Alguém me ajuda?
Maria deseja comprar um espelho para se maquiar. Ela quer que sua imagem seja ampliada 1,5 vezes quando ela estiver a 20cm do espelho. As características que devem ter esse espelho são
a) côncavo com raio de curvatura igual a 24cm
b) côncavo com raio de curvatura igual a 120cm
c) convexo com raio de curvatura igual a 120cm
d) convexo com foco igual a 12cm
e) côncavo com foco igual a 12cm
Com cálculo
Gab: b
Krikor:
Tem alguma coisa errada
Soluções para a tarefa
Respondido por
48
Olá!

Descobrindo o foco:

Logo:

O espelho é concavo porque a imagem é ampliada.
Espero ter ajudado!
Descobrindo o foco:
Logo:
O espelho é concavo porque a imagem é ampliada.
Espero ter ajudado!
Anexos:

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