Q. $F_{1}$and$F_{2}$ are focal lengths of objective and eyepiece respectively of the telescope. The angular magnification for the given telescope is equal to
A) $\frac{F_{1}}{F_{2}}$
B) $\frac{F_{2}}{F_{1}}$
C) $\frac{F_{1}F_{2}}{F_{1} + F_{2}}$
D) $\frac{F_{1} + F_{2}}{F_{1}F_{2}}$
Explanation
(a) Magnifying power of a telescope having objective of focal length ($f_{0}$) and image distance $(u_{e})$ is
$M = - \frac{f_{0}}{u_{e}}$
To see with relaxed eye final image should be formed at infinity.
The distance between the objective and eyepiece is so adjusted the image is formed at the focus of the eyepiece.
Substituting $v_{e} = f_{e},$ we get
$|M| = \frac{f_{0}}{f_{e}} = \frac{F_{1}}{F_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.