{"id":2273,"date":"2016-06-20T03:30:22","date_gmt":"2016-06-20T03:30:22","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2273"},"modified":"2024-09-02T14:04:41","modified_gmt":"2024-09-02T14:04:41","slug":"estudo-analitico-das-lentes-esfericas-resolucao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/optica\/optica-geometrica\/estudo-analitico-das-lentes-esfericas\/estudo-analitico-das-lentes-esfericas-resolucao\/","title":{"rendered":"Estudo Anal\u00edtico das Lentes Esf\u00e9ricas – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

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

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

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

01-<\/b><\/span><\/span><\/span>\u00a0a) Lente convergente \u2013 lentes de vidro no ar, de bordas finas s\u00e3o convergentes ou, observe que , ap\u00f3s se refratarem na lente os raios de luz convergem para o eixo principal.<\/b><\/span><\/span><\/p>\n

b) P=15cm\u00a0 —\u00a0 P\u2019=10cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/15 + 1\/10\u00a0 —\u00a0 1\/f=(10 + 15)\/150\u00a0\u00a0 —\u00a0 f=150\/25\u00a0 —\u00a0\u00a0f=6cm<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0a) largura do quadro da fita=tamanho do objeto\u00a0 —\u00a0 o=35mm\u00a0 —\u00a0 o=35.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 largura da tela=tamanho da imagem\u00a0 —<\/b><\/span><\/span><\/p>\n

i= – 10,5m ( negativo, pois toda imagem real \u201cprojetada\u201d \u00e9 invertida)\u00a0 —\u00a0 P\u2019=30m\u00a0 —\u00a0 A=i\/o= – 10,5\/35.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0A= – 300\u00a0(a imagem \u00e9 300 vezes maior que o objeto e \u00e9 invertida)<\/b><\/span><\/span><\/p>\n

b) A=-P\u2019P\/P\u00a0 —\u00a0 -300=-30\/P\u00a0\u00a0—\u00a0 P=10cm\u00a0(a fita est\u00e1 a 10cm da lente)<\/b><\/span><\/span><\/p>\n

c) 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/10 + 1\/30\u00a0 —\u00a0 1\/f=(3 + 1)\/30\u00a0 —\u00a0f=7,5cm<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0a) A lente \u00e9 convergente, pois a imagem \u00e9 projetada (real e invertida) e, na tela ela aprece como direita, pois o slide \u00e9 colocado invertido\u00a0 —\u00a0 f=5cm\u00a0 —\u00a0 P=6cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/5=1\/6 + 1\/P\u2019\u00a0 —\u00a0 1\/5 \u2013 1\/6=1\/P\u2019\u00a0 —\u00a0 1\/P\u2019=(6 \u2013 5)\/30\u00a0 —\u00a0\u00a0P\u2019=30cm<\/b><\/span><\/span><\/p>\n

b) A=-P\u2019\/P=-30\/6\u00a0 —\u00a0\u00a0A= -5\u00a0(a imagem \u00e9 ampliada 5 vezes e \u00e9 invertida)<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0a) 1\/50=1\/P + 1\/52\u00a0 —\u00a0 1\/50 \u2013 1\/52=1\/P\u00a0 —\u00a0 1\/P=(52 \u2013 50)\/2600\u00a0 —\u00a0\u00a0P=1.300mm=1,3m<\/b><\/span><\/span><\/p>\n

b) i= -36mm (negativa, pois \u00e9 invertida)\u00a0 —\u00a0\u00a0 i\/o=-P\u2019\/P\u00a0 —\u00a0 -36\/o=-52\/1.300\u00a0\u00a0—\u00a0 o=900mm=90cm<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Na \u00e1gua\u00a0 —\u00a0 f=65cm\u00a0 —\u00a0 P=40cm\u00a0 —\u00a0 1\/65=1\/40 + 1\/P\u2019\u00a0 —\u00a0 1\/P\u2019=(40 \u2013 65)\/2.600\u00a0 —\u00a0 P\u2019= – 104cm (negativa, imagem virtual)\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 —\u00a0 i\/0=-(-104)\/40\u00a0 —\u00a0 i=2,6.o\u00a0 (a imagem \u00e9 direita e 2,6 vezes maior que o objeto)\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0A e b) A lente \u00e9 convergente porque a imagem \u00e9 maior que o objeto e porque \u00e9 projetada e consequentemente real e invertida\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i= – 4.o\u00a0 —\u00a0 i\/o=-P\u2019p\/P\u00a0 —\u00a0 – 4.o\/o = -P\u2019\/P\u00a0 —\u00a0 P\u2019=4P\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/12=1\/P + 1\/4P\u00a0 —\u00a0 1\/12=(4 + 1)\/4P\u00a0 —\u00a0 4P=60\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P=15cm (dist\u00e2ncia do objeto \u00e0 lente) —\u00a0\u00a0P\u2019=4P\u00a0 —\u00a0 P\u2019=4.15\u00a0 —\u00a0\u00a0P\u2019=60cm (dist\u00e2ncia da imagem \u00e0 lente)<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0o=15cm\u00a0 —\u00a0 i=+3cm (positiva porque toda imagem virtual \u00e9 direita)\u00a0 —\u00a0 P=30cm\u00a0 — i\/o=-P\u2019\/P\u00a0 —\u00a0 3\/15=- P\u2019\/30\u00a0 —\u00a0\u00a0P\u2019= – 6cm\u00a0 —\u00a0 m\u00f3dulo P\u2019=6cm<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span>\"\"<\/p>\n

