A telescope has an objective of focal length 90 cm and an eyepiece of focal length 5.0 cm…

Step 1. Normal adjustment, for comparison. $|m| = \dfrac{90\ \text{cm}}{5.0\ \text{cm}} = 18$ and the length is $95\ \text{cm}$.

Step 2. The magnifying power at the near point. $|m| = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right) = 18 \times \left(1 + \dfrac{5.0\ \text{cm}}{25\ \text{cm}}\right) = 18 \times 1.2 = 21.6$.

Step 3. The distance of the first image from the eyepiece. $|u_e| = \dfrac{Df_e}{D + f_e} = \dfrac{25 \times 5.0}{25 + 5.0}\ \text{cm} = 4.17\ \text{cm}$.

Step 4. The distance between the lenses. $f_o + |u_e| = 90\ \text{cm} + 4.17\ \text{cm} = 94.2\ \text{cm}$.