74. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is :

74. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is :

(a) \( \dfrac{L}{l} - 1 \)

(b) \( \dfrac{L + l}{L - 1} \)

(c) \( \dfrac{L}{l} \)

(d) \( \dfrac{L}{I} + 1 \)

Answer :  (c) \( \dfrac{L}{I} \)

Concept / Approach: Using the information given for eye piece, we can find a relation between L, l, $f_o, f_e$ and then we can find the magnification of telescope. Also remember from the construction of telescope that length of telescope is $f_o + f_e$ which is object distance for eye piece in this problem.

\( M_{\text{lens}} = \dfrac{h'}{h} =\dfrac{v}{u}\)

\( \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \)

\( M_{\text{lens}} = \dfrac{f}{f+u} \)

\( M_e= \dfrac{f_e}{f_e - [f_e + f_o ] } \)

\( \dfrac{-l}{+L} = \dfrac{f_e}{-f_o} \)

\( \dfrac{f_o}{f_e} = \dfrac{L}{l} \)

In normal adjustment, angular magnification of telescope,

\( M_{\text{telescope}} = \dfrac{f_o}{f_e} = \dfrac{L}{l} \)