1\/15=1\/S + 1\/(80 \u2013 S)\u00a0 — 1\/15=(80 \u2013 S) + S\/S.(80 \u2013 S)\u00a0 —\u00a0 1200=80S \u2013 S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 80S + 1200=0\u00a0 —\u00a0 \u0394=B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>-4.A.C\u00a0 —\u00a0 \u0394=40\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

S= – B \u00b1\u221a \u0394\/2.A\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=(80 + 40)\/2\u00a0 —\u00a0\u00a0S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=60cm\u00a0e\u00a0P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019=80 \u2013 60=20cm\u00a0\u00a0—\u00a0 S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=(80 \u2013 40)\/2\u00a0 —\u00a0\u00a0S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=20cm\u00a0e\u00a0P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2019=80 \u2013 20=60cm<\/b><\/span><\/span><\/p>\n

Observe que, para que a imagem seja real e n\u00edtida sobre a tela existem duas\u00a0 posi\u00e7\u00f5es entre elas 20cm e 60cm, mas a dist\u00e2ncia d<\/b><\/span><\/span><\/p>\n

entre essas duas posi\u00e7\u00f5es \u00e9 a mesma e vale d=60 \u2013 20\u00a0 —\u00a0\u00a0d=40cm<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0a) o=0,6cm\u00a0 —\u00a0 P=20cm\u00a0 —\u00a0\u00a0 se objeto e imagem est\u00e3o do mesmo lado, a imagem \u00e9 virtual e\u00a0 P\u2019= <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

– 100cm\u00a0\u00a0 —\u00a01\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/25 + 1\/-100\u00a0 —\u00a0\u00a0 1\/f=3\/100\u00a0 —\u00a0 f=100\/3cm, mas a converg\u00eancia C ser\u00e1 em dioptrias se f estiver em metros\u00a0 —\u00a0\u00a0 C=1\/f=1\/(1\/3)\u00a0 —\u00a0\u00a0C=3 di<\/b><\/span><\/span><\/p>\n

b) i\/o=-P\u2019\/P\u00a0 —\u00a0 i\/0,6=-(-100)\/25\u00a0 —\u00a0\u00a0i=2,4cm<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0a) 1\/f=1\/180 + 1\/36\u00a0 —\u00a0 1\/f=(1 + 5)\/180\u00a0 —\u00a0\u00a0f=30cm<\/b><\/span><\/span><\/p>\n

b) O comprimento da imagem da l\u00e2mpada \u00e9 de – 24 cm. A representa\u00e7\u00e3o geom\u00e9trica est\u00e1 representada na figura adiante.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Semelhan\u00e7a de tri\u00e2ngulos\u00a0 —\u00a0 120\/A\u2019B\u2019=180\/36\u00a0 —\u00a0\u00a0A\u2019B\u2019=24cm<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da altura da imagem formada pela lente L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/1,5=1\/2 + 1\/P\u2019\u00a0 —\u00a0 1\/1,5 \u2013 1\/2= 1\/P\u2019\u00a0 —\u00a0 P\u2019=6cm\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 —\u00a0 i\/1=-6\/2\u00a0 —\u00a0 i= – 3cm (negativa, pois \u00e9 invertida)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Lente L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 O=3cm\u00a0 i=-6cm (deve ser direita em rela\u00e7\u00e3o ao objeto, portanto invertida em rela\u00e7\u00e3o a i) —\u00a0 i\/O=-P\u2019\/P\u00a0 — – 6\/3=<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

-P\u2019\/P\u00a0 —\u00a0 P\u2019=2P\u00a0 —\u00a0 1\/1,5=1\/P + 1\/2P\u00a0 —\u00a0 P=2,25cm\u00a0 —\u00a0 x=6 + 2,25\u00a0 —\u00a0\u00a0x=8,25cm\u00a0<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0Lente L<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 f=-10cm (divergente)\u00a0 —\u00a0 P=20cm\u00a0 —\u00a0 1\/-10=1\/20 + 1\/P\u2019\u00a0 —\u00a0 -1\/10 \u2013 1\/20=1\/P\u2019\u00a0 —\u00a0 P\u2019=-20\/3cm (virtual, atr\u00e1s da lente)<\/b><\/span><\/span><\/p>\n

