In an astronomical telescope in normal adjustment a straight black… | NEET 2015 (AIPMT)

Q42NEET 2015 · AIPMTDifficult

Optical Instruments & the Eye

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

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Answer: (b)

The eyepiece images the objective (the line), which is at $u = f_o + f_e$ from it.
$\dfrac{1}{v} = \dfrac{1}{f_e} - \dfrac{1}{f_o + f_e} \Rightarrow v = \dfrac{f_e(f_o + f_e)}{f_o}$ .
Linear magnification $\dfrac{I}{L} = \dfrac{v}{u} = \dfrac{f_e}{f_o}$ .
Telescope magnification $M = \dfrac{f_o}{f_e} = \dfrac{L}{I}$ .
So the answer is (b) $\dfrac{L}{I}$ .

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