\/span><\/span>figura II<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>ponto objeto real P est\u00e1 no ar (menos refringente, menor \u00edndice de refra\u00e7\u00e3o)<\/b><\/span><\/span>\u00a0e emitindo\u00a0<\/b><\/span><\/span>dois raios de luz, um vertical que n\u00e3o sofre<\/b><\/span><\/span>\u00a0desvio e o\u00a0<\/b><\/span><\/span>outro obl\u00edquo que, ao refratar-se para\u00a0 a \u00e1gua aproxima-se da normal<\/b><\/span><\/span>\u00a0atingindo os\u00a0<\/b><\/span><\/span>olhos do observador e determina a imagem virtual P\u2019, acima do objeto real P.<\/b><\/span><\/span><\/p>\n

Observe que,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>em ambos os casos a imagem est\u00e1 sempre acima do objeto.<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

20- <\/b><\/span><\/span><\/span>Veja na figura da esquerda os \u00e2ngulos que confirmam que o \u00e2ngulo de incid\u00eancia do raio de luz na interface ar \u2013 l\u00edquido (ponto A) vale i = 60<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando a lei de\u00a0<\/b><\/span><\/span>Snell \u2013 Descartes\u00a0<\/b><\/span><\/span>na figura da direita acima no\u00a0<\/b><\/span><\/span>ponto A<\/b><\/span><\/span>da\u00a0<\/b><\/span><\/span>interface ar \u2013 l\u00edquido<\/b><\/span><\/span><\/p>\n

Nar.seni = nl\u00edquido.senr\u00a0\"\"\u00a0<\/b><\/span><\/span>1.sen60<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>\u00a0= nl\u00edquido.sen30<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a01.\"\"\u00a0= nl\u00edquido.\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>nl\u00edquido\u00a0=\"\"<\/b><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

21- <\/b><\/span><\/span><\/span>Podemos<\/b><\/span><\/span>\u00a0resolver esta\u00a0<\/b><\/span><\/span>quest\u00e3o<\/b><\/span><\/span>\u00a0atrav\u00e9s da Lei de\u00a0<\/b><\/span><\/span>Snell-Descartes\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Temos, ent\u00e3o, que\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span><\/p>\n

R-<\/b><\/span><\/span>\u00a0Portanto, a resposta correta \u00e9 a alternativa A.<\/b><\/span><\/span><\/p>\n

22- <\/span><\/span><\/span>
\n\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Os\u00a0<\/b><\/span><\/span>pontos<\/b><\/span><\/span>\u00a0do\u00a0<\/b><\/span><\/span>tubo retil\u00edneo luminescente\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>podem ser vistos por pessoas fora da piscina\u00a0<\/b><\/span><\/span>s\u00e3o aqueles\u00a0<\/b><\/span><\/span>localizados abaixo do ponto M\u00a0<\/b><\/span><\/span>j\u00e1 que os\u00a0<\/b><\/span><\/span>pontos acima de M ser\u00e3o impedidos de se refratarem pela boia circular opaca.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

23- <\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando em\u00a0<\/b><\/span><\/span>(II) a lei de Snell-Descartes<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>n<\/b><\/span><\/span>poliestireno<\/b><\/span><\/span>.seni = n<\/b><\/span><\/span>ar\u00a0<\/b><\/span><\/span>.senr<\/b><\/span><\/span>\u00a0\"\"\u00a0n<\/b><\/span><\/span>poliestireno<\/b><\/span><\/span>.sen25<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>\u00a0= 1.sen40<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span> \"\"\u00a0<\/b><\/span><\/span>n<\/b><\/span><\/span>poliestireno<\/b><\/span><\/span>.0,4 = 1.0,6\u00a0\"\"\u00a0<\/b><\/span><\/span>n<\/b><\/span><\/span>poliestireno<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>n<\/b><\/span><\/span>poliestireno<\/b><\/span><\/span>\u00a0= 1,5.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

24- <\/span><\/span><\/span>
\n\u00cdndice de refra\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na\u00a0<\/b><\/span><\/span>express\u00e3o acima<\/b><\/span><\/span>\u00a0que, sendo a\u00a0<\/b><\/span><\/span>velocidade c da luz constante, <\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>velocidade v<\/b><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span>propaga\u00e7\u00e3o da luz no meio\u00a0<\/b><\/span><\/span>(no caso, o vidro) \u00e9 <\/b><\/span><\/span>inversamente proporcional ao \u00edndice absoluto de refra\u00e7\u00e3o da luz nesse meio.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>dispers\u00e3o luminosa\u00a0ocorre artificialmente\u00a0<\/b><\/span><\/span>quando a\u00a0luz se\u00a0<\/b><\/span><\/span>dispersa como no interior de um prisma<\/b><\/span><\/span>\u00a0(ou no\u00a0<\/b><\/span><\/span>interior da esfera de vidro\u00a0<\/b><\/span><\/span>do exerc\u00edcio) ou quando\u00a0<\/b><\/span><\/span>passa do ar para a \u00e1gua,\u00a0sendo\u00a0mais desviada a luz monocrom\u00e1tica violeta\u00a0(menor velocidade) e\u00a0menos desviada a luz monocrom\u00e1tica vermelha (maior velocidade).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como, pelo enunciado, o \u00edndice\u00a0<\/b><\/span><\/span>refra\u00e7\u00e3o absoluto do vidro \u00e9 maior para a cor azul do que para a<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