Lente L<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 f=10cm\u00a0 —\u00a0 P=20cm\u00a0 —\u00a0 1\/10=1\/20 + 1\/P\u2019\u00a0 —\u00a0 1\/10 \u2013 1\/20=1\/P\u2019\u00a0 —\u00a0 1\/P\u2019=(2 \u2013 1)\/20\u00a0 —\u00a0 P\u2019=20cm<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

D=60,0 \u2013 6,6\u00a0 —\u00a0 d=53,4cm\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Lente L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 A\u2019B\u2019\u00a0 —\u00a0 lente L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 A\u2019\u2019B\u2019\u2019<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A imagem final \u00e9 real (intersec\u00e7\u00e3o dos pr\u00f3prios raios luminosos), direita e maior em rela\u00e7\u00e3o ao objeto original.<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Espelho c\u00f4ncavo\u00a0 —\u00a0 1\/30=1\/60 + 1\/P\u2019\u00a0 —\u00a0 1\/30 \u2013 1\/60=1\/P\u2019\u00a0 —\u00a0 P\u2019=60cm (A\u2019B\u2019)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Lente convergente\u00a0 —\u00a0 a imagem i(E) conjugada pelo espelho funciona como objeto para a lente\u00a0 —\u00a0 P=15cm\u00a0 —\u00a0 f=12cm\u00a0 —\u00a0 1\/12=1\/15 + 1\/P\u2019\u00a0 — 1\/ 12 \u2013 1\/15=1\/P\u2019\u00a0 —\u00a0 1\/P\u2019=(5 \u2013 4)\/60\u00a0 —\u00a0 P\u2019=60cm\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

15-<\/b><\/span><\/span><\/span>\u00a0a) Trata-se de uma lente convergente (bordas finas) de \u00edndice de refra\u00e7\u00e3o em rela\u00e7\u00e3o ao ar \u2013 n<\/b><\/span><\/span>lente<\/b><\/span><\/span><\/sub>\/n<\/b><\/span><\/span>ar<\/b><\/span><\/span><\/sub>=1,35 — face convexa \u2013 Rc=2,5.10<\/b><\/span><\/span>-3\u00a0<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0\u00a0 face plana Rp=\u221e\u00a0 —\u00a0 equa\u00e7\u00e3o dos fabricantes de lentes C=1\/f=((n<\/b><\/span><\/span>lente<\/b><\/span><\/span><\/sub>\/n<\/b><\/span><\/span>ar<\/b><\/span><\/span><\/sub>\u00a0\u2013 1).(1\/R<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ 1\/R<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 C=1,35 \u2013 1,00).(1\/2,5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0+1\/\u221e)\u00a0 —\u00a0 C=(0,35).(0,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0+ 0)\u00a0 —\u00a0\u00a0C=140di<\/b><\/span><\/span><\/p>\n

b) Se a imagem \u00e9 direita\u00a0 —\u00a0 i=50.O\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0 50.O\/O=-P\u2019\/P\u00a0 —\u00a0 P\u2019=-50P\u00a0 —\u00a0 1\/f=140m\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 140=1\/P – 1\/50P\u00a0 —\u00a0 140=49\/50P\u00a0 —\u00a0P=0,007m=7mm<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span><\/span>\u00a0a) f=20cm\u00a0 —\u00a0 P=60cm\u00a0 —\u00a0 1\/20=1\/60 + 1\/P\u2019\u00a0 —\u00a0 1\/20 \u2013 1\/60 =1\/P\u2019\u00a0 —\u00a0 1\/P\u2019=(3 \u2013 1)\/60\u00a0 —\u00a0\u00a0P\u2019=30cm<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

Como a imagem \u00e9 projetada, ela \u00e9 invertida (troca cima por baixo) e reversa (troca direita pela esquerda) Figura C)<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0n<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>=1,5\u00a0 —\u00a0 n<\/b><\/span><\/span>ar<\/b><\/span><\/span><\/sub>=1,0\u00a0 —\u00a0 R<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 R<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=\u221e\u00a0 —\u00a0 equa\u00e7\u00e3o dos fabricantes de lente\u00a0 —\u00a0 1\/f=C=(n<\/b><\/span><\/span>v\u00a0<\/b><\/span><\/span><\/sub>– n<\/b><\/span><\/span>ar<\/b><\/span><\/span><\/sub>).(1\/R<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ 1\/R<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

1\/f=(1,5 \u2013 1,0).(1\/2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0+ 0)\u00a0 —\u00a0 1\/f=25m\u00a0 —\u00a0 P\u2019=-36cm=-36.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m (negativa-imagem virtual-lente divergente) —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 25=1\/P \u2013 1\/36.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 25 + 1\/36.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=1\/P\u00a0 —\u00a0 P=0,04m\u00a0 —\u00a0\u00a0 i=20.10<\/b><\/span><\/span>-2\u00a0<\/b><\/span><\/span><\/sup>m\u00a0(positiva \u2013 direita)\u00a0 —\u00a0 i\/O=-P\u2019\/P\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

