[tex]Dane:\\p = \frac{1}{2}\\x = 10 \ cm\\Szukane:\\f = ?[/tex]
Rozwiązanie
Z powiększenia:
[tex]p = \frac{y}{x}\\\\\frac{1}{2} =\frac{y}{x}\\\\2y = x \ \ /:2\\\\\underline{y = \frac{x}{2}}[/tex]
Z równania soczewki:
[tex]\frac{1}{f} = \frac{1}{x}+\frac{1}{y}\\\\y = \frac{x}{2}\\\\\frac{1}{f} = \frac{1}{x}+\frac{2}{x}\\\\\frac{1}{f} = \frac{3}{x}\\\\f = \frac{x}{3} = \frac{10 \ cm}{3}\\\\\boxed{f = 3\frac{1}{3} \ cm\approx3,3 \ cm} \ - \ dlugosc \ ogniskowej[/tex]
