 # Two light waves having the same wavelength `
Question:

Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first wave travels a path $L_{1}$ through a medium of refractive index $n_{1}$ while the second wave travels a path of length $L_{2}$ through a medium of refractive index $n_{2}$. After this the phase difference between the two waves is :

1. $\frac{2 \pi}{\lambda}\left(\frac{L_{2}}{n_{1}}-\frac{L_{1}}{n_{2}}\right)$

2. $\frac{2 \pi}{\lambda}\left(\frac{L_{1}}{n_{1}}-\frac{L_{2}}{n_{2}}\right)$

3. $\frac{2 \pi}{\lambda}\left(n_{1} L_{1}-n_{2} L_{2}\right)$

4. $\frac{2 \pi}{\lambda}\left(n_{2} L_{1}-n_{1} L_{2}\right)$

Correct Option: , 3

Solution:

(3) The distance traversed by light in a medium of refractive index $\mu$ in time $t$ is given by

$d=v t$               ....(1)

where $v$ is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

$\Delta=c t=c \times \frac{d}{v}$    [from equation (i)]

$=d \frac{c}{v}=\mu d$               ...(ii) $\quad\left(\because \mu=\frac{c}{v}\right)$

This distance is the equvalent distance in vacuum and is called optical path.

Optical path for first ray which travels a path $L_{1}$ through a medium of refractive index $n_{1}=n_{1} L_{1}$

Optical path for second ray which travels a path $L_{2}$ through a medium of refractive index $n_{2}=n_{2} L_{2}$

Path difference $=n_{1} L_{1}-n_{2} L_{2}$

Now, phase difference

$=\frac{2 \pi}{\lambda} \times$ path difference $=\frac{2 \pi}{\lambda} \times\left(n_{1} L_{1}-n_{2} L_{2}\right)$