Q 12-10-130JEE MainJEE Main 2020 (3 Sep, Shift 2)Easy
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:
Answer: (C) $\dfrac{2\pi}{\lambda}(n_1L_1 - n_2L_2)$
Optical path lengths are $n_1L_1$ and $n_2L_2$. Phase difference $= \dfrac{2\pi}{\lambda}\times$ (optical path difference) $= \dfrac{2\pi}{\lambda}(n_1L_1 - n_2L_2)$.
Solution by Sreeraj P, M.Sc Physics