Question

An astronomical telescope of ten-fold angular magnification has a length of 44cm. The focal length of the objective is:
(A) 44cm
(B) 440cm
(C) 40cm
(D) 4cm

Answer 1

Hint: The length of a telescope is given by the sum of the focal lengths of the eyepiece and the objective. The focal length of the eyepiece is expressed in terms of the focal length of the objective. We can compute this value in the formula for angular magnification to get the answer.

Formula Used: The formulae used in the solution are given here.
 $ L = \left| {{f_o}} \right| + \left| {{f_e}} \right| $ where $ L $ is the length of the telescope, $ {f_0} $ is the distance of the focus of objective from origin and $ {f_e} $ is the focal length of the eyepiece.
Angular magnification of the telescope is given by, $ \left| m \right| = \dfrac{{{f_0}}}{{{f_e}}} $ where $ {f_0} $ is the distance of the focus of objective from origin and $ {f_e} $ is the focal length of the eyepiece.

Complete step by step answer:
An astronomical telescope is one that has an objective with a long focal length and an eyepiece with a short focal length, usually used for observing celestial bodies.
The astronomical telescope makes use of two positive lenses: the objective, which forms the image of a distant object at its focal length, and the eyepiece, which acts as a simple magnifier with which to view the image formed by the objective. Its length is equal to the sum of the focal lengths of the objective and eyepiece, and its angular magnification is given by, $ \left| m \right| = \dfrac{{{f_0}}}{{{f_e}}} $ where $ {f_0} $ is the distance of the focus of objective from origin and $ {f_e} $ is the focal length of the eyepiece.
The length of the telescope is given by $ L = \left| {{f_o}} \right| + \left| {{f_e}} \right| $ where $ L $ is the length of the telescope, $ {f_0} $ is the distance of the focus of objective from origin and $ {f_e} $ is the focal length of the eyepiece.
It has been given that an astronomical telescope of ten-fold angular magnification has a length of 44cm.
Since length of the telescope is 44cm, we have,
 $ \left| {{f_o}} \right| + \left| {{f_e}} \right| = L = 44cm $
 $ \Rightarrow 44cm = \left| {{f_o}} \right| + \left| {{f_e}} \right| $
The focal length of the eyepiece is given by, $ \left| {{f_e}} \right| = 44 - \left| {{f_o}} \right| $ .
Angular magnification of the telescope is tenfold. Thus,
 $ 10 = \dfrac{{{f_0}}}{{{f_e}}} $
Substituting the value of $ {f_e} $ in the equation for angular magnification, we get,
 $ 10 = \dfrac{{{f_0}}}{{44 - {f_0}}} $
Simplifying the equation and rearranging the terms, we get,
 $ \Rightarrow 440 - 10{f_0} = {f_0} $
 $ \Rightarrow {f_0} = \dfrac{{440}}{{11}} = 40cm $
Hence, the focal length of the objective is 40cm. The correct answer is Option C.

Note:
The astronomical telescope can be used for terrestrial viewing, but seeing the image upside down is a definite inconvenience. Viewing stars upside down is no problem. Another inconvenience for terrestrial viewing is the length of the astronomical telescope, equal to the sum of the focal lengths of the objective and eyepiece lenses. A shorter telescope with upright viewing is the Galilean telescope.