Q. In Ramsden eyepiece, the two planoconvex lenses each of focal length $f$are separated by a distance 12 cm. The equivalent focal length (in cm) of the eyepiece is
A) 10.5
B) 12.0
C) 13.5
D) 15.5
Explanation
(c) $d = \frac{2f}{3}$
$or\ f = \frac{3d}{2} = \frac{3 \times 12}{2} = \ 18\ cm$
Equivalent focal length is
$f' = \frac{f_{1}f_{2}}{f_{1} + f_{2}} + \frac{f}{4} = \frac{18 \times 18}{18 + 18} + \frac{18}{4}$
= 9 + 4.5 = 13.5 cm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.