{"id":2231,"date":"2016-05-15T00:08:36","date_gmt":"2016-05-15T00:08:36","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2231"},"modified":"2016-05-15T00:09:03","modified_gmt":"2016-05-15T00:09:03","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-estudo-analitico-dos-espelhos-esfericos","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/estudo-analitico-dos-espelhos-esfericos\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-estudo-analitico-dos-espelhos-esfericos\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Estudo Anal\u00edtico dos Espelhos Esf\u00e9ricos"},"content":{"rendered":"

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

Estudo Anal\u00edtico dos Espelhos Esf\u00e9ricos<\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span> <\/b>a) P\u2019=+40cm (real)\u00a0 —\u00a0 f=+30cm (c\u00f4ncavo)\u00a0\u00a0\u00a0 i= -3cm (invertida)\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/30=1\/P + 1\/40\u00a0 —\u00a0 1\/P = 1\/30 \u2013 1\/40\u00a0 —\u00a0 (4 \u2013 3)\/120 = 1\/P\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=120cm<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>f=-40cm espelho convexo)\u00a0 —\u00a0 espelho convexo a imagem \u00e9 direita e\u00a0<\/b><\/span><\/span>i<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>O<\/b><\/span><\/span>\u00a0tem mesmo sinal\u00a0 —\u00a0 O=20i\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i\/20i=-P\u2019\/P\u00a0 —\u00a0 P=-20P\u2019\u00a0 —\u00a0 1\/f = 1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/-40 = 1\/-20P\u2019 + 1\/P\u2019\u00a0 — \u00a0-1\/40 = 19\/20P\u2019\u00a0 —\u00a0 20P\u2019P=-760\u00a0 —\u00a0<\/b><\/span><\/span><\/span> P\u2019=-38cm\u00a0 —\u00a0 P=-20P\u2019\u00a0 —\u00a0 P=(-20).(-38)=760cm\u00a0 —\u00a0 P=7,6m\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> <\/b>a) O espelho, para obter a maior intensidade de radia\u00e7\u00e3o solar poss\u00edvel, deve ter 60m de raio de curvatura, de modo que o navio fique em seu foco quando estiver a 30m da praia.<\/b><\/span><\/span><\/p>\n

b) Intensidade de radia\u00e7\u00e3o = Pot\u00eancia\/\u00e1rea\u00a0 —\u00a0 I=P\/S\u00a0 —\u00a0 60% de I=0,6.500\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 I=300W\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>cada espelho tem \u00e1rea<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

S=0,5.1,0\u00a0 —\u00a0 S=0,5m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como s\u00e3o 60 espelhos\u00a0 —\u00a0 S<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=60.0,5\u00a0 —\u00a0 S<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=30m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 I=P\/S<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 300=P\/30\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 P=9.000W ou P=9kW<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>O=h\u00a0 — P=15cm\u00a0 —\u00a0 sendo a imagem direita e menor que o objeto a imagem \u00e9 virtual e o espelho \u00e9<\/b><\/span><\/span>\u00a0convexo<\/b><\/span><\/span>\u00a0e i=h\/5\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i\/O=-P\u2019\/P\u00a0 —\u00a0 (h\/5)\/h=-P\u2019\/15\u00a0 —\u00a0 P\u2019=-3cm\u00a0 —\u00a0 1\/f = 1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/15 + 1\/-3 \u00a0— f=-3,75cm\u00a0 —\u00a0 R=2.(-3,75)\u00a0 —<\/b><\/span><\/span><\/p>\n

