{"id":2212,"date":"2016-05-14T23:46:23","date_gmt":"2016-05-14T23:46:23","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2212"},"modified":"2024-09-02T13:44:29","modified_gmt":"2024-09-02T13:44:29","slug":"dioptro-plano-resolucao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/dioptro-plano\/dioptro-plano-resolucao\/","title":{"rendered":"Dioptro Plano – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0A imagem est\u00e1 sempre acima do objeto e quanto mais afastado estiver o observador, mais pr\u00f3xima ainda da superf\u00edcie ela estar\u00e1.\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>R-B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0A imagem est\u00e1 sempre acima do objeto e na vertical que os une<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=3cm\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=?\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=1,5\u00a0 —\u00a0 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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.1,5=3.1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=2cm (a altura aparente, dist\u00e2ncia da imagem da letra at\u00e9 o topo da placa de vidro)\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo e veja que a imagem est\u00e1 sempre acima do objeto e na mesma reta vertical:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10m\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=?\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1,3\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0 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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.1=10.1,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=13m\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0A imagem est\u00e1 sempre acima do objeto\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0d<\/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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.1,3=13.1,0\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=10cm\u00a0 —\u00a0 esta express\u00e3o s\u00f3 \u00e9 v\u00e1lida para posi\u00e7\u00f5es onde o observador esteja bem pr\u00f3ximo \u00e0 vertical que une objeto e imagem (posi\u00e7\u00e3o O<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) e, \u00e0 medida que o observador for se afastando (posi\u00e7\u00e3o O<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>), a imagem vai subindo\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0d<\/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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.1,33=10.1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=7,5m\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0Continuar\u00e1 vendo os 12 azulejos mas mais curtos, de altura (lado vertical) menor e valendo\u00a0 —\u00a0 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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=12.12=144cm\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=4\/3\u00a0 —\u00a0 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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.3\/3=144.1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=108cm\/12=9cm\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2,0m\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=4\/3\u00a0 — 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>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.4\/3=2.1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=1,5m<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0Dioptro A-B\u00a0 —\u00a0 c\u00e1lculo de d\u2019, eleva\u00e7\u00e3o de P para P\u2019 em rela\u00e7\u00e3o a S\u00a0 —\u00a0 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>\u00a0 —\u00a0 d\u2019.1,4=28.1,3\u00a0 —\u00a0 d\u2019=26cm<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Diopro B-ar\u00a0 —\u00a0 P\u2019 torna-se objeto fornecendo uma imagem P\u2019\u2019, em rela\u00e7\u00e3o a S\u2019, que \u00e9vista pelo observador\u00a0 —\u00a0 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>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

1,3=1.(26 + 39)\u00a0 —\u00a0 d\u2019\u2019=65\/1,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>d\u2019\u2019=50cm<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0(01) Correta \u2013 a luz sai do nariz do jacar\u00e9, sofre refra\u00e7\u00e3o na interface \u00e1gua-ar aproximando-se da normal e atingindo os olhos do pato, que o enxerga na posi\u00e7\u00e3o N.<\/b><\/span><\/span><\/span><\/p>\n

(02) Correta \u2013 Correta, a imagem est\u00e1 sempre acima do objeto.<\/b><\/span><\/span><\/span><\/p>\n

(04) Correta, pois \u00e9 o mesmo meio (\u00e1gua).<\/b><\/span><\/span><\/span><\/p>\n

(08) Errada \u2013 mais refringente o meio implica em maior \u00edndice de refra\u00e7\u00e3o e menor velocidade.<\/b><\/span><\/span><\/span><\/p>\n

