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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 :
Option: 1
$\frac{\text{L}} {\text{I}} - 1$

Option: 2
$\frac{{\text{L} + \text{I}}} {{\text{L} - \text{I}}}$

Option: 3
$\frac{\text{L}} {\text{I}}$

Option: 4
$\frac{\text{L}} {\text{I}} + 1$

Answers (1)

best_answer

As we learnt in

Length of Astronomical Telescope -

$L=f_{o}+frac{f_{e}, D}{f_{e}+ D}$

- wherein

$f_{o}=$ Focal length of objective.

$f_{e}=$ Focal length of eyepiece

Magnification of telescope is

$M=\frac{fo}{fe}$

Here

$\frac{fe}{fe+u}$ $=\frac{-l}{L}$

$=> \frac{fe}{fe-\left ( fo+fe \right )}$ $=\frac{-l}{L}$ $(Since\ u =-fo+fe)$

$or \frac{-fe}{fo}=\frac{-l}{L}$

$or \: \frac{fe}{fo}=\frac{l}{L}$

$=>M=\frac{fo}{fe}=\frac{L}{l}$

Correct option is 3.

Posted by

shivangi.shekhar

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