Optical Path

The phase (and hence the interference behaviour) of a light wave depends not on the raw physical DISTANCE it has travelled, but on how many wavelengths' worth of PHASE it has accumulated along the way -- and since wavelength itself SHRINKS inside a medium of refractive index $n > 1$ (Section 7.6, $\lambda = \lambda_0/n$), the same physical distance travelled through a denser medium produces a LARGER phase change than travelling the identical physical distance through vacuum or air.

This motivates the definition of OPTICAL PATH: for a wave travelling a physical distance $\Delta x$ through a medium of refractive index n, the optical path is defined as $n\,\Delta x$ -- the distance the wave would have needed to travel in VACUUM to accumulate that same amount of phase change. Formally, since the wave vector in a medium is $k' = \omega/v = \omega n/c$, a physical distance $\Delta x$ produces a phase change $\Delta\phi' = k'\Delta x = (\omega/c)(n\,\Delta x)$ -- exactly the SAME phase change a distance $n\,\Delta x$ would produce travelling in vacuum (where $k=\omega/c$). So $n\,\Delta x$ is, in this precise sense, EQUIVALENT (for phase purposes) to a distance $n\,\Delta x$ in vacuum, which is exactly the optical path. Optical path can equally be understood as the corresponding distance light would cover in vacuum in the SAME amount of TIME it actually takes to cross the medium: since time = distance/speed, the time to cross a physical distance $d_{medium}$ at speed $v = c/n$ is $d_{medium}/v = n\,d_{medium}/c$, and light travelling in vacuum for that same time covers a distance $c \times (n\,d_{medium}/c) = n\,d_{medium}$ -- the identical optical path. In vacuum itself, since $n=1$, the optical path is always exactly equal to the actual physical distance travelled. …