vermelha, a luz azul, no interior do vidro sofre maior desvio ao atravessar a esfera (veja figura acima).<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

25- <\/b><\/span><\/span><\/span>A) Veja a equa\u00e7\u00e3o de Gauss abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo a\u00a0<\/b><\/span><\/span>lente a mesma,\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>dist\u00e2ncia focal f tamb\u00e9m \u00e9 a mesma\u00a0<\/b><\/span><\/span>e a<\/b><\/span><\/span>amplia\u00e7\u00e3o A\u00a0<\/b><\/span><\/span>(aumento linear transversal) \u00e9 fornecida pela\u00a0<\/b><\/span><\/span>express\u00e3o A = \u2013\"\"\u00a0<\/b><\/span><\/span>que mostra que\u00a0<\/b><\/span><\/span>A depende apenas das dist\u00e2ncias do objeto e da imagem \u00e0 lente.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se a\u00a0<\/b><\/span><\/span>amplia\u00e7\u00e3o fosse a mesma nas duas posi\u00e7\u00f5es, P1\u00a0seria igual \u00e0 P1\u2019, P2seria igual \u00e0 P2\u2019\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>\u2502A\u2502= \u25021\u2502(objeto e imagem teriam o mesmo tamanho)<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe na\u00a0<\/b><\/span><\/span>figura\u00a0<\/b><\/span><\/span>que, com o\u00a0<\/b><\/span><\/span>objeto mais pr\u00f3ximo da lente P1\u2019 > P1<\/b><\/span><\/span>\u00a0e<\/b><\/span><\/span>\u2502A1\u2502> \u25021\u2502(tamanho aumentado).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Da mesma maneira<\/b><\/span><\/span>, com o objeto mais afastado da lente\u00a0P2\u2019 < P2\u00a0e \u2502A2\u2502< \u25021\u2502(tamanho diminu\u00eddo).<\/b><\/span><\/span><\/span><\/span><\/p>\n

B)\u00a0<\/b><\/span><\/span>Est\u00e1 sendo\u00a0<\/b><\/span><\/span>pedida\u00a0a dist\u00e2ncia entre\u00a0<\/b><\/span><\/span>o objeto e a lente<\/b><\/span><\/span>\u00a0na situa\u00e7\u00e3o em que est\u00e3o mais pr\u00f3ximos entre si (posi\u00e7\u00e3o P), ou seja, est\u00e1 pedindo P1.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Nessas condi\u00e7\u00f5es a\u00a0<\/b><\/span><\/span>dist\u00e2ncia do objeto \u00e0 lente ser\u00e1 P1\u00a0<\/b><\/span><\/span>e da\u00a0<\/b><\/span><\/span>imagem \u00e0 lente P1\u2019 = L \u2013 P1\u00a0= 10 \u2013 P1<\/b><\/span><\/span>\u00a0<\/b><\/span>e <\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

na<\/b><\/span><\/span> outra posi\u00e7\u00e3o<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>dist\u00e2ncia do objeto \u00e0 lente ser\u00e1 P2\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>da imagem \u00e0 lente P2\u2019 = L \u2013 P2\u00a0= 10 \u2013 P2.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe ainda, na\u00a0<\/b><\/span><\/span>figura acima, que P2\u00a0\u2013 P1\u00a0= 6 (I).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

26- <\/b><\/span><\/span><\/span>Primeiramente precisamos\u00a0<\/b><\/span><\/span>encontrar o valor da dist\u00e2ncia da imagem ao espelho<\/b><\/span><\/span>, para isso\u00a0<\/b><\/span><\/span>vamos utilizar a equa\u00e7\u00e3o do aumento linear<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

A \u00e9 o aumento linear<\/b><\/span><\/span><\/span><\/span><\/p>\n

P\u2019 \u00e9 a dist\u00e2ncia da imagem ao espelho<\/b><\/span><\/span><\/span><\/span><\/p>\n

