A compound microscope has f_o = 0.6 cm, f_e = 2.0 cm, tube length L = 16 cm, D = 25 cm. (a) Find M with the final image at the near point. (b) Find M with the final image at infinity. (c) Where is the specimen relative to the objective in case (b)? Step 1: m_o ≈ −L/f_o = −16/0.6 ≈ −26.67. M_e = 1 + D/f_e = 1 + 25/2 = 13.5. M = −26.67 × 13.5 ≈ −360×. Step 2: M_e = D/f_e = 12.5 → M = −26.67 × 12.5 ≈ −333×. (Infinity setting: slightly less M, relaxed viewing — standard for photomicrography.) Step 3: Final image at infinity means the intermediate image lies exactly at the eyepiece's focal point, 16 cm above the objective's image plane... precisely: v_o = L = 16 cm (approx). 1/u_o = 1/v_o − 1/f_o → 1/u_o = 1/16 − 1/0.6 = (0.6 − 16)/9.6 → u_o ≈ −0.623 cm. The specimen sits just 0.623 cm below the objective — just outside f_o, as theory demands.