• # question_answer In Ramsden eyepiece, the two planoconvex lenses each of focal length $f$ are separated by a distance$12cm$. The equivalent focal length (in $cm$) of the eyepiece is A)  10.5                                       B)  12.0 C)  13.5                                       D)  15.5

$d=\frac{2f}{3}$ or            $f=\frac{3d}{2}=\frac{3\times 12}{2}=18cm$ Equivalent focal length is $f=\frac{{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}}=\frac{18\times 18}{18+18}+\frac{18}{4}$      $=9+4.5$      $=13.5\,\,cm$