How to calculate the focal length of spherical shell (zero meniscus lens) I can get my head around the optics of a spherical shell (a meniscus lens that neither diverge nor converge, being neither positive nor negative etc). 
I don’t see how I can describe in a formula how rays would diverge or converge towards some focus through such a lens. The lens maker formula 1/f = (n-1)(1/R1 - 1/R2) certainly suggests a focal length when plugging in R1 and R2 being inner and outer radius of the spherical shell.
I guess my question is if the lens maker formula is valid? And also is it so that rays effectively translate as opposed to change direction, i.e., all parallel rays entering also exists in parallel? 
 A: I know that this post is quite old. However, the answer is maybe interesting for you. Directly applying lensmaker's equation doesn't work due to the air inside the lens.
I would rather suggest the following:
Since the middle of such a shell consists of air we can conceptually cut it in two halves. 
These two halves are two identical lenses (just turned by 180°).
The focal length of one half is:
$$ \frac{1}{f} = (n-1) \left(\frac1{R_1} - \frac1{R_2} + \frac{(n-1) d_\text{lens}}{n R_1 R_2}\right)$$
Note that both $R_1$ and $R_2$ are positive. In our special case $R_2 < R_1$ and so this lens acts as a diverging lens. $d_\text{lens} = R_1 -R_2$ is the thickness of our lens.
Now we've got a lens system of two such lenses and we have to combine those. This done via:
$$\frac{1}{f_{res}} = \frac1{f_1} + \frac1{f_2} - \frac{d}{f_1 f_2}$$
where $d$ is the distance between this lenses (two spherical shell halves). In our case it is $d=2\cdot \frac{R_1+R_2}{2}$.
To finally answer your question. In the case of a zero meniscus lens we get $R_1 = R_2$ and also $d_\text{lens} = 0$. This means that the focal length is 0. So you won't see any effect of the spherical shell.
edit: this whole calculation neglects spherical aberrations. So this is only valid for rays close to the center of the shell.

