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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
$\frac{L}{I}$
B
$\frac{L}{I} + 1$
C
$\frac{L}{I} - 1$
D
$\frac{L + I}{L - I}$
Detailed Solution
In normal adjustment the objective is at distance $u = -(f_0 + f_e)$ from the eyepiece.
Magnification by the eyepiece: $\frac{I}{L} = \frac{f_e}{f_e + u} = \frac{f_e}{f_e - (f_0 + f_e)} \Rightarrow \frac{I}{L} = \frac{f_e}{f_0}$ (in magnitude)
$M = \frac{f_0}{f_e} = \frac{L}{I}$