Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray…
Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path $L_1$ through a medium of refractive index $\mu_1$, while the second ray travels a path of length $L_2$ through a medium of refractive index $\mu_2$. The two waves are then combined to observe interference. The phase difference between the two waves is
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 $\mathrm{t}$ is given by
\(d=v t \quad (1)\)
where $v$ is the velocity of light in the medium. The distance traversed by light in a vacuum in this time,
$\Delta=c t$
$=c \cdot \frac{d}{v}[$ from equation (1)]
This distance is the equivalent distance in vacuum and is called the optical path.
Here, the optical path for first ray $=\mu_1 L_1$
The optical path for second ray $=\mu_2 L_2$
Path difference $=\mu_1 L_1-\mu_2 L_2$
Now, phase difference $=\frac{2 \pi}{\lambda} \times$ path difference $=\frac{2 \pi}{\lambda} \times\left(\mu_1 L_1-\mu_2 L_2\right)$
This distance is the equivalent distance in vacuum and is called the optical path.
Here, the optical path for first ray $=\mu_1 L_1$
The optical path for second ray $=\mu_2 L_2$
Path difference $=\mu_1 L_1-\mu_2 L_2$
Now, phase difference $=\frac{2 \pi}{\lambda} \times$ path difference $=\frac{2 \pi}{\lambda} \times\left(\mu_1 L_1-\mu_2 L_2\right)$