(16) Errada \u2013 quanto mais afastado o pato estiver, maior ser\u00e1 o \u00e2ngulo de incid\u00eancia que, para que ocorra reflex\u00e3o total, dever\u00e1 superar o \u00e2ngulo limite.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Soma=01 + 02 + 04 = 07<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0 D<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0deve ser aumentada, pois a m\u00e1quina deve fotografar com nitidez a imagem, que est\u00e1 acima do objeto, de menos de 5 cm pois, ela ainda dever\u00e1 estar no interior da placa de vidro\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Tomando um ponto P no centro da moeda e, com o recipiente cheio de l\u00edquido, o observador deve ver a imagem do ponto P em P\u2019, na posi\u00e7\u00e3o indicada na figura.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo PYX\u00a0 —\u00a0 (PX)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (0,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (1,0)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 PX=\u221a(1,25)\u00a0 —\u00a0 seni=0,5\/\u221a(1,25)\u00a0 —\u00a0 aplicando Snell-Descartes no ponto X\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.seni=n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.senr\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/span>\u00a0n<\/b><\/span><\/span><\/span>l\u00edquido.<\/b><\/span><\/span><\/sub><\/span>0,5\/\u221a(1,25) = 1.sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.0,5\/\u221a(1,25=1.\u221a2\/2\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>=\u221a2,5\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0d<\/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>\u00a0 —\u00a0 3.1,33=d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4,0m\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0R- B\u00a0 —\u00a0 \u00e0 medida que\u00a0 a luz vai se distanciando da interface \u00e1gua-ar, o raio refratado vai se afastando cada vez mais da normal.<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0R- A\u00a0 (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0(01) Falsa \u2013 a luz comporta-se como onda e part\u00edcula<\/b><\/span><\/span><\/span><\/p>\n

(02) Falsa \u2013 a velocidade de propaga\u00e7\u00e3o da luz depende do meio<\/b><\/span><\/span><\/span><\/p>\n

(04) Verdadeira \u2013 ao se refratar na interface entre dois meios transparentes diferentes, a luz sofre desvio, mudando sua dire\u00e7\u00e3o de propaga\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

(16) Verdadeira –<\/b><\/span><\/span><\/span><\/p>\n

Soma=(04) + (16) = 20<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0(01) \u2013 Falsa \u2013 quando a luz policrom\u00e1tica branca se refrata do ar para a \u00e1gua, quem sofre menor desvio \u00e9 a cor vermelha, pois sua velocidade \u00e9 maior na \u00e1gua<\/b><\/span><\/span><\/span><\/p>\n

(02) Correta \u2013 \u00e0 medida que ele se afasta para a esquerda, o \u00e2ngulo de incid\u00eancia da luz emitida pela cabe\u00e7a torna-se maior que o da cauda, tornando-se maior que o \u00e2ngulo limite, fazendo com que a cabe\u00e7a sofra reflex\u00e3o total primeiro e que desapare\u00e7a primeiro que a cauda<\/b><\/span><\/span><\/span><\/p>\n

(04) Falsa \u2013 claro que acontece, se ela n\u00e3o se refratasse do ar para a \u00e1gua, ela n\u00e3o se dispersaria.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

(08) Correta \u2013 maior \u00edndice de refra\u00e7\u00e3o, maior desvio<\/b><\/span><\/span><\/span><\/p>\n

(16) Correta) \u2013 tg30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=3\/comprimento do tubar\u00e3o c)\u00a0 —\u00a0\u00a0 \u221a3=3\/c\u00a0 —\u00a0 c=5,3m<\/b><\/span><\/span><\/span><\/p>\n

Soma=(02 + 08 + 16) = 26<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0A resposta s\u00f3 pode estar entre as alternativas A, D ou C, pois D<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0capta a luz emitida na vertical onde n\u00e3o sofre desvio (incid\u00eancia normal), assim o indicador s\u00f3 pode estar na vertical. Quando se refrata do vidro (\u00e1gua) para o ar a luz sofre desvio, se\u00a0<\/b><\/span><\/span><\/span>afastando<\/b><\/span><\/span><\/span>\u00a0da normal, at\u00e9 atingir D<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>. A \u00fanica alternativa que satisfaz \u00e9 a\u00a0<\/b><\/span><\/span><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Lei de Snell\u00a0 —\u00a0 nsen\u03b2=constante\u00a0 —\u00a0 se o \u00edndice de refra\u00e7\u00e3o aumenta, o \u00e2ngulo diminui e vice \u2013 versa\u00a0 —\u00a0 observe nas figuras a seguir que:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 \u03b1 \u2013 \u00e2ngulo de incid\u00eancia\u00a0 —\u00a0 (90 \u2013 \u03b2) \u2013 \u00e2ngulo de refra\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) Da figura, no tri\u00e2ngulo ret\u00e2ngulo APC\u00a0 —\u00a0 tg\u03b1=0,9\/1=0,9\u00a0 —\u00a0 da tabela fornecida\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b1 = 42<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span><\/p>\n

