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Answer in brief: Explain what is the optical path length. How is it different from actual path length?

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Consider, a light wave with an angular frequency of w and a wave vector of k travelling in the x-direction through a vacuum. The phase of this wave is (kx - ωt). In a vacuum, light speed at c, but in a medium, it speeds at v.
$k =\frac{2 \pi}{\lambda}=\frac{2 \pi v}{v \lambda}=\frac{\omega}{v}$ as $\omega=2 \pi v$ and $v = v \lambda$, where $v$ is the frequency of light.
If the wave travels a distance $\Delta x$, its phase changes by $\Delta \Phi= k \Delta x =\omega$ $\Delta x / v$.
Similarly, if the wave is travelling in vacuum,
$k =\omega / c \text { and } \Delta \Phi=\omega \Delta x / c$
Now, consider a wave travelling a distance $\Delta x$ in the medium, the phase difference generated is,
$\Delta \Phi^{\prime}=k^{\prime} \Delta x=\omega n \Delta x / c=\omega \Delta x^{\prime} / c$
where $\Delta x ^{\prime}= n \Delta x$
The distance $n \Delta x$ is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time $t$ in vacuum (with the speed c).
The optical path length in a medium is the corresponding path in a vacuum that light traverses at the same time as it does in the medium.
Now, speed $=\frac{\text { distance }}{\text { time }}$
$ \therefore \text { time }=\frac{\text { distance }}{\text { speed }}$
$\therefore t =\frac{ d _{\text {medium }}}{ v _{\text {medium }}}=\frac{ d _{\text {vaccum }}}{ v _{\text {vaccum }}} $
Hence, the optical path $= d _{\text {vacuum }}$
$ =\frac{ v _{\text {vaccum }}}{ v _{\text {medium }}} \times d _{\text {medium }}$
$= n \times d _{\text {medium }} $
Thus, a distance d travelled in a medium of refractive index n introduces a path difference $= nd - d = d (n - 1)$ over a ray travelling equal distance through vacuum.

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