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 focal length of lenses are

Step-by-step solution

For a telescope adjusted for parallel rays, the magnifying power M is given by M = f_o f_e , where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece. Given M = 9 , we have f_o f_e = 9 , which implies f_o = 9f_e . The distance between the objective and the eyepiece for parallel rays is L = f_o + f_e . Given L = 20 cm , we substitute f_o = 9f_e into the equation: 9f_e + f_e = 20 cm 10f_e = 20 cm f_e = 2 cm Substituting f_e = 2 cm back into f_o = 9f_e : f_o = 9 2 cm = 18 cm . An