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

Step-by-step solution

Magnifying power of telescope,m = 9 array ll & f₀ f_e =9 & f₀=9 f_e (i) array Distance between objective and eyepiece is given as aligned & f₀+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), f₀=9 2=18 ~cm & f₀ f_e = 18 2 =9 aligned