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 . The magnification of the telescope is

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Solution

Q.
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 $l$ . The magnification of the telescope is [2015]

1 $\frac{L + I}{L - I}$

2 $\frac{L}{I}$

3 $\frac{L}{I} + 1$

4 $\frac{L}{I} - 1$

Ans.
(2)

For eye-piece lens,

$m = \frac{f}{f + u} = \frac{h_{1}}{h_{0}} \Rightarrow \frac{f_{e}}{f_{e} - (f_{o} + f_{e})} = \frac{I}{L}$

$\Rightarrow \frac{f_{o}}{f_{e}} = - \frac{L}{I} = Magnification of the telescope$

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