Question

when two lenses of focal length +10cm and -10cm are placed in contact then find the net power?

Answer 1

$f_1=+10\ cm\\ f_2=-10\ cm\\f=focal\ length\ of\ the\ combination\\ \\ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \\ \\ \Rightarrow \frac{1}{f} = \frac{1}{10} + \frac{1}{-10} \\ \\ \Rightarrow \frac{1}{f} = \frac{1}{10} - \frac{1}{10} \\ \\ \Rightarrow \frac{1}{f} =0 \\ \\ Since\ Power= \frac{1}{f};\ Power\ is\ 0.$