It would appear that a thick, planar optical glass window is also capable of lengthening back focus - and would probably introduce less new problems.
Why is this not done in practice?
Because it won't work. It won't lengthen the back focus enough. That is, a flat optical element shifts the focus by an amount less than the thickness of that flat element.
In the following diagram, a marginal ray (the red ray) from the edge of the lens's exit pupil of diameter \$d\$ is focused at the back-focus distance \$b\$. That ray approaches at an angle \$\theta_1.\$
But that lens is shifted too far from the image sensor by an amount \$t\$, so that the lens can't achieve infinity focus. So inserting a flat element of thickness \$t\$ with refractive index \$n_2\$ will cause a shift in the back focus distance towards the image sensor by an amount \$s,\$ because of Snell's Law: \$n_1\sin\theta_1 = n_2\sin\theta_2.\$
For convenience's sake, I will normalize the refractive index \$n_2\$ with respect to air (\$n_1\$) with: \$n = n_2/n_1.\$
- The dashed red ray is the original path if the optical flat were not present, or alternately, is the path the ray would follow if the glass had the same refractive index as the air surrounding it (if \$n = 1\$).
- The black ray is the desired path that would shift the back focus distance by \$t.\$ But in order to achieve that, the refractive index would have to be infinite. This is undesirable:
- A material with infinite refractive index has a reflectivity of 1, meaning that it is a mirror, and admits no light through the air/material interface.
- This can also be understood by remembering that the refractive index of a material is defined as the ratio of the speed of light in a vacuum to the phase velocity of light in that material: \$n = c/v.\$ In order for \$n = \infty,\$ \$v\$ must equal zero. Thus, light does not travel through the material.
- The blue ray is the light path for typical optical materials (\$n > 1,\$ usually in the neighborhood of 1.5–1.6, up to around 1.7 or 1.8 for lead glass). Geometrically, you can see that for finite (and realistic, positive) values of \$n,\$ the focus shift distance \$s\$ will always be less than \$t.\$
Mathematically, the amount of shift \$s\$ is given by:
$$\begin{align*}
s &= t\left(1-\frac{\cos{\theta_1}}{n\cos{\theta_2}}\right) \\
\end{align*}$$
For paraxial rays (that is, \$\cos\theta \approx 1\$ for small values of \$\theta\$), this simplifies to
$$s = t\left(1-{1\over n}\right)$$
Thus, you can see that the entire expression in the parentheses that multiplies \$t\$ is positive and less than 1. So the shift \$s\$ is always less than \$t\$ for real world materials.
(See: Ray, Sidney F., 2003. Applied Photographic Optics, 3rd ed., pp. 22–23.) Ray notes that (bold emphasis mine),
The [rear] plate has unit magnification, so image size is unchanged. With convergent beams, astigmatism and other aberrations are introduced.
Lenses of critical performance or assembly difficulty are often designed with a built-in 'turret' of filters so that one is always in the light path, thus maintaining optical performance.
When using filters of high density or opaque filters, critical visual focusing is done using a dummy or blank filter of clear glass of identical thickness.
I keep saying "real world materials", because there are some metamaterials that have been built that have a negative refractive index, which would theoretically replace the need for concave (diverging) elements. Most of the negative-index metamaterials constructed so far are not transparent to visible light.