Explanation

(b) The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index $\mu$ in time t is given by
$d = vt\ \ \ \ \ \ \ldots\ldots\ldots.(i)$
Where v is velocity of light in the medium. The distance traversed by light in a vacuum in this time.
$\Delta = ct$
$= c\ .\frac{d}{v}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack From\ Eq.(i) \right\rbrack$
$= d\frac{c}{v} = \mu d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)$
$\left( Since,\ \mu = \frac{c}{v} \right)$
This distance is the equivalent distance in vacuum and is called optical path.
Here, optical path for first ray $= n_{1}L_{1}$
Optical path for second ray $= n_{2}L_{2}$
Path difference $= n_{1}L_{1} - n_{2}L_{2}$
Now, phase difference
$= \frac{2\pi}{\lambda} \times$ path difference
$= \frac{2\pi}{\lambda}\ \times \left( n_{1}L_{1} - n_{2}L_{2} \right)$