R=-7,5cm e o sinal negativo significa que o espelho \u00e9 convexo\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Situa\u00e7\u00e3o inicial (t=0)<\/b><\/span><\/span>\u00a0 —\u00a0 P=50cm\u00a0 —\u00a0 f=10cm\u00a0 —\u00a0 1\/f=1\/P+1\/P\u2019\u00a0 —\u00a0 1\/10=1\/50+1\/P\u2019\u00a0 —\u00a0 1\/10 \u2013 1\/50=1\/P\u2019\u00a0 —\u00a0 4\/50=1\/P\u2019\u00a0 —\u00a0 P\u2019=12,5cm (posi\u00e7\u00e3o inicial da imagem)<\/b><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o final (t=5s)\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 4= \u0394S\/5\u00a0 —\u00a0 \u0394S=20cm\u00a0 —\u00a0 como ele se aproxima do espelho sua nova dist\u00e2ncia do v\u00e9rtice do espelho ser\u00e1\u00a0 —\u00a0 P=50 \u2013 20\u00a0 —\u00a0 P=30cm\u00a0 —\u00a0 f=10cm\u00a0 —\u00a0 1\/10 = 1\/30 + 1\/P\u2019\u00a0 —\u00a0 P\u2019= 15cm (nova posi\u00e7\u00e3o da imagem)\u00a0 —\u00a0 A imagem percorreu \u0394S=15 \u2013 12,5\u00a0 —\u00a0 \u0394S=2,5cm\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 R- E<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>O=1m\u00a0 —\u00a0 espelho convexo \u2013 a imagem \u00e9 virtual,\u00a0<\/b><\/span><\/span>direita e menor<\/b><\/span><\/span>\u00a0que o objeto\u00a0 —\u00a0 O=4i\u00a0\u00a0 —\u00a0 P=1,2m\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0 i\/4i=-P\u2019\/1,2\u00a0 —\u00a0 P\u2019= – 0,3m\u00a0 —\u00a0 1\/f = 1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/1,2 + 1\/-0,3\u00a0 —\u00a0 f= -1,2\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>f= – 0,4m<\/b><\/span><\/span>\u00a0(o sinal negativo significa que o espelho \u00e9 convexo)\u00a0<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> <\/b>a) Espelho convexo e a imagem \u00e9 virtual, direita e menor que o objeto. (tem a vantagem de aumentar o campo visual).<\/b><\/span><\/span><\/p>\n

b) f=-0,25m (espelho convexo)\u00a0 —\u00a0 O=1,6m\u00a0 —\u00a0 P=2,25m\u00a0 —1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 -1\/0,25=1\/2,25 + 1\/P\u2019\u00a0 —\u00a0 P\u2019= -0,225m\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i\/O=-P\u2019\/P\u00a0 —\u00a0 i\/1,6=-(-0,225)\/2,25\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 i=0,16m=16cm<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Convexo\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 f=-10cm\u00a0 —\u00a0 P=6cm\u00a0 —\u00a0 1\/-10=1\/6 + 1\/P\u2019\u00a0 —\u00a0 P\u2019= -3,75cm\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0 1,5\/O=-(-3,75)\/6\u00a0 —\u00a0 O=2,4cm\u00a0 —<\/b><\/span><\/span>\u00a0c\u00f4ncavo<\/b><\/span><\/span>\u00a0 —\u00a0 f=10cm\u00a0 —\u00a0 i=-1,5m (imagem invertida)\u00a0 —\u00a0 O=2,4cm\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0 -1,5\/2,4=-P\u2019\/P\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P\u2019=0,625P\u00a0 —\u00a0 1\/10=1\/P + 1\/0,625P\u00a0 —\u00a0 P=16,25\/0,625\u00a0 —\u00a0 P=26cm\u00a0\u00a0<\/b><\/span><\/span>R- C\u00a0<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> <\/b>Situa\u00e7\u00e3o inicial (t=0)\u00a0 —\u00a0 como o objeto est\u00e1 sobre C, a imagem estar\u00e1 sob C e P\u2019=6m (figura abaixo)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Situa\u00e7\u00e3o final (t=2s)\u00a0 —\u00a0 V=d\/t\u00a0 —\u00a0 1=d\/2\u00a0 —\u00a0 d=2m \u2013 o objeto se deslocou 2m e est\u00e1 a P=6-2=4m do v\u00e9rtice P=4cm\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/3=1\/4 + 1\/P\u2019\u00a0 —\u00a0 P\u2019=12m (figura abaixo)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A imagem, que estava a 6m do espelho, se afastou 6m do mesmo\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span><\/span> <\/b>A=-P\u2019\/P\u00a0 —\u00a0 2=-P\u2019\/P\u00a0 —\u00a0 P\u2019=-2P\u00a0 —\u00a0 1\/1=1\/P \u2013 1\/2P\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=0,5m=50cm<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span> <\/b>a) P=30m\u00a0 —\u00a0 f=-6m (convexo)\u00a0 —\u00a0 1\/f = 1\/P + 1\/P\u2019\u00a0 —\u00a0 -1\/6=1\/30 + 1\/P\u2019\u00a0 —\u00a0 P\u2019=-5m (observe que para um objeto a 30m do espelho convexo, a imagem seria vista como se estivesse a 5m do mesmo, aparentando assim, ser menor).<\/b><\/span><\/span><\/p>\n

