Why does focus change when object is behind a glass slide? I have an object that I am observing and focusing on, on a microscope slide. Half of the object is naked and half is covered with another microscope slide (not a coverslip).
Why do I have, in order to focus on the object, to bring the covered part further away from the objective than the naked part ?
 A: In short, because the glass slide has a bigger index of refraction and therefore adds optical path, so the object looks like it is further away than it is.
If you want a bit more detail, you can see what's going on using a rough ABCD-matrix analysis, where the matrices for the flat air-glass interface, the in-glass propagation, and the flat glass-air interface are
$$
\begin{pmatrix} 1 & 0 \\ 0 & 1/n \end{pmatrix},
\begin{pmatrix} 1 & d \\ 0 & 1 \end{pmatrix},\ \text{and} \ 
\begin{pmatrix} 1 & 0 \\ 0 & n \end{pmatrix},
$$
respectively, so the total transfer matrix, 
$$
\begin{pmatrix} 1 & 0 \\ 0 & 1/n \end{pmatrix}
\begin{pmatrix} 1 & d \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & n \end{pmatrix}
=
\begin{pmatrix} 1 & n\:d \\ 0 & 1 \end{pmatrix}, 
$$
is identical to free-space propagation over a larger distance. (Or possibly smaller ─ don't trust me on the details.)
A: The rays from the object emerging from the glass slide appear to come from an image that is closer to the glass-air interface than the object is. This is because of the refraction that takes place at the glass-air interface, which is also the reason why a swimming pool filled with water appears shallower than it really is. Unlike an image formed through a spherical lens though, this "image" location depends on where you are viewing it from, which complicates things a bit. It is however quite easy to estimate the apparent distance from the interface when the rays that form the image are roughly perpendicular to the glass-air interface.
Consider a point in (or just underneath) the glass slide a distance $h$ away from the interface, and two rays emanating from it: one normal to the interface, defining the optical axis, and one that reaches the interface with a small angle of incidence $\theta$. By Snell's law,
$$ \sin \theta' = n \sin \theta $$
or
$$ \theta' \approx n\theta$$
where $n$ is the glass refractive index, $\theta'$ is the angle of refraction, and the second equation uses the fact that $\sin \theta \approx \theta $ for small $\theta$. The image location is found by extrapolating the refracted ray back into the glass, and finding the point at which it intersects the optical axis. By doing some geometry, you will find that this point is located a distance
$$ h' = h \frac{\tan \theta}{\tan \theta'} \approx h \frac{\theta}{\theta'} \approx \frac{h}{n} $$
from the interface. The image thus appears closer to the viewer by a distance of
$$ h - h' = \left(1 - \frac{1}{n}\right) h $$
compared to the case without the glass slide.
