Question

aIn an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is :

• A. $\frac{L}{I}$
• B. $\frac{L}{I} + 1$
• C. $\frac{L}{I} - 1$
• D. $\frac{L + I}{L - I}$

Answer 1

Answer: (A)

$\frac{L}{I}$

Answer 2

: Let $f_{o}$and $f_{e}$ be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) $= f_{0} + f_{e}$ Treating the line on the objective as the object, and the eyepiecs as the lens,

$u = - (f_{o} + f_{e})andf = f_{e}$

$\frac{1}{v} - \frac{2}{- (f_{o} + f_{e})} = \frac{1}{f_{e}}$

or,$\frac{1}{v} = \frac{1}{f_{e}} - \frac{1}{f_{o} + f_{e}} = \frac{f_{o}}{(f_{o} + f_{e})f_{e}}$

or, $v = \frac{(f_{o} + f_{e})f_{e}}{f_{o}}$

Magnification, $= \left| \frac{v}{u} \right| = \frac{f_{e}}{f_{o}} = \frac{imagesize(izfrfcEcdkvkdkj)}{object{si}ze(oLrqdkvkdkj)} = \frac{I}{O}$

$\therefore\frac{f_{o}}{f_{e}} = \frac{L}{l} =$ Magnification of telescope in normal adjustment.