b) No espelho plano a imagem \u00e9 sim\u00e9trica (objeto e imagem s\u00e3o eq\u00fcidistantes do espelho) e de mesmo tamanho do objeto \u2013 30m<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span> <\/b>a) V=d\/t\u00a0 —\u00a0 5=d\/2 \u00a0—\u00a0 d=10cm<\/b><\/span><\/span><\/p>\n

Para t = 2,0 s \u00eb x = 10 cm.\u00a0 —\u00a0\u00a0 em t = 2,0 s, o objeto estar\u00e1 a 40 cm do v\u00e9rtice do espelho, ou seja, ele estar\u00e1 antes do centro de curvatura C do espelho.<\/b><\/span><\/span><\/p>\n

Para um objeto que se encontra antes do centro de curvatura de um espelho c\u00f4ncavo, as caracter\u00edsticas da imagem formada s\u00e3o: real, invertida e menor.<\/b><\/span><\/span><\/p>\n

b) Para que a imagem se forme no infinito (imagem impr\u00f3pria) o objeto deve se encontrar no foco do espelho. Portanto, ele dever\u00e1 percorrer d=40 cm\u00a0 —\u00a0 V=d\/t\u00a0 —\u00a0 5=40\/t\u00a0 —\u00a0 t=8s.<\/b><\/span><\/span><\/p>\n

c) Dist\u00e2ncia percorrida pelo objeto em 7 s:<\/b><\/span><\/span><\/p>\n

V=d\/t\u00a0 —\u00a0 5=d\/7\u00a0 —\u00a0 d=35cm\u00a0 —\u00a0 logo a posi\u00e7\u00e3o do objeto ser\u00e1: p = 15 cm.\u00a0 —\u00a0 calculando a posi\u00e7\u00e3o da imagem formada usando a rela\u00e7\u00e3o: (1\/p) + (1\/p’) = 1\/f\u00a0 — utilizamos o fato de que f = R\/2\u00a0 —p’ = 30 cm<\/b><\/span><\/span><\/p>\n

Em t = 7,0 s o objeto se encontra entre o foco e o Centro de Curvatura e, portanto, sua imagem ser\u00e1 real, maior e invertida.<\/b><\/span><\/span><\/p>\n

O c\u00e1lculo do tamanho da imagem formada pode ser realizado utilizando a equa\u00e7\u00e3o para amplia\u00e7\u00e3o da imagem, dada por:<\/b><\/span><\/span><\/p>\n

A = i\/o = p’\/p\u00a0 —\u00a0 i\/10 = – [(30)\/15]\u00a0 —\u00a0 i = – 20 cm<\/b><\/span><\/span><\/p>\n

Nesta equa\u00e7\u00e3o i e o s\u00e3o os tamanhos da imagem e do objeto, respectivamente. O sinal negativo indica que a imagem formada \u00e9 invertida.<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span> <\/b>Quando observava cravos e espinhas seu rosto estava entre o foco (1,5m) e o espelho, que fornecia uma imagem virtual, direita e maior. Quando colou seu rosto a 2m do espelho, ele ficou entre o foco e o centro de curvatura e, nesse caso, a imagem fica maior e invertida.\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span> <\/b>a) P=4m\u00a0 —\u00a0 P\u2019=-0,2m (imagem virtual)\u00a0 —\u00a0 1\/f = 1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/4 \u2013 1\/0,2\u00a0 —\u00a0 f=-4\/19m\u00a0 —\u00a0 R=2f=2.(-4\/10)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R= (8\/19) m<\/b><\/span><\/span><\/p>\n

