[tex]dane:\\f = 8 \ cm\\p = 2\\szukane:\\x = ?\\\\Rozwiazanie\\\\p = \frac{y}{x}\\\\2 = \frac{y}{x}\\\\y = 2x\\\\\\\frac{1}{f} = \frac{1}{x}+\frac{1}{y}\\\\y = 2x\\\\\frac{1}{f}=\frac{1}{x}+\frac{1}{2x}\\\\\frac{1}{f} = \frac{2}{2x}+\frac{1}{2x}\\\\\frac{1}{f} = \frac{3}{2x}\\\\2x = 3f \ \ /:2\\\\x = \frac{3}{2}f\\\\x = \frac{3}{2}\cdot 8 \ cm\\\\x = 12 \ cm[/tex]
Odp. Przedmiot należy ustawić w odległości 12 cm od zwierciadła.
