The focal length of the objective and the eyepiece of a telescope are 50 ~cm and 5 ~cm respectively. If the telescope is focussed for distinct vision on a scale distant 2 ~m from its objective, then its magnifying power will be :

Step-by-step solution

Given: f ₀=50 ~cm , f _ e =5 ~cm d =25 ~cm , u ₀=-200 ~cm Magnification M = ? As 1 v ₀ - 1 u ₀ = 1 f ₀ 1 v ₀ = 1 f ₀ + 1 u ₀ = 1 50 - 1 200 = 4-1 200 = 3 200 or v ₀= 200 3 ~cm Now v _ e = d =-25 ~cm From, 1 v_e - 1 u_e = 1 f_e aligned & - 1 u_e = 1 f_e - 1 v_e & = 1 5 + 1 25 = 6 25 aligned or, v _ e = -25 6 ~cm aligned & Magnification M = M ₀ M _ e & = v ₀ u ₀ v _ e u _ e = -200 / 3 200 -25 -25 / 6 & =- 1 3 6=-2 aligned