The magnifying power of a telescope is 9 . When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm . The ratio of focal length of objective lens to focal length of eyepiece is found to be m, then the value of m is
C. 9
Magnifying power of telescope,m = 9 array ll & f_0 f_e =9 \\ & f_0=9 f_e (i) array Distance between objective and eyepiece is given as aligned & f_0+f_e=20 (ii)\\ & 9 f_e+f_e=20 \\ & [from Eq. (i)] \\ & 10 f_e=20 \\ & f_e=2 ~cm aligned aligned & From Eq. (i), f_0=9 2=18 ~cm \\ & f_0 f_e = 18 2 =9 aligned
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