Whenever I see a 2D drawing of dispersion occurring when light travels through a solid prism, I see the rays get bent downwards on entry and downwards on exit again. For example here: https://www.wikiwand.com/en/Dispersion_(optics)
To my understanding of optics when entering a medium with a higher optical density, the ray should get bent towards the normal of the surface, rotated CW and CCW when entering one with a lower IOR. However, the drawings suggest that it gets bent in the same direction upon entry and exit.
[Physics] Why does the light passing through a prism get bent in the same direction twice
dispersionopticsrefraction
Related Solutions
In general, reflection and refraction happen when light passes from one medium to another. You can see this if you see your own reflection in a window. Now, as a light ray approaches the critical angle, not only does the refracted ray get closer to the surface, but the amount of light transmitted gets less and less. At the critical angle, the refracted ray would be along the surface, but the amount of light is zero.
The graph below shows how much light is reflected at various angles of incidence. The graph on the right shows your situation of a ray of light moving from a dense medium to a lighter medium (glass to air, for example). At the critical angle, 100% of light is reflected, leaving none to propagate along the surface.
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.
Best Answer
The normals in consideration for the incident and emergent rays are different.
For simplicity, take a monochromatic beam of light incident on a prism, as shown in this figure:
In reference to this figure, the incident ray should bend towards the normal, which would mean a clockwise rotation ($\phi_1 < \theta_1$)
And the ray within the prism would bend away from the new normal at the new interface, corresponding to another clockwise rotation. ($\phi_2 < \theta_2$)
For a beam of light, dispersion will cause different wavelengths of light to bend in different angles, but they will all bend in the same sense.
Hope this helps.
Image source.