Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1895 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \pi}{\lambda}\left(\mu_{1} L_{1}-\mu_{2} L_{2}\right)$
First ray optical path $=\mu_{1} \mathrm{~L}_{1}$ second ray optical path $=\mu_{2} \mathrm{~L}_{2}$
So, phase difference
$\begin{aligned}
\Delta \phi &=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x} \\
\therefore \quad \Delta \phi &=\frac{2 \pi}{\lambda}\left(\mu_{1} \mathrm{~L}_{1}-\mu_{2} \mathrm{~L}_{2}\right)
\end{aligned}$
So, phase difference
$\begin{aligned}
\Delta \phi &=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x} \\
\therefore \quad \Delta \phi &=\frac{2 \pi}{\lambda}\left(\mu_{1} \mathrm{~L}_{1}-\mu_{2} \mathrm{~L}_{2}\right)
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.