Q115 / 120 · VITEEE 2015 · Physics
Two light rays having the same wavelength in vacuum are in phase initially. Then, the first ray travels a path $L_{I}$ through a medium of refractive index $\mu_{1}$ while the second ray travels a path $L_{2}$ 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
- A$\frac{2 \pi}{\lambda}\left(\frac{L_{1}}{\mu_{1}}-\frac{L_{2}}{\mu_{2}}\right)$
- B$\frac{2 \pi}{\lambda}\left(L_{2}-L_{1}\right)$
- C$\frac{2 \pi}{\lambda}\left(\mu_{2} L_{1}-\mu_{1} L_{2}\right)$
- Correct answerD$\frac{2 \pi}{\lambda}\left(\mu_{1} L_{1}-\mu_{2} L_{2}\right)$
Solution
First ray optical path $=\mu_{1} \mathrm{~L}_{1}$ second ray optical path $=\mu_{2} \mathrm{~L}_{2}$
So, phase difference
$\begin{aligned}
\Delta \phi &=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x} \\
\therefore \quad \Delta \phi &=\frac{2 \pi}{\lambda}\left(\mu_{1} \mathrm{~L}_{1}-\mu_{2} \mathrm{~L}_{2}\right)
\end{aligned}$
Hence, the correct answer is option: 'D'.