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

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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

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Magnifying power of telescope,m = 9 ⇒ ⇒ ​ f e ​ f 0 ​ ​ = 9 f 0 ​ = 9 f e ​ … ( i ) ​ Distance between objective and eyepiece is given as ​ f 0 ​ + f e ​ = 20 … ( ii ) ⇒ 9 f e ​ + f e ​ = 20 [from Eq. (i)] ⇒ 10 f e ​ = 20 ⇒ f e ​ = 2 cm ​ ​ ∴ From Eq. (i), f 0 ​ = 9 × 2 = 18 cm ∴ f e ​ f 0 ​ ​ = 2 18 ​ = 9 ​

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