An eye is capable of distinguishing between two points on an object...

The focal lengths of the objective and eyepiece are given by:

$f_o = \dfrac{100}{P_o} = \dfrac{100}{20} = 5 \text{ cm}$

$f_e = \dfrac{100}{P_e} = \dfrac{100}{10} = 10 \text{ cm}$

For the eyepiece, the final image is formed at the near point, so $v_e = -25 \text{ cm}$. Using the lens formula:

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

$\dfrac{1}{-25} - \dfrac{1}{u_e} = \dfrac{1}{10}$

$\dfrac{1}{u_e} = -\dfrac{1}{25} - \dfrac{1}{10} = -\dfrac{7}{50} \implies u_e = -\dfrac{50}{7} \text{ cm}$

The distance between the objective and the eyepiece is $L = 20 \text{ cm}$. The image distance for the objective is:

$v_o = L - |u_e| = 20 - \dfrac{50}{7} = \dfrac{90}{7} \text{ cm}$

Using the lens formula for the objective:

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

$\dfrac{7}{90} - \dfrac{1}{u_o} = \dfrac{1}{5}$

$\dfrac{1}{u_o} = \dfrac{7}{90} - \dfrac{18}{90} = -\dfrac{11}{90} \implies u_o = -\dfrac{90}{11} \text{ cm}$

The linear magnification of the objective is:

$m_o = \left| \dfrac{v_o}{u_o} \right| = \dfrac{90/7}{90/11} = \dfrac{11}{7}$

The angular magnification of the eyepiece is:

$m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{10} = \dfrac{7}{2}$

The total magnifying power of the compound microscope is:

$M = m_o \times m_e = \dfrac{11}{7} \times \dfrac{7}{2} = 5.5$

The unaided eye can distinguish a minimum separation of $d = 0.22 \text{ mm}$ at a distance of $25 \text{ cm}$. When viewed through the microscope, the minimum separation d_\min on the object that can be distinguished is reduced by the magnification factor:

d_\min = \dfrac{d}{M} = \dfrac{0.22 \text{ mm}}{5.5} = 0.04 \text{ mm}

Answer: $0.04 \text{ mm}$