Question

In an astronomical telescope in normal adjustment a straight black line of length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $I$ . The magnification of the telescope is

• A. $\frac{L+I}{ L-I}$
• B. $\frac {L}{I}$
• C. $\frac{L}{I}+ I$
• D. $\frac{L}{I}- I$

Answer 1

Answer: (B)

$\frac {L}{I}$

Answer 2

(b): The situation is shown in the figure.
Objective
Let $f _o \, \, and \, \, f _e$ be the focal lengths of the objective
and eyepiece respectively.
For normal adjustment distance of the objective
from the eyepiece (tube length) = $f _o + / f_e$
Treating the line on the objective as the object
and eyepiece as the lens.
$\therefore \, \, u = -(f_o + f_c ) \, \, and \, \, f= f_e$
As $\frac{1}{v} + \frac{1}{u} =\frac{1}{f}$
$\therefore \, \, \, \, \frac{1}{v}- \frac{1}{- (f_o+f_e)}=\frac{1}{f_e}$
$\frac{1}{v} = \frac{1}{f_e} - \frac{1}{ f_o+ f_e} = \frac{f_o +f_e -f_e}{ f_e(f_o+f_e)}= \frac{f_o}{f_e(f_o+f_e)}$
or $v= \frac{ f_e(f_o+f_e)}{f_o}$
Thus $\frac{I}{L} = \frac{v}{u} = \frac{f_o}{ f_o +f_e} =\frac{f_e}{f_o}$
or $\frac{f_o}{f_e} = \frac{L}{I}$ \hspace30mm .......(i)
$\therefore$ The magnification of the telescope in normal
adjustment is
$m= \frac{ f_o}{f_e} = \frac{ L}{I}$ \hspace30mm (using (i))