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:

We start with two monochromatic light waves that are initially in phase. Their wavelength in vacuum is given to be $$\lambda$$.

For any wave that moves through a medium, the phase it accumulates depends on the distance travelled and on the speed of propagation in that medium. The standard relation connecting phase $$\Phi$$, path length $$L$$ and wavelength is

$$\Phi = \frac{2\pi}{\lambda_{\text{medium}}}\,L,$$

where $$\lambda_{\text{medium}}$$ is the wavelength of the light inside the medium. Because the frequency of light does not change when it enters a medium, the wavelength in a medium of refractive index $$n$$ is shortened according to the well-known formula

$$\lambda_{\text{medium}} = \frac{\lambda}{n}.$$

Substituting this value of $$\lambda_{\text{medium}}$$ into the phase formula, we obtain

$$\Phi \;=\; \frac{2\pi}{\lambda_{\text{medium}}}\,L \;=\; \frac{2\pi}{\dfrac{\lambda}{n}}\;L \;=\; \frac{2\pi n}{\lambda}\,L \;=\; \frac{2\pi}{\lambda}\,(nL).$$

Thus a wave travelling a distance $$L$$ in a medium of refractive index $$n$$ accumulates the phase

$$\Phi = \frac{2\pi}{\lambda}\,nL.$$

Now, let us apply this to our two waves separately:

The first wave travels a distance $$L_1$$ in the medium with refractive index $$n_1$$, so its phase after emerging is

$$\Phi_1 = \frac{2\pi}{\lambda}\,n_1L_1.$$

The second wave travels a distance $$L_2$$ in the medium with refractive index $$n_2$$, so its phase after emerging is

$$\Phi_2 = \frac{2\pi}{\lambda}\,n_2L_2.$$

The phase difference $$\Delta\Phi$$ between the two waves after they have completed their respective journeys is simply the algebraic difference of these two phases. Taking the first wave minus the second, we have

$$\Delta\Phi = \Phi_1 \;-\; \Phi_2 = \frac{2\pi}{\lambda}\,n_1L_1 \;-\; \frac{2\pi}{\lambda}\,n_2L_2.$$

Factoring out the common factor $$\frac{2\pi}{\lambda}$$ gives

$$\Delta\Phi = \frac{2\pi}{\lambda}\;\bigl(n_1L_1 - n_2L_2\bigr).$$

This expression matches exactly the form given in Option C.

Hence, the correct answer is Option C.