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

The Correct Option is B

(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))