A compound microscope is composed of an objective with a focal...

Using the lens formula for the objective lens:

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

Substituting $u_o = -0.5 \text{ cm}$ and $f_o = 1 \text{ cm}$:

$\dfrac{1}{v_o} - \dfrac{1}{-0.5} = \dfrac{1}{1}$

$\dfrac{1}{v_o} + 2 = 1 \Rightarrow \dfrac{1}{v_o} = -1 \Rightarrow v_o = -1 \text{ cm}$

The image formed by the objective is virtual and is located $1 \text{ cm}$ in front of the objective.

For the eyepiece, the final image is real and is formed on a screen $30 \text{ cm}$ behind it. Using the lens formula for the eyepiece:

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

Substituting $v_e = 30 \text{ cm}$ and $f_e = 5 \text{ cm}$:

$\dfrac{1}{30} - \dfrac{1}{u_e} = \dfrac{1}{5}$

$\dfrac{1}{u_e} = \dfrac{1}{30} - \dfrac{1}{5} = \dfrac{1 - 6}{30} = -\dfrac{5}{30} = -\dfrac{1}{6}$

$u_e = -6 \text{ cm}$

The intermediate image acts as the object for the eyepiece and is located $6 \text{ cm}$ in front of it.

Let the separation between the lenses be $L$. Since the intermediate image is $1 \text{ cm}$ in front of the objective, its total distance from the eyepiece is $L + 1 \text{ cm}$.

Therefore, $L + 1 = 6 \Rightarrow L = 5 \text{ cm}$.