b) H=O=1,6m\u00a0 —\u00a0 i=h\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0 h\/1,6=-(-0,2)\/4\u00a0 —\u00a0 h=0,08m=8cm<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span> <\/b>P=90cm\u00a0 —\u00a0 O=10i\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0 i\/10i=-P\u2019\/90\u00a0 —\u00a0 P\u2019=-9cm\u00a0 —\u00a0 1\/f=1\/90-1\/9\u00a0 — f=-10cm\u00a0 —\u00a0 em m\u00f3dulo f=10cm\u00a0\u00a0\u00a0<\/b><\/span><\/span>\u00a0R- D\u00a0<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span> <\/b>Antes\u00a0 —\u00a0 i=3O\/4\u00a0 —\u00a0 i\/O=-P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019\/P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 (3O\/4)\/O = -P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019\/P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019= – 3P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 1\/f=1\/P<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>+ 1\/P\u2019<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/f= 1\/P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0-4\/3P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

f=-3P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0(I)\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 i=O\/4\u00a0\u00a0 —\u00a0\u00a0 i\/O= -P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2019\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 (O\/4)\/O=-P\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 — \u00a0p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2019=-P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 1\/f=1\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>+ 1\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2019<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 1\/f=1\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 4\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0—\u00a0<\/b><\/span><\/span>f= -P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/3 (II)\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 igualando I com II\u00a0 —\u00a0 -3P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=-P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> <\/b>Sendo a dist\u00e2ncia do Sol \u00e0 Terra muito grande, os raios de luz emitidos por ele chegam \u00e0 Terra como feixes de raios paralelos e a imagem P\u2019 se forma no foco\u00a0 —\u00a0 P\u2019= f =1m —\u00a0 P=250d\u00a0 —\u00a0 O=d<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

i\/O = -P\u2019\/P\u00a0 —\u00a0 i\/d = – 1\/250d\u00a0 —\u00a0 i=-1\/250=4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 –\u00a0 R- E<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> <\/b>C\u00e1lculo da imagem A\u2019 de A, conjugada pelo espelho esf\u00e9rico c\u00f4ncavo\u00a0 —\u00a0 P=30cm\u00a0 —\u00a0 f=20cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/20 \u2013 1\/30=1\/P\u2019\u00a0 —\u00a0 P\u2019=60cm\u00a0 —\u00a0 observe na figura abaixo que os raios de luz\u00a0 emitidos por A, sofrem reflex\u00e3o no espelho<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c\u00f4ncavo, incidem no espelho, onde s\u00e3o refletidos mas,a imagem se forma em A\u2019, que se comporta como objeto virtual para o espelho plano. Os raios de luz que atingem o espelho plano devem sofrer reflex\u00e3o e retornar a A.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como no espelho plano objeto e imagem s\u00e3o sim\u00e9tricos ao espelho as dist\u00e2ncias AE e AE\u2019 s\u00e3o iguais e de 15cm cada uma. Assim, a dist\u00e2ncia pedida d vale30 + 15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=45cm<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span> <\/b>Observe na figura abaixo que, como os raios solares incidem paralelamente ao eixo principal, eles se refletem de modo que o prolongamento dos raios refletidos se encontre no foco F\u2019, onde est\u00e1 a imagem do Sol, que vai servir como objeto virtual para o espelho c\u00f4ncavo E\u2019\u2019.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, para E\u2019\u2019, temos\u00a0 —\u00a0 P=24cm\u00a0 —\u00a0 P\u2019 (dist\u00e2ncia da imagem do Sol projetada na tela por E\u2019\u2019)\u00a0 —\u00a0 f=15cm\u00a0 —\u00a0 1\/f=1\/P+1\/P\u2019\u00a0 —\u00a0 1\/15=1\/24 + 1\/P\u2019\u00a0 —\u00a0 1\/P\u2019=(8 \u2013 5)\/120\u00a0 —\u00a0 P\u2019=40cm \u00a0—\u00a0 d=40 \u2013 30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=10cm<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Ambos os espelhos s\u00e3o c\u00f4ncavos, possuindo, ent\u00e3o, abscissas focais positivas\u00a0 —\u00a0 equa\u00e7\u00e3o dos pontos conjugados\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 P\u00b4=P.f\/(P \u2013 f)\u00a0 —\u00a0 espelho da esquerda (1)\u00a0 —\u00a0 f<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2m\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1m\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019=1 x 2\/(1 \u2013 2)\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019= – 2m (imagem virtual e atr\u00e1s do espelho)\u00a0 —\u00a0 espelho da direita (2)\u00a0 —\u00a0 f<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5m\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1m\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2019=1 x 5\/(1 \u2013 5)\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2019= – 5\/4 m (imagem virtual e atr\u00e1s do espelho)\u00a0 —\u00a0 a figura mostra as localiza\u00e7\u00f5es objeto e imagens em rela\u00e7\u00e3o aos espelhos\u00a0 —\u00a0 observe que a dist\u00e2ncia D entre a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

