A compound microscope has an objective of focal length 1.2 cm and an eyepiece of focal length 3.0…

Step 1. The first image. $\dfrac{1}{v_o} = \dfrac{1}{f_o} + \dfrac{1}{u_o} = \dfrac{1}{1.2\ \text{cm}} - \dfrac{1}{1.5\ \text{cm}} = 0.1667\ \text{cm}^{-1}$. So $v_o = 6.0\ \text{cm}$.

Step 2. The objective. $m_o = \dfrac{v_o}{u_o} = \dfrac{6.0\ \text{cm}}{-1.5\ \text{cm}} = -4.0$.

Step 3. The eyepiece. The final image is at the near point, so $m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25\ \text{cm}}{3.0\ \text{cm}} = 9.33$. Then $M = m_om_e = -4.0 \times 9.33 = -37.3$.

Step 4. The distance between the lenses. The first image must be at $|u_e| = \dfrac{Df_e}{D + f_e} = \dfrac{25 \times 3.0}{25 + 3.0}\ \text{cm} = 2.68\ \text{cm}$ from the eyepiece. So the lenses are $6.0\ \text{cm} + 2.68\ \text{cm} = 8.68\ \text{cm}$ apart.