[tex]dane:\\Z = 2 \ D\\y = 6 \ m\\szukane:\\x = ?\\\\Rozwiazanie\\\\Z = \frac{1}{f} \ \ \rightarrow \ \ f = \frac{1}{Z}\\\\f = \frac{1}{2} \ m = 0,5 \ m\\\\\\Z \ rownania \ soczewki\\\\\frac{1}{f} = \frac{1}{x}+\frac{1}{y}\\\\\frac{1}{x} = \frac{1}{f}-\frac{1}{y}\\\\\frac{1}{x} = \frac{y-f}{fy}\\\\x = \frac{fy}{y-f} = \frac{0,5 \ m\cdot6 \ m}{6 \ m - 0,5 \ m} =\frac{3 \ m^{2}}{5,5 \ m}\\\\x = 0,(54) \ m \approx0,55 \ m[/tex]