imagem da esquerda (i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) e a da direita i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0vale\u00a0 —\u00a0 D=2 + 2 + 5\/4\u00a0 —\u00a0 D=21\/4 m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>p = 10 cm\u00a0 —\u00a0 A = 5\u00a0 —\u00a0 A= – P\u2019\/P\u00a0 —\u00a0 5= – P\u2019\/10\u00a0 —\u00a0 P\u2019=-50cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/10 + 1\/(-50)\u00a0 —\u00a0 1\/f=(5 \u2013 1)\/50\u00a0 –f=50\/4\u00a0 —\u00a0 f=12,5cm\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>h =o= 8 cm\u00a0 —\u00a0 \u00a0p = 80 cm\u00a0 —\u00a0 \u00a0h\u2019 = i=1,6 cm\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 — 1,6\/8= – P\u2019\/80\u00a0 —\u00a0 P\u2019= – 16cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/80 + 1\/(-16)\u00a0 —\u00a0 f= – 20cm\u00a0 —\u00a0 valor absoluto\u00a0 —\u00a0 |f| = 20 cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que, pelo enunciado, o farolete deve emitir raios de luz paralelos quando a l\u00e2mpada estiver localizada a 10cm do espelho esf\u00e9rico c\u00f4ncavo, ou seja, ela deve estar sobre o foco, pois os raios de luz, nessa posi\u00e7\u00e3o emergem paralelos\u00a0 —\u00a0 f=+10cm (I)\u00a0 —<\/b><\/span><\/span><\/p>\n

Como a imagem \u00e9 projetada e ampliada 5 vezes, a imagem \u00e9 invertida em rela\u00e7\u00e3o ao objeto e, consequentemente o aumento linear transversal \u00e9 negativo\u00a0 —\u00a0 A=-5\u00a0 —\u00a0 A=-P\u2019\/P\u00a0 —\u00a0 -5=-P\u2019\/P\u00a0 —\u00a0 P\u2019=5P (II)\u00a0 —\u00a0 substituindo (I) e (II) na equa\u00e7\u00e3o dos pontos conjugados\u00a0 —\u00a0 1\/f=1\/p + 1\/P\u2019\u00a0 —\u00a0 1\/10=1\/P + 1\/5P\u00a0 —\u00a0 P=12cm \u00a0—\u00a0 portanto a l\u00e2mpada encontra-se 2cm adiante do foco\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a situa\u00e7\u00e3o descrita ilustrada na figura abaixo\u00a0 —\u00a0 como o raio refratado incide no <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

espelho passando pelo foco, ele reflete paralelo ao eixo principal\u00a0 —\u00a0 quando um \u00e2ngulo \u00e9 pequeno (\u03b8 < 10\u00b0), voc\u00ea pode fazer a aproxima\u00e7\u00e3o\u00a0 —\u00a0 sen \u03b8 = tg \u03b8 = \u03b8 \u00a0\u00a0(radiano)\u00a0 —\u00a0 como nesse caso i = 7\u00b0 e r < i (\u00e2ngulos pequenos), voc\u00ea pode trocar o seno pela tangente na lei de Snell\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

n<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.tg i = n<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.tg r\u00a0 —\u00a0 1.(0,12)=1,2.tg r\u00a0 —\u00a0\u00a0 tg r = 0,12\/1,2\u00a0 —\u00a0 tg r = 0,1\u00a0 —\u00a0 observe no tri\u00e2ngulo OPD hachurado da figura\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

tg r = d\/3\u00a0 —\u00a0 0,1=d\/3\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=0,3m<\/b><\/span><\/span><\/p>\n