20.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/O=-(-36.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)\/4.10<\/b><\/span><\/span>-2\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 O=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 — O=2cm\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a01- Correta \u2013 os raios de luz paralelos provenientes do Sol, se refratam na lupa (lente convergente) e convergem para um \u00fanico ponto que \u00e9 o foco.<\/b><\/span><\/span><\/p>\n

2- Falsa \u2013 \u00e9 direita e virtual<\/b><\/span><\/span><\/p>\n

3- Correta \u2013 f=20cm\u00a0 —\u00a0 P=10cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/20 \u2013 1\/10 =1\/P\u2019\u00a0 —\u00a0 P\u2019= -20cm\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 —\u00a0 i\/o=-(-20)\/10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i\/o=2\u00a0 —\u00a0 i=2.o\u00a0<\/b><\/span><\/span><\/p>\n

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

19-<\/b><\/span><\/span><\/span>\u00a0Uma lente convergente funciona como lupa somente se o objeto estiver entre o foco e a lente\u00a0 —\u00a0 C=1\/f\u00a0 —\u00a0 5=1\/f\u00a0 —\u00a0 f=0,2m\u00a0 —\u00a0 f=20cm\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0Inicialmente o alvo est\u00e1 no foco da lente, pois a imagem \u00e9 n\u00edtida\u00a0 —\u00a0 aproxima-se a vela da lente at\u00e9\u00a0 P=3f\/2\u00a0 —\u00a0 P\u2019=?\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/1,5f +1\/P\u2019\u00a0 —\u00a0 1\/f \u2013 1\/1,5f=1\/P\u2019\u00a0 —\u00a0\u00a0 0,5\/1,5f=1\/P\u2019\u00a0 —\u00a0 P\u2019=3f\u00a0\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/p>\n

21-\u00a0a) Como a imagem est\u00e1 projetada no anteparo, ela \u00e9 real. O objeto tamb\u00e9m \u00e9 real. Conclu\u00edmos que se trata de uma lente convergente.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

1\/f=1\/P + 1\/P\u2019\u00a0\u00a0 —\u00a0 1\/f=1\/60 + 1\/20\u00a0 —\u00a0 1\/f= (1 + 3)\/60\u00a0\u00a0—\u00a0 f=15cm\u00a0<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a01\/f=1\/08f + 1\/P\u2019\u00a0 —\u00a0 1\/f \u2013 1\/08f =1\/P\u2019\u00a0 —\u00a0 (0,8 \u2013 1,0)\/0.8f=1\/P\u2019\u00a0 —\u00a0 P\u2019=-4f\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 —\u00a0 i\/1,6=-(-4f)\/0,8f\u00a0 —\u00a0\u00a0i=8mm<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Como a imagem \u00e9 projetada, ela \u00e9 invertida\u00a0 —\u00a0 A=-20=-P\u2019\/P\u00a0 —\u00a0 P\u2019=20P\u00a0 —\u00a0 1\/f=1\/P + 1\/20P\u00a0 —\u00a0 1\/d=1\/P + 1\/20P\u00a0 —\u00a0 20P=21d\u00a0\u00a0 —\u00a0 P\u2019=20P=21d\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0a) i=2.o\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 —\u00a0 2.o\/o=-P\u2019\/P\u00a0 —\u00a0 P\u2019=-2P\u00a0 —\u00a0 1\/f=1\/P \u2013 1\/2P\u00a0 —\u00a0 1\/f=(2 \u2013 1)\/2P\u00a0 —\u00a0 f=2P\u00a0 —\u00a0\u00a0f\/P=2<\/b><\/span><\/span><\/p>\n

b) Com \u00e1gua\u00a0 —\u00a0 f<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=2,5f\u00a0 —\u00a0 P \u00e9 o mesmo e vale\u00a0 —\u00a0\u00a0 P=f\/2\u00a0 —\u00a0 1\/f<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=1\/P + 1\/P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u2019\u00a0 —\u00a0 1\/2.5f=1\/(f\/2) + 1\/P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u2019\u00a0 —\u00a0\u00a0 1\/2,5f \u2013 2\/f=1\/P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u2019\u00a0 —\u00a0 1\/P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u2019 =(1 \u2013 5)\/2,5f\u00a0 —\u00a0 P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u2019 = – 0,625f\u00a0 —\u00a0 A= – P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u2019\/P\u00a0 —\u00a0 A= – (-0,625f)\/f\/2\u00a0 —\u00a0\u00a0A=1,25\u00a0<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0Dados\u00a0 —\u00a0\u00a0 P = 30 cm\u00a0 —\u00a0 f = 10 cm\u00a0 —\u00a0\u00a0 o = 6 cm.<\/b><\/span><\/span><\/p>\n

