In a compound microscope, the separation between the objective and the eyepiece is adjustable between 9.8 cm and 11.8 cm . Given that the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, determine the range of its magnifying power, provided the final image is always required at a distance of 24 cm from the eye.

Step-by-step solution

For the eyepiece, using the lens formula 1 v_e - 1 u_e = 1 f_e with v_e = -24 cm and f_e = 6 cm : 1 -24 - 1 u_e = 1 6 1 u_e = - 1 24 - 1 6 = - 5 24 u_e = -4.8 cm The magnification produced by the eyepiece is m_e = v_e u_e = -24 -4.8 = 5 . The distance between the objective and the eyepiece is L = v_o + |u_e| = v_o + 4.8 . For the minimum separation L = 9.8 cm : v_o = 9.8 - 4.8 = 5.0 cm Using the lens formula for the objective, 1 v_o - 1 u_o = 1 f_o : 1 5 - 1 u_o = 1 1 1 u_o = 1 5 - 1 = - 4 5 u_o = -1.25 cm The magn