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Physics
If f0 and fe are the focal lengths of the objective and eye-piece respectively of a telescope, then its magnifying power will be
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Q. If $f_0$ and $f_e$ are the focal lengths of the objective and eye-piece respectively of a telescope, then its magnifying power will be
KCET
KCET 2000
Ray Optics and Optical Instruments
A
$f_0 + f_e$
13%
B
$f_0 \times f_e$
13%
C
$f_0 / f_e$
51%
D
$ \frac{1}{2} (f_0 + f_e)$
23%
Solution:
Magnification of a telescope is given by $m = \frac{f_0}{f_e},$ where $f_0$ is the focal length of the objective lens and $f_e$ is the focal length of eye-piece.