a) 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/10=1\/30 + 1\/P\u2019\u00a0 —\u00a0 (3 \u2013 1)\/30=1\/P\u2019\u00a0 —\u00a0\u00a0P\u2019=15cm\u00a0 —\u00a0 essa imagem real (p\u2019 > 0) da vela funciona como objeto real para o espelho plano, que fornece uma segunda imagem, virtual e sim\u00e9trica conforme voc\u00ea pode observar na figura<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0fornecida onde a dist\u00e2ncia D da imagem final da vela at\u00e9 a mesma vale\u00a0 —\u00a0 \u00a0D = 30 + 20 + 5\u00a0 —\u00a0 \u00a0D = 55 cm.<\/b><\/span><\/span><\/p>\n

b) O altura da imagem da vela fornecida pelo espelho plano \u00e9 igual a altura da imagem fornecida pela lente, pois a imagem formada no espelho plano tem o mesmo tamanho que o objeto\u00a0 —\u00a0 equa\u00e7\u00e3o do aumento linear transversal\u00a0 —\u00a0 i\/o=-P\u2019\/P\u00a0 —<\/b><\/span><\/span><\/p>\n

i\/6=-15\/30\u00a0 —\u00a0 i= – 3cm\u00a0 —\u00a0\u00a0a imagem \u00e9 invertida e tem altura de 3 cm.\u00a0<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0A figura mostra a constru\u00e7\u00e3o da imagem:<\/b><\/span><\/span>\"\"<\/p>\n

Observe na figura acima que os tri\u00e2ngulos sombreados s\u00e3o semelhantes\u00a0 —\u00a0 2h\/h=P\/P\u2019\u00a0 —\u00a0 P=2P\u2019\u00a0 —\u00a0 P + P\u2019+=90\u00a0 —\u00a0 2P\u2019 + P\u2019+=90\u00a0 —\u00a0 P\u2019=30cm\u00a0 —\u00a0 P=60cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/60 + 1\/30\u00a0 —\u00a0 f=60\/3\u00a0 —\u00a0 f=20cm\u00a0 —\u00a0 C=1\/f\u00a0 —\u00a0 C=1\/0,2\u00a0 —\u00a0 C=5di<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0Uma lente esf\u00e9rica \u00e9 delgada quando sua espessura \u00e9 desprez\u00edvel, em rala\u00e7\u00e3o aos raios de curvatura das faces. Ela pode ser de bordas delgadas (finas) ou de bordas grossas, como representado abaixo.<\/b><\/span><\/span>\"\"<\/p>\n

(01) Errada\u00a0 —\u00a0 se o \u00edndice de refra\u00e7\u00e3o do material que constitui a lente \u00e9 maior que o \u00edndice de refra\u00e7\u00e3o do meio, lentes delgadas de bordas delgadas s\u00e3o convergentes e lentes delgadas de bordas grossas s\u00e3o divergentes.<\/b><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0 trata-se do Teorema da Verg\u00eancia.<\/b><\/span><\/span><\/p>\n

(04) Correta.<\/b><\/span><\/span><\/p>\n

(08) Correta.<\/b><\/span><\/span><\/p>\n

(16) Correta.\u00a0<\/b><\/span><\/span><\/p>\n

R- (02 + 04 + 08 + 16) = 30<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0a) Falsa \u00a0—\u00a0 a lente usada para proje\u00e7\u00f5es de imagens (de objetos reais) \u00e9 convergente, e para corre\u00e7\u00e3o de miopia utiliza-se lente divergente.<\/b><\/span><\/span><\/p>\n

b) Falsa\u00a0 — iImagens virtuais n\u00e3o s\u00e3o projet\u00e1veis.<\/b><\/span><\/span><\/p>\n

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

d) Falsa\u00a0 —\u00a0 as faces dos prismas s\u00e3o espelhos planos, fornecendo imagens de mesmo tamanho.<\/b><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 A lupa fornece imagem virtual, n\u00e3o podendo ser projetada. \u00a0<\/b><\/span><\/span><\/p>\n

