First slide

Ray optics

Question

In an astronomical telescope in normal adjustment, straight black line of length L is drawn on the objective. The eye piece forms a real image of this line. The length of this image is l. The magnification of the telescope is

Difficult

A

$\frac{L}{l}$
B

$\frac{L}{l} + 1$
C

$\frac{L}{l} - 1$
D

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

Solution

$\frac{v_{e}}{u_{e}} = \frac{1}{L} and u_{e} = f_{0} + f_{e} \frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}} \frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{- (f_{0} + f_{e})} \frac{1}{v_{e}} = \frac{1}{f_{e}} - \frac{1}{(f_{0} + f_{e})} = \frac{f_{0} + f_{e} - f_{e}}{(f_{0} + f_{e}) f_{e}} = \frac{f_{0}}{f_{e} (f_{0} + f_{e})} \frac{1}{L} = \frac{[\frac{f_{e} (f_{0} + f_{e})}{f_{0}}]}{f_{0} + f_{e}} = \frac{f_{e}}{f_{0}}$
magnification of telescope = $\frac{f_{0}}{f_{e}} = \frac{L}{l}$

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