P \u00e9 a dist\u00e2ncia do objeto ao espelho<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando P\u2019:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Substituindo, sabendo que o\u00a0<\/b><\/span><\/span>A foi dado no texto, quando ele diz que o aumento \u00e9 de 5 vezes<\/b><\/span><\/span>.\u00a0<\/b><\/span><\/span>Deixaremos tudo em cm, pois a resposta final est\u00e1 em cm<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Agora que j\u00e1\u00a0<\/b><\/span><\/span>temos a dist\u00e2ncia da imagem, podemos substituir na f\u00f3rmula de gauss<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

f \u00e9 a dist\u00e2ncia focal<\/b><\/span><\/span><\/span><\/span><\/p>\n

p \u00e9 a dist\u00e2ncia do objeto ao espelho<\/b><\/span><\/span><\/span><\/span><\/p>\n

p\u2019 \u00e9 a dist\u00e2ncia da imagem ao espelho<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo os valores,\u00a0<\/b><\/span><\/span>j\u00e1 jogando o sinal negativo (-100 cm) para fora da fra\u00e7\u00e3o<\/b><\/span><\/span>, ficando sinal negativo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Multiplicando tudo por -1,\u00a0<\/b><\/span><\/span>ou seja, invertendo as fra\u00e7\u00f5es<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/span><\/p>\n

27- <\/b><\/span><\/span><\/span>A lupa \u00e9 uma lente convergente e todos os raios de luz que incidem paralelamente ao eixo principal (caso do sol a pino) emergem da lente convergindo para o foco f. Assim, no caso do exerc\u00edcio, f = h.<\/b><\/span><\/span><\/p>\n

Veja\u00a0<\/b><\/span><\/span>a constru\u00e7\u00e3o geom\u00e9trica da imagem\u00a0<\/b><\/span><\/span>em uma lupa\u00a0<\/b><\/span><\/span>(lente convergente) em que o objeto tem que estar entre f<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0= f = h e O.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Natureza\u00a0\"\"\u00a0Virtual\u00a0<\/b><\/span><\/span>(obtida no\u00a0<\/b><\/span><\/span>cruzamento dos prolongamentos\u00a0<\/b><\/span><\/span>dos raios luminosos).<\/b><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o\u00a0\"\"\u00a0Antes de fo<\/b><\/span><\/span><\/p>\n

Tamanho e orienta\u00e7\u00e3o\u00a0\"\"\u00a0maior\u00a0que o objeto e\u00a0direita\u00a0em rela\u00e7\u00e3o a ele.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

28- <\/b><\/span><\/span><\/span>Uma das imagens i\u00a0<\/b><\/span><\/span>do objeto que\u00a0<\/b><\/span><\/span>est\u00e1 antes do ponto antiprincipal Ao (200 cm)\u00a0<\/b><\/span><\/span>\u00e9 formada diretamente pela\u00a0<\/b><\/span><\/span>lente biconvexa<\/b><\/span><\/span>\u00a0(convergente) que tem as\u00a0<\/b><\/span><\/span>seguintes caracter\u00edsticas<\/b><\/span><\/span>, geometricamente localizadas na\u00a0<\/b><\/span><\/span>figura abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Objeto O antes de Ao<\/b><\/span><\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas da imagem\u00a0i:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Natureza<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>Real\u00a0<\/b><\/span><\/span>(<\/b><\/span><\/span>obtida no\u00a0<\/b><\/span><\/span>cruzamento do pr\u00f3prio raio luminoso<\/b><\/span><\/span>\u00a0(linha cheia)<\/b><\/span><\/span>).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Localiza\u00e7\u00e3o<\/b><\/span><\/span>\u00a0\"\"\u00a0entre\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>i<\/b><\/span><\/span>\u00a0e A<\/b><\/span><\/span>i<\/b><\/span><\/span>.<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Tamanho e orienta\u00e7\u00e3o<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>menor que o objeto<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>invertida\u00a0<\/b><\/span><\/span>em rela\u00e7\u00e3o ao mesmo<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>outra imagem i\u2019\u00a0<\/b><\/span><\/span>captada pelo\u00a0<\/b><\/span><\/span>espelho plano\u00a0<\/b><\/span><\/span>(<\/b><\/span><\/span>real, invertida e maior que o objeto, pois o <\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

objeto<\/b><\/span><\/span>\u00a0est\u00e1 a 80cm do espelho, entre o foco F = 60cm e o centro de curvatura C = 120 cm)\u00a0<\/b><\/span><\/span>funciona como objeto para\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>lente biconvexa\u00a0<\/b><\/span><\/span>(convergente) e\u00a0<\/b><\/span><\/span>sua <\/b><\/span><\/span>imagem i\u2019\u2019<\/b><\/span><\/span>,<\/b><\/span><\/span>\u00a0vista pelo observador ser\u00e1\u00a0<\/b><\/span><\/span>direita em rela\u00e7\u00e3o ao objeto, real e maior que o mesmo,\u00a0<\/b><\/span><\/span>conforme constru\u00e7\u00e3o geom\u00e9trica na figura acima.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