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

29-<\/b><\/span><\/span><\/span>\u00a0Dados\u00a0 —\u00a0 \u00a0y = 4 cm\u00a0 —\u00a0\u00a0 y\u2019 = 1 cm\u00a0 —\u00a0 \u00a0p = d = 3 cm\u00a0 —\u00a0 \u00a0p\u2019 = D = 150 cm\u00a0 —\u00a0\u00a0 T = 250 \u00b0C\u00a0 —\u00a0 calculando o aumento linear transversal (em m\u00f3dulo), antes do aquecimento.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Depois do aquecimento, o aumento linear \u00e9 o mesmo, pois n\u00e3o se alteram as posi\u00e7\u00f5es do objeto e da imagem\u00a0 —\u00a0 os novos comprimentos da imagem e do objeto s\u00e3o, respectivamente: (y\u2019 + \u2206y\u2019) e (y + \u2206y)\u00a0 —\u00a0 aplicando novamente a equa\u00e7\u00e3o do aumento\u00a0 —\u00a0 A=(Y\u2019+ \u2206Y)\/(Y + \u2206Y)\u00a0 —\u00a0 substituindo valores\u00a0 —\u00a0 50=(200 + 1)\/(4 + \u2206Y)\u00a0 —\u00a0 \u2206y=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>cm\u00a0 —\u00a0 \u2206y \u00e9 a dilata\u00e7\u00e3o sofrida pelo objeto\u00a0 —\u00a0\u00a0 \u2206y=Y\u03b1 \u2206T\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=4\u03b1250\u00a0 —\u00a0\u03b1=2,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/p>\n

30-<\/b><\/span><\/span><\/span>Determina\u00e7\u00e3o gr\u00e1fica da imagem\u00a0 —\u00a0 LM \u2013 objeto\u00a0 —\u00a0 L\u2019M\u2019 \u2013 imagem:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

C\u00e1lculo da coordenada P\u2019 das imagens M\u2019 e L\u2019 pela equa\u00e7\u00e3o dos pontos conjugados\u00a0 —\u00a0 1\/f=1\/P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0+ 1\/P\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/-20=1\/25 + 1\/P\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>= – 100\/9cm\u00a0 —\u00a0 1\/f=1\/P<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0+ 1\/P\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 1\/-20=1\/35 + 1\/P\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>= – 140\/11cm\u00a0 —\u00a0 c\u00e1lculo da coordenada Y\u2019 das imagens M\u2019 e L\u2019 pela equa\u00e7\u00e3o do aumento linear transversal\u00a0 —\u00a0 Y\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/Y<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>= – P\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Y\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/5= – ( – 140\/11)\/35\u00a0 —\u00a0 Y\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=20\/11cm\u00a0 —<\/b><\/span><\/span><\/p>\n

Y\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\/Y<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>= – P\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Y\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\/15= – ( – 100\/9)\/25\u00a0 —\u00a0 Y\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=20\/3cm\u00a0 —\u00a0 Comprimento da imagem da barra\u00a0 —\u00a0 c<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= (P\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 P\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (Y\u2019<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 Y\u2019<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 c<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>={ – 100\/9 \u2013 ( – 140\/11)}<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (20\/3 \u2013 20\/11)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 c<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u2248256.000\/9.800\u00a0 —\u00a0 c<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u224826\u00a0 —\u00a0\u00a0c\u22485,1cm<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0I. Falsa\u00a0 —\u00a0 a moeda n\u00e3o est\u00e1 na posi\u00e7\u00e3o vista aparentemente devido ao fen\u00f4meno da refra\u00e7\u00e3o, que desvia os raios luminosos.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 voc\u00ea pode acender o palito de f\u00f3sforo colocando a cabe\u00e7a dele no foco, ponto de encontro dos raios solares refratados pela lente convergente.<\/b><\/span><\/span><\/p>\n

III. Correta.<\/b><\/span><\/span><\/p>\n

IV. Correta\u00a0 —\u00a0 o n\u00famero de imagens (n) fornecidas pela associa\u00e7\u00e3o de dois espelhos planos \u00e9 dado por:<\/b><\/span><\/span><\/p>\n

N=360\/\u03b8 – 1, sendo \u03b8 o \u00e2ngulo formado entre os espelhos\u00a0 —\u00a0 se os espelhos s\u00e3o colocados paralelamente entre si, \u03b8= 0\u00ba\u00a0 —\u00a0 ent\u00e3o n tende para infinito.\u00a0<\/b><\/span><\/span><\/p>\n

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

32-<\/b><\/span><\/span><\/span>\u00a0Observe na figura abaixo que P=7,5cm\u00a0 —\u00a0 f=+5cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/5=1\/7,5 + 1\/P\u2019\u00a0 —\u00a0 P\u2019=37,5\/2,5\u00a0 —\u00a0 P\u2019=15cm\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P\u2019 > 0\u00a0 —\u00a0 imagem real\u00a0 —\u00a0 aumento linear transversal\u00a0 —\u00a0 A= – P\u2019\/P= – 15\/7,5\u00a0 —\u00a0 A= – 2\u00a0 —\u00a0 como A < 0 a imagem \u00e9 invertida e como \u2502A\u2502=2, a altura da imagem \u00e9 o dobro da altura do objeto\u00a0 —\u00a0 observe a constru\u00e7\u00e3o geom\u00e9trica abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0Lente convergente\u00a0 —\u00a0 f=+4cm\u00a0 —\u00a0 a imagem \u00e9 direita (aumento positivo) e de tamanho tr\u00eas vezes maior que o do objeto\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