 <\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0Dados: f=R\/2=1\/2=0,5m=50cm\u00a0 —\u00a0 P=50 + 10=60cm\u00a0 —\u00a0 equa\u00e7\u00e3o dos pontos conjugados de <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Gauss\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a01\/50=1\/60 + 1\/P\u2019\u00a0 —\u00a0 1\/50 \u2013 1\/60=1\/P\u2019\u00a0 —\u00a0 (6 \u2013 5)\/300=1\/P\u2019\u00a0 —\u00a0 P\u2019=300cm\u00a0 —\u00a0 dist\u00e2ncia entre objeto e imagem\u00a0 —\u00a0 d=300 \u2013 60=240cm\u00a0 —\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span> A dist\u00e2ncia focal do espelho \u00e9 igual \u00e0 metade de seu raio de curvatura\u00a0 —\u00a0 f=+1\/2m\u00a0 —\u00a0 f= + 50cm (positiva porque o espelho \u00e9 c\u00f4ncavo)\u00a0 —\u00a0 quando o l\u00e1pis estiver a 10cm do espelho \u2013 P=10cm\u00a0 —\u00a0 equa\u00e7\u00e3o dos pontos conjugados de Gauss\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/50 \u2013 1\/10=1\/P\u2019\u00a0 —\u00a0 (1 \u2013 5)\/50=1\/P\u2019\u00a0 —\u00a0 4P\u2019=-50\u00a0 —\u00a0 P\u2019= – 12,5cm (a imagem \u00e9 virtual e est\u00e1 atr\u00e1s do espelho e a<\/b><\/span><\/span><\/p>\n

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

12,5cm dele)\u00a0 —\u00a0 A= -P\u2019\/P=-(-12,5)\/10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>A=1,25<\/b><\/span><\/span>\u00a0(a imagem \u00e9 direita A>0 e 1,25 vezes maior que o objeto).<\/b><\/span><\/span><\/p>\n

Segunda etapa\u00a0 — dist\u00e2ncia (P) em que o l\u00e1pis deveria estar do v\u00e9rtice do espelho, para que sua imagem fosse direita e ampliada cinco vezes\u00a0 —\u00a0 A=+5 (imagem direita A>0)\u00a0 —\u00a0 A= – P\u2019\/P\u00a0 —\u00a0 5= – P\u2019\/P\u00a0 —\u00a0 P\u2019=-5P\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/50=1\/P + 1\/(-5P).<\/b><\/span><\/span><\/p>\n

1\/50=1\/P \u2013 1\/5P\u00a0 —\u00a0 1\/50=4\/5P\u00a0 —\u00a0 5P=200\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=40cm<\/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 Estudo Anal\u00edtico dos Espelhos Esf\u00e9ricos 01- a) P\u2019=+40cm (real)\u00a0 —\u00a0 f=+30cm (c\u00f4ncavo)\u00a0\u00a0\u00a0 i= -3cm (invertida)\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/30=1\/P + 1\/40\u00a0 —\u00a0 1\/P = 1\/30 \u2013 1\/40\u00a0 —\u00a0 (4 \u2013 3)\/120 = 1\/P\u00a0 —\u00a0\u00a0P=120cm b) 02- f=-40cm espelho convexo)\u00a0 —\u00a0 espelho convexo a imagem \u00e9 direita e\u00a0i\u00a0e\u00a0O\u00a0tem mesmo sinal\u00a0 —\u00a0 O=20i\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2227,"menu_order":0,"comment_status":"open","ping_status":"open","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-2231","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2231","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=2231"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2231\/revisions"}],"predecessor-version":[{"id":2233,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2231\/revisions\/2233"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2227"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2231"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}