The focal lengths of the objective and the eyepiece of a compound microscope are $1\text{cm}$ and $5.0\text{cm}$ respectively. An object, placed at a distance of $1.1\text{cm}$ from the objective, has its final image formed at infinity. The distance between the lenses will be $\left(\text{in cm}\right)$.

For the objective,
$u_0=-1.1\text{cm},f_0=+1.0\text{cm},v_0=?$

From the lens formula,
$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$

$\Rightarrow \frac{1}{v_0}=\frac{1}{u_0}+\frac{1}{f_0}=\frac{-1}{1.1}+\frac{1}{1.0}$

$\Rightarrow \frac{1}{v_0}=+\frac{0.1}{1.1}=+\frac{1}{11}$

$\Rightarrow v_0=+11\text{cm}$

The image of the object is formed by the objective at $11\text{cm}$ from the objective between the objective and the eyepiece.

When the final image is formed by the eyepiece is at infinity, the image formed by the objective will be at the focus of the eyepiece, that is, at a distance $5.0\text{cm}$ in front of the eyepiece.

Hence, the distance between the lens is,

$d=11\text{cm}+5.0\text{cm}=16\text{cm}$

Hence, $16$ is the correct answer.