There is a problem given as
A light signal travels from $A$ to $B$ within a glass slab which is moving (in same direction as that of light) with speed 0.3c with respect to an external observer.If the refractive index of slab is 1.5, then what is the speed of the signal as observered by the observer?
The problem is easy. We know that
$$\begin{align} \text{Velocity of light in a medium} \,\,=&\,\, \frac{\text{Velocity of light in vaccuum}}{\text{Refractive index of the material}}\\\text{Velocity of light in glass slab} \,\,=&\,\, \frac{c}{1.5}\end{align}$$
This velocity of signal comes out in the frame in which the glass slab is stationary. But as the signal is light signal, so I think that the speed of signal will also appear same to the observer for whom the glass slab appears to move.
But in the solution, the relativistic addition of velocity is also used ,i.e.,
$$\displaystyle u_x=\frac{u_x'+v_{frame}}{1+\frac{u_x'v_{frame}}{c^2}}$$
$$\displaystyle\implies u_x=\frac{\frac{c}{1.5}+0.3c}{1+\frac{0.3}{1.5}}=0.81c$$
This is the final answer given.
But I have a doubt that in relativity it is assumed that speed of light in vaccuum remains constant in all inertial frames. But doesn't then speed of light in any medium should also remain constant in any other inertial frame (irrespective of material of those other frames)?
Please explain why velocity transformation equations are used after getting speed of light in the frame of glass slab. The observer should also observe the velocity of light to be $\frac{c}{1.5}$.
Please clasrify the doubt!