A=+3<\/b><\/span><\/span><\/p>\n

a) A= – P\u2019\/P\u00a0 —\u00a0 3= – P\u2019\/P\u00a0 —\u00a0 P\u2019= – 3P\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0\u00a0 1\/4 = 1\/P \u2013 1\/3P\u00a0 —\u00a0 1\/4 = 2\/3P\u00a0 —\u00a0\u00a0P=8\/3 cm (dist\u00e2ncia do objeto \u00e0 lente)<\/b><\/span><\/span><\/p>\n

b) P\u2019 = – 3P\u00a0 —\u00a0 P\u2019= – 3.(8\/3)\u00a0 —\u00a0 P\u2019= – 8 cm\u00a0 —\u00a0 como a imagem \u00e9 virtual, ela \u00e9 direita e se encontra a 8cm do centro \u00f3ptico da lente, e do mesmo lado que o objeto\u00a0 —\u00a0 imagem direita\u00a0 —\u00a0 A > 0\u00a0 —\u00a0 A= i\/o\u00a0 —\u00a0 3=i\/0,7\u00a0 —\u00a0\u00a0i=+2,1cm (tamanho da imagem)<\/b><\/span><\/span><\/p>\n

—\u00a0 observe a constru\u00e7\u00e3o geom\u00e9trica da imagem na figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

34-<\/b><\/span><\/span><\/span> R- (01 + 04 + 16)=21\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

 <\/p>\n

35-<\/b><\/span><\/span><\/span> Observe na figura abaixo\u00a0 —\u00a0 o=altura do objeto=1cm\u00a0 —\u00a0 i=altura da imagem= – 0,4cm (invertida em rela\u00e7\u00e3o ao<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

objeto)\u00a0 —\u00a0 dist\u00e2ncia do objeto \u00e0 lente=P=d\u00a0 —\u00a0 dist\u00e2ncia da imagem \u00e0 lente=P\u2019\u00a0 —\u00a0 P + P\u2019 = 56cm\u00a0 —\u00a0 P=56 \u2013 P\u2019\u00a0 — i\/o = – P\u2019\/P\u00a0 —\u00a0\u00a0 – 0,4\/1 = – P\u2019\/(56 \u2013 P\u2019)\u00a0 —\u00a0 0,4.56 \u2013 0,4P\u2019=P\u2019\u00a0 —\u00a0 1,4P\u2019=22,4\u00a0 —\u00a0 P\u2019=16cm\u00a0 —\u00a0 P + P\u2019=56\u00a0 —\u00a0P + 16=56\u00a0 —\u00a0\u00a0P=d=40cm.<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span> \u00a0Lupa \u2013\u00a0Tamb\u00e9m chamada de lente de aumento \u00e9 uma simples lente convergente que fornece de um objeto colocado entre seu foco F e seu centro \u00f3ptico O uma imagem virtual, direita e maior que o objeto observado\u00a0 —\u00a0 Observe no esquema abaixo a<\/b><\/span><\/span><\/p>\n

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

forma\u00e7\u00e3o da imagem A\u2019B\u2019 de um objeto AB em uma lupa\u00a0 —\u00a0\u00a0 observe\u00a0 que a imagem \u00e9\u00a0virtual\u00a0e assim, nas equa\u00e7\u00f5es\u00a0 —\u00a0\u00a0 1\/f =<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

1\/P + 1\/P\u2019\u00a0 —\u00a0 i\/o = -P\u2019\/P\u00a0 —\u00a0\u00a0 A = i\/o = -P\u2019\/P\u00a0 —\u00a0 P\u2019 deve e ser substitu\u00edda com sinal negativo, \u00a0pois\u00a0P\u2019<\u00a00 (a imagem \u00e9 virtual)\u00a0 — dados do exerc\u00edcio\u00a0 —\u00a0 f=+10cm (lente convergente)\u00a0 —\u00a0 i=+100 (toda imagem virtual \u00e9 direita)\u00a0 —\u00a0 i\/o = – P\u2019\/P\u00a0 —\u00a0 100.o\/o = – P\u2019\/P\u00a0 —\u00a0 P\u2019 = – 10P\u00a0 —\u00a0 equa\u00e7\u00e3o dos pontos conjugados de Gauss\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/10=1\/P + 1\/(- 10P)\u00a0 —\u00a0 1\/10=1\/P \u2013 1\/10P\u00a0 —<\/b><\/span><\/span><\/p>\n

