An amateur astronomer designs an astronomical telescope in normal adjustment. The total tube length of the telescope is 105 ~cm and its magnifying power has a magnitude of 20 . What are the optical powers of the objective and the eyepiece lenses respectively?

Step-by-step solution

For an astronomical telescope in normal adjustment, the magnitude of magnifying power is given by: |m| = f_o f_e = 20 f_o = 20 f_e The tube length is given by: L = f_o + f_e = 105 ~cm Substituting f_o : 20 f_e + f_e = 105 ~cm 21 f_e = 105 ~cm f_e = 5 ~cm = 0.05 ~m Then, the focal length of the objective is: f_o = 20 5 = 100 ~cm = 1 ~m The optical power of a lens is P = 1 f( in meters ) . Power of the objective: P_o = 1 1 = +1 ~D Power of the eyepiece: P_e = 1 0.05 = +20 ~D Therefore, the powers of the objective and