An astronomical telescope is to be designed with a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm , determine the powers of the objective and the eyepiece.

Step-by-step solution

For an astronomical telescope in normal adjustment, the magnifying power is m = f_o f_e and the tube length is L = f_o + f_e . Given m = 50 , we have f_o f_e = 50 f_o = 50 f_e . Given L = 102 cm , substituting f_o gives 50 f_e + f_e = 102 cm . 51 f_e = 102 cm f_e = 2 cm = 0.02 m . The focal length of the objective is f_o = 50 2 cm = 100 cm = 1 m . The power of the objective is P_o = 1 f_o = 1 1 = 1 D . The power of the eyepiece is P_e = 1 f_e = 1 0.02 = 50 D .