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

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$\frac{L}{l}$

$\frac{L}{l} + 1$

$\frac{L}{l} - 1$

$\frac{L + l}{L - l}$

answer is A.

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Detailed Solution

The angular magnification is given by m= $\frac{f_{0}}{f_{e}}$

$\begin{array}{l} u = - (f_{0} + f_{e}) \\ S o, u s i n g l e n s f o r m u l a, \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{- (f_{0} + f_{e})} = \frac{1}{f_{e}} \\ v = \frac{f_{e} (f_{0} + f_{e})}{f_{0}} F u r t h e r, \frac{h_{i}}{h_{0}} = \frac{v}{u} \\ \Rightarrow \frac{- l}{L} = \frac{\frac{f_{e} (f_{0} + f_{e})}{f_{0}}}{- (f_{0} + f_{e})} \Rightarrow \frac{f_{0}}{f_{e}} = \frac{L}{l} \Rightarrow m = \frac{L}{l} \end{array}$