Determine the maximum magnifying power of a compound microscope...

Focal length of objective lens, $f_o = \dfrac{100}{P_o} = \dfrac{100}{25} = 4 \text{ cm}$

Focal length of eyepiece lens, $f_e = \dfrac{100}{P_e} = \dfrac{100}{5} = 20 \text{ cm}$

For maximum magnifying power, the final image is formed at the least distance of clear vision, $v_e = -25 \text{ cm}$.

Using the lens formula for the eyepiece:

$\dfrac{1}{v_e} - \dfrac{1}{u_e} = \dfrac{1}{f_e}$

$\dfrac{1}{-25} - \dfrac{1}{u_e} = \dfrac{1}{20}$

$\dfrac{1}{u_e} = -\dfrac{1}{25} - \dfrac{1}{20} = -\dfrac{9}{100} \Rightarrow |u_e| = \dfrac{100}{9} \text{ cm}$

The separation between the lenses is $L = v_o + |u_e| = 30 \text{ cm}$.

$v_o = 30 - \dfrac{100}{9} = \dfrac{170}{9} \text{ cm}$

Using the lens formula for the objective:

$\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$

$\dfrac{9}{170} - \dfrac{1}{u_o} = \dfrac{1}{4}$

$\dfrac{1}{u_o} = \dfrac{9}{170} - \dfrac{1}{4} = \dfrac{18 - 85}{340} = -\dfrac{67}{340} \Rightarrow |u_o| = \dfrac{340}{67} \text{ cm}$

The magnifying power of the objective is:

$m_o = \dfrac{v_o}{|u_o|} = \dfrac{170/9}{340/67} = \dfrac{67}{18}$

The magnifying power of the eyepiece is:

$m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{20} = \dfrac{9}{4}$

The maximum magnifying power of the compound microscope is:

$M = m_o \times m_e = \dfrac{67}{18} \times \dfrac{9}{4} = \dfrac{67}{8} = 8.375 \approx 8.4$