P=10 \u2013 1\u00a0 —\u00a0 P = 9cm\u00a0 (dist\u00e2ncia entre a lupa e o objeto que \u00e9 a impress\u00e3o digital) —\u00a0\u00a0R- A\u00a0<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span> Como os raios de luz incidem paralelamente na superf\u00edcie da lupa (lente convergente), eles se refratam convergindo para o foco f formando o ponto luminoso\u00a0 —\u00a0 f=20cm\u00a0 —\u00a0 o celular \u00e9 o objeto que est\u00e1 a 15cm da lupa\u00a0 —\u00a0 P=15cm\u00a0 —\u00a0 1\/20 = 1\/15 + 1\/P\u2019\u00a0 —\u00a0 (3 \u2013 4)\/60=1\/P\u2019\u00a0 —\u00a0 P\u2019= -60cm (imagem virtual P\u2019<0)\u00a0 —\u00a0 i\/o= – P\u2019\/P\u00a0 —\u00a0 i\/o= – (-60)\/15\u00a0 —\u00a0 i\/o=4\u00a0 —\u00a0 A=4\u00a0 —\u00a0 imagem direita (A>0) e 4 vezes maior que o objeto\u00a0 —\u00a0\u00a0R- C\u00a0\u00a0—\u00a0 observa\u00e7\u00e3o\u00a0 —\u00a0 como o objeto est\u00e1 entre o foco e a lente convergente<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(lupa), voc\u00ea poderia tra\u00e7ar os raios de luz e caracterizar a imagem obtida\u00a0 — natureza:\u00a0virtual (obtida no cruzamento dos prolongamentos dos raios luminosos)\u00a0 —\u00a0 localiza\u00e7\u00e3o:\u00a0antes do foco\u00a0 —\u00a0 tamanho e orienta\u00e7\u00e3o:\u00a0maior que o objeto e direita em rela\u00e7\u00e3o a ele.<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span> Observando o gr\u00e1fico fornecido voc\u00ea notar\u00e1 que, quando o objeto estiver a 20cm do centro \u00f3ptico da lente (P=20cm), a imagem estar\u00e1 tamb\u00e9m a 20cm do centro \u00f3ptico da lente (P\u2019=20cm)\u00a0\u00a0 —\u00a0 essas duas posi\u00e7\u00f5es correspondem aos dois pontos antiprincipais da lente, que correspondem ao dobro da dist\u00e2ncia focal\u00a0 —\u00a0 2f=20\u00a0 —\u00a0 f=10cm\u00a0 —\u00a0 analisando cada alternativa:<\/b><\/span><\/span><\/p>\n

I. Falsa\u00a0 —\u00a0 a verg\u00eancia C de uma lente corresponde ao inverso de sua dist\u00e2ncia focal, medida em metros\u00a0 —\u00a0 C=1\/f=1\/0,1\u00a0 — C=10 dioptrias (di)<\/b><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 quando o objeto estiver entre 0 e 10cm do centro \u00f3ptico da lente, ele estar\u00e1 entre o foco e a lente e a imagem<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

conjugada ser\u00e1 virtual, direita e maior em rela\u00e7\u00e3o ao objeto, como voc\u00ea pode observar na figura acima.<\/b><\/span><\/span><\/p>\n

III. Verdadeira\u00a0 —\u00a0 dados\u00a0 —\u00a0 P=50cm\u00a0 —\u00a0 f=10cm\u00a0 —\u00a0 P\u2019=?\u00a0 —\u00a0 equa\u00e7\u00e3o dos pontos conjugados\u00a0 —\u00a0 1\/f = 1\/P + 1\/P\u2019\u00a0 —<\/b><\/span><\/span><\/p>\n

1\/10 = 1\/50 + 1\/P\u2019\u00a0 —\u00a0 1\/p\u2019<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>= 4\/50\u00a0 —\u00a0 p\u2019 = 12,5cm\u00a0 —\u00a0 aumento linear transversal\u00a0 —\u00a0 A= – P\u2019\/P= – 12,5\/50\u00a0 —\u00a0 A= – 1\/4 ( o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sinal negativo mostra que a imagem \u00e9 invertida em rela\u00e7\u00e3o ao objeto\u00a0 —\u00a0 veja figura acima\u00a0 —\u00a0\u00a0R- B<\/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 das Lentes Esf\u00e9ricas \u00a0 \u00a0 01-\u00a0a) Lente convergente \u2013 lentes de vidro no ar, de bordas finas s\u00e3o convergentes ou, observe que , ap\u00f3s se refratarem na lente os raios de luz convergem para o eixo principal. b) P=15cm\u00a0 —\u00a0 P\u2019=10cm\u00a0 —\u00a0 1\/f=1\/P + 1\/P\u2019\u00a0 —\u00a0 1\/f=1\/15 + 1\/10\u00a0 —\u00a0 1\/f=(10<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2269,"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-2273","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2273","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=2273"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2273\/revisions"}],"predecessor-version":[{"id":10937,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2273\/revisions\/10937"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2269"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2273"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}