[tex]dane:\\x = 36 \ cm\\p = \frac{1}{3}\\szukane:\\f = ?\\Z = ?[/tex]
Rozwiazanie
Długość ogniskowej :
[tex]p = \frac{y}{x}\\\\\frac{1}{3} = \frac{y}{x}\\\\3y = x \ \ /:3\\\\y = \frac{x}{3}\\\\\\\frac{1}{f} = \frac{1}{x} + \frac{1}{y}\\\\y = \frac{x}{3}\\\\\frac{1}{f} = \frac{1}{x} + \frac{3}{x}\\\\\frac{1}{f} = \frac{4}{x}\\\\4f = x \ \ /:4\\\\f = \frac{1}{4}x = \frac{1}{4}\cdot 36 \ cm\\\\f = 9 \ cm = 0,09 \ m[/tex]
Zdolność skupiająca ogniskowej Z:
[tex]Z = \frac{1}{f}\\\\Z = \frac{1}{0,09 \ m}\\\\Z = 11,1 \ D \ \ (dioptrii)[/tex]