b) Lei de Snell\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1 = n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.sen(90 \u2013 \u03b2)\u00a0 —\u00a0 (1,3)x(0,67) = (1)xsen(90 \u2013 \u03b2)\u00a0 —\u00a0 sen(90 \u2013 \u03b2) = 0,87\u00a0 —\u00a0 da tabela fornecida\u00a0 —\u00a0 (90 \u2013 \u03b2) = 60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b2 = 30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span><\/p>\n

c) Da figura acima, no tri\u00e2ngulo ABI\u00a0 —\u00a0 tg\u03b2 = y\/x\u00a0 —\u00a0 tg30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=y\/0,9\u00a0 —\u00a0 0,58=y\/0.9 \u00a0—\u00a0<\/b><\/span><\/span><\/span>y=0,52m<\/b><\/span><\/span><\/span><\/p>\n

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

22-<\/b><\/span><\/span><\/span> Dados: sen 48,6<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0= 0,75, cos 48,6\u00ba = 0,66.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0a) Observe na figura 1, nos dois tri\u00e2ngulos hachurados), que o \u00e2ngulo de incid\u00eancia da luz na superf\u00edcie de separa\u00e7\u00e3o \u00e1gua-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

vidro vale \u03b8\u00a0 — observe na figura 2 que o \u00e2ngulo de refra\u00e7\u00e3o na interface \u00e1gua-vidro \u00e9 \u03c6 e na interface vidro-ar, o \u00e2ngulo de incid\u00eancia \u00e9 \u03c6 e o de refra\u00e7\u00e3o \u00e9 \u03b2\u00a0 —\u00a0 lei de Snell-Descartes na interface \u00e1gua-vidro\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.sen\u03b8 = n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>.sen\u03c6\u00a0\u00a0 —\u00a0 lei de Snell-Descartes na interface vidro-ar\u00a0 —n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>.sen\u03c6 = n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.sen\u03b2\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>.sen\u03c6 = 1.sen\u03b2\u00a0\u00a0 —\u00a0 \u00e2ngulo limite (L) para que ocorra reflex\u00e3o total na interface vidro-ar\u00a0 —\u00a0 senL=sen\u03c6=n<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span>=n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>=1\/n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 sen\u03c6=1\/n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>.1\/n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1.sen\u03b2 (II)\u00a0 —\u00a0 sen\u03b2=1 (IV)\u00a0 —\u00a0 \u03b2=45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.sen\u03b8<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= n<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span>.sen\u03c6 (I)=n<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.sen\u03b2=1 —\u00a0 n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.sen\u03b8<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=1\/sen\u03b8<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0= c\/V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1\/sen \u03b8<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= c\/V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1\/0,75=3.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2,25.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>km\/s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

23-<\/b><\/span><\/span><\/span> Na figura \u00a0o ponto objeto real P est\u00e1 dentro da \u00e1gua e emitindo dois raios de luz, um vertical que n\u00e3o sofre desvio e<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

o outro obl\u00edquo que, ao se refratar para o ar, afasta-se da normal atingindo os olhos do observador e determina a imagem, vista pelo observador e que \u00e9 virtual (P\u2019), e est\u00e1 acima do objeto real P.<\/b><\/span><\/span><\/p>\n

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

24-<\/b><\/span><\/span><\/span> Como a imagem est\u00e1 acima do observador, ela estar\u00e1 mais pr\u00f3xima do mesmo e, devido ao campo visual maior dar\u00e1 ao observador a impress\u00e3o de ser maior que o objeto\u00a0\u00a0—\u00a0\u00a0R- D.\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Dioptro Plano 01-\u00a0A imagem est\u00e1 sempre acima do objeto e quanto mais afastado estiver o observador, mais pr\u00f3xima ainda da superf\u00edcie ela estar\u00e1.\u00a0\u00a0\u00a0\u00a0R-B 02-\u00a0A imagem est\u00e1 sempre acima do objeto e na vertical que os une do=3cm\u00a0 —\u00a0 no=1\u00a0 —\u00a0 di=?\u00a0 —\u00a0 ni=1,5\u00a0 —\u00a0 di.ni=do.no\u00a0 —\u00a0 di.1,5=3.1\u00a0 —\u00a0 di=2cm (a altura aparente, dist\u00e2ncia<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2208,"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-2212","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2212","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=2212"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2212\/revisions"}],"predecessor-version":[{"id":10933,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2212\/revisions\/10933"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2208"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}