In a compound microscope, the separation between the objective and...

For the eyepiece, using the lens formula $\dfrac{1}{v_e} - \dfrac{1}{u_e} = \dfrac{1}{f_e}$ with $v_e = -24 \text{ cm}$ and $f_e = 6 \text{ cm}$:

$\dfrac{1}{-24} - \dfrac{1}{u_e} = \dfrac{1}{6}$

$\dfrac{1}{u_e} = -\dfrac{1}{24} - \dfrac{1}{6} = -\dfrac{5}{24}$

$u_e = -4.8 \text{ cm}$

The magnification produced by the eyepiece is $m_e = \dfrac{v_e}{u_e} = \dfrac{-24}{-4.8} = 5$.

The distance between the objective and the eyepiece is $L = v_o + |u_e| = v_o + 4.8$.

For the minimum separation $L = 9.8 \text{ cm}$:

$v_o = 9.8 - 4.8 = 5.0 \text{ cm}$

Using the lens formula for the objective, $\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$:

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

$\dfrac{1}{u_o} = \dfrac{1}{5} - 1 = -\dfrac{4}{5} \Rightarrow u_o = -1.25 \text{ cm}$

The magnification of the objective is $m_o = \dfrac{v_o}{u_o} = \dfrac{5}{-1.25} = -4$.

The magnifying power is $|M| = |m_o \times m_e| = |-4 \times 5| = 20$.

For the maximum separation $L = 11.8 \text{ cm}$:

$v_o = 11.8 - 4.8 = 7.0 \text{ cm}$

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

$\dfrac{1}{u_o} = \dfrac{1}{7} - 1 = -\dfrac{6}{7} \Rightarrow u_o = -\dfrac{7}{6} \text{ cm}$

The magnification of the objective is $m_o = \dfrac{v_o}{u_o} = \dfrac{7}{-\dfrac{7}{6}} = -6$.

The magnifying power is $|M| = |m_o \times m_e| = |-6 \times 5| = 30$.

Therefore, the range of its magnifying power is $20$ to $30$.