29- \"\"<\/b><\/span><\/span><\/span><\/p>\n

Nesse\u00a0<\/b><\/span><\/span>microsc\u00f3pio de luz,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>velocidade<\/b><\/span><\/span>\u00a0de propaga\u00e7\u00e3o\u00a0<\/b><\/span><\/span>da luz<\/b><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span>constante e vale V = c = 3.10<\/b><\/span><\/span>8<\/b><\/sup><\/span>\u00a0m\/s,<\/b><\/span><\/span>\u00a0e assim, pela\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o fundamental da ondulat\u00f3ria V = constante =c = \u03bb.f,<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>comprimento de onda \u03bb \u00e9 inversamente proporciona \u00e0 frequ\u00eancia f.<\/b><\/span><\/span><\/p>\n

Como pelo enunciado, o intervalo\u00a0<\/b><\/span><\/span>de frequ\u00eancias do espectro de luz vis\u00edvel<\/b><\/span><\/span>\u00a0est\u00e1\u00a0<\/b><\/span><\/span>compreendido\u00a0<\/b><\/span><\/span>entre 4<\/b><\/span><\/span>,0.10<\/b><\/span><\/span>14<\/b><\/sup><\/span>\u00a0Hz e 7,5.10<\/b><\/span><\/span>14<\/b><\/sup><\/span>\u00a0Hz<\/b><\/span><\/span>, a\u00a0<\/b><\/span><\/span>menor estrutura celular<\/b><\/span><\/span>\u00a0(menor \u03bb) que se\u00a0<\/b><\/span><\/span>poderia observar<\/b><\/span><\/span>\u00a0nesse\u00a0<\/b><\/span><\/span>microsc\u00f3pio de luz<\/b><\/span><\/span>\u00a0corresponde \u00e0\u00a0<\/b><\/span><\/span>maior frequ\u00eancia nesse intervalo que \u00e9 de \u00e9 de f = 7,5.10<\/b><\/span><\/span>14<\/b><\/sup><\/span>\u00a0Hz.<\/b><\/span><\/span><\/p>\n

c = \u03bb.f<\/b><\/span><\/span>\u00a0\"\"\u00a03.10<\/b><\/span><\/span>8<\/b><\/sup><\/span>\u00a0= \u03bb.7,5.10<\/b><\/span><\/span>14<\/b><\/sup><\/span>\u00a0\"\"\u00a0\u03bb =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>\u03bb = 0,4.10<\/b><\/span><\/span>-6<\/b><\/sup><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>\u03bb = 4.10<\/b><\/span><\/span>-7<\/b><\/sup><\/span>\u00a0m = 400 nm.<\/b><\/span><\/span><\/p>\n

Portanto,\u00a0<\/b><\/span><\/span>observando a figura<\/b><\/span><\/span>\u00a0voc\u00ea percebe que a\u00a0<\/b><\/span><\/span>menor estrutura celular que voc\u00ea pode perceber<\/b><\/span><\/span>\u00a0nesse\u00a0<\/b><\/span><\/span>microsc\u00f3pio<\/b><\/span><\/span>\u00a0\u00e9 o\u00a0<\/b><\/span><\/span>Ret\u00edculo Endoplasm\u00e1tico dimens\u00e3o 420 nm.<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

30- <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00f5es das quest\u00f5es de Vestibulares recentes por assunto \u2013 2017 \u2013 2016 \u2013 2015 \u2013 \u00d3ptica 01- A\u00a0velocidade das nanosondas corresponde a 20% da velocidade da luze ter\u00e1\u00a0valor\u00a0\u00a0V = 20% de 3.108\u00a0m\/s\u00a0\u00a0V = 0,2\u00d73.108\u00a0= 0,6.108\u00a0m\/s\u00a0\u00a0V = 6.107\u00a0m\/s. Um\u00a0ano-luz \u00e9 a dist\u00e2ncia que a luz\u00a0pode viajar em\u00a0um ano com velocidade de 3.108\u00a0m\/s.\u00a0Por exemplo,\u00a0em\u00a0um ano-luz, a dist\u00e2ncia percorrida\u00a0seria de aproximadamente 9.460.800.000.000 (nove\u00a0trilh\u00f5es,<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3431,"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-3433","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3433","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=3433"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3433\/revisions"}],"predecessor-version":[{"id":3435,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3433\/revisions\/3435"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3431"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3433"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}