First slide

Ray optics

Question

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

Difficult

A

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

$\frac{L}{I}$
C

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

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

Solution

The situation is shown in the figure.
Let $f_{0}$ and $f_{e}$ be the focal lengths of the objective and eyepiece respectively.
For normal adjustment distance of the objective from the eyepiece (tube length) $f_{0}$ + $f_{e}$
Treating the line on the objective as the object and eyepiece as the lens.
$∴ u = - (f_{o} + f_{e}) and f = f_{e}$
$As \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$∴ \frac{1}{v} - \frac{1}{- (f_{o} + f_{e})} = \frac{1}{f_{e}}$
$\frac{1}{v} = \frac{1}{f_{e}} - \frac{1}{f_{o} + f_{e}} = \frac{f_{o} + f_{e} - f_{e}}{f_{e} (f_{o} + f_{e})} = \frac{f_{o}}{f_{e} (f_{o} + f_{e})}$
$or v = \frac{f_{e} (f_{o} + f_{e})}{f_{o}}$
$Thus, \frac{I}{L} = \frac{v}{| u |} = \frac{\frac{f_{e} (f_{o} + f_{e})}{f_{o}}}{(f_{o} + f_{e})} = \frac{f_{e}}{f_{o}}$
$or \frac{f_{o}}{f_{e}} = \frac{L}{I}$
$∴$ The magnification of the telescope in normal adjustment is
$m = \frac{f_{o}}{f_{e}} = \frac{L}{I} . . . . (i)$

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