An astronomical telescope has an objective of focal length 150 cm and an eyepiece of focal length 5 cm . It is initially focused on a distant star in normal adjustment. The observer then adjusts the eyepiece to refocus the telescope so that the final image forms at the least distance of distinct vision ( 25 cm ). What is the magnitude and direction of the displacement of the eyepiece?

Step-by-step solution

In normal adjustment, the final image is formed at infinity. The object for the eyepiece (the intermediate image formed by the objective) lies exactly at its principal focus. u_ e1 = f_e = 5 cm The initial tube length is L₁ = f_o + f_e = 150 + 5 = 155 cm . When the telescope is adjusted so the final image is at the least distance of distinct vision ( v_e = -25 cm ), we use the lens formula for the eyepiece to find the new object distance u_ e2 : 1 v_e - 1 u_ e2 = 1 f_e 1 -25 - 1 u_ e2 = 1 5 1 u_ e2 = - 1 25 - 1 5 =