tion:Given The objective of an ASTRONOMICAL telescope has the diameter of 150 MM and the focal LENGTH of 4 m.The eye piece has focal length of 25 mm.Calculate the   magnifying and resolving power of the telescope. What is the distance between the objective and the eye piece. Take wavelength 6000 a°Given diameter d = 150 mm = 150 x 10^-3 m, fo = 4.0 m, fe = 25 mm = 25 x 10^-3 m, λ = 6000 = 6000 x 10^-7 m Now resolving power of telescope = d / 1.22 λ                                                            = 150 x 10^-3 / 1.22 x 6 x 10^-7                                                            = 2.05 x 10^5 Now magnifying power m = focal length of objective / focal length of eye piece = fo / fe                                                                                                                                       = 4 / 25 x 10^-3                                                                                                                                        = 160 Now distance between objective and eyepiece will be = fo + fe      = 4.0 + 25 x 10^-3      = 4.025 m REFERENCE link will bebrainly.in/question/8865670