In Concepts of Physics by H.C.Verma, I got the definition of optical path. However, one thing that I couldn't understand is why only the spatial phase change is considered in its definition.
As he writes:
Consider a light wave travelling in a medium of refractive index $\mu$. Its equation may be written as $$E = E_0\sin \omega\left(t - \frac{x}{v}\right ) = E_0\sin \omega\left(t - \frac{\mu x}{c}\right )$$. If the light wave travels a distance $\Delta x$, the phase changes by $$\delta_1= \mu\frac{\omega}{c} \Delta x\tag{I}$$. Instead, if the light wave travels in vacuum, the equation will be $$E = E_0\sin \omega\left(t - \frac{x}{v}\right )$$ If the light travels through a distance $\mu \Delta x$, the phase changes by $$\delta_2= \omega \frac{(\mu \Delta x)}{c}= \mu \frac{\omega}{c} \Delta x \tag{II}$$. From $\text{I}\, \& \, \text{II}$ , we see that a wave travelling through a distance $\Delta x$ in a medium of refractive index $\mu$ suffers the same phase change as when it travels a distance $\mu \Delta x$ in vacuum. ... The quantity $\mu \Delta x$ is the optical path of the light.
Apparently, it was an easy reading. But then I wondered why the phase change was $\mu \dfrac{\omega}{c} \Delta x$. I noticed the words If the light wave travels a distance $\Delta x$; now doesn't travelling needs time? Say, it needs time of $\Delta t$ to come at $x + \Delta x$; hence the phase-change should be $\omega \left((t +\Delta t) - \mu\dfrac{x + \Delta x}{c} \right) -\omega\left(t - \dfrac{\mu x}{c}\right ) = \underset{\text{phase-change}}{\boxed{\omega\left(\Delta t -\mu\frac{\Delta x}{c}\right)}}$ . This should be the phase-change which is not at all equal to $\delta_1 \, \text{or}\, \delta_2 =\mu \dfrac{\omega}{c} \Delta x $.
So, my questions are:
1) Why is time not taken into account, after all travelling needs time?
2) Even if time is ignored, phase-change is not equal to $\mu \dfrac{\omega}{c} \Delta x$ but $-\mu \dfrac{\omega}{c} \Delta x$. Where did the '$-$' sign go then?