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The magnifying power of telescope is 9. when it is adjusted for parallel rays, the distance between objective and eye piece is found to be 20 cm. The focal length of lenses are

$(a)\;18 cm, 2 cm \\ (b)\;11 cm, 9 cm \\ (c)\;10 cm, 10 cm \\ (d)\;15 cm, 5 cm $

1 Answer

magnification $m=9$
$m= \large\frac{f_o}{f_e}$
$9=\large\frac{f_0}{f_e}$
$f_0= 9 f_e$
$f_0+f_e= 20$
$10\;f_e=20$
$f_e=2$
$f_0 = 2 \times 9 $
$\qquad= 18 \;cm$
Hence a is the correct answer.
answered Jan 23, 2014 by meena.p
 

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