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Question: Answered & Verified by Expert
Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path $l_{1}$ through a medium of refractive index $n_{1}$ while the second ray travels a path of length $l_{2}$ through a medium of refractive index $n_{2}$. The two waves are then combined to observe interference. The phase difference between the two waves is
PhysicsElectrostaticsVITEEEVITEEE 2010
Options:
  • A $\frac{2 \pi}{\lambda}\left(l_{2}-l_{1}\right)$
  • B $\frac{2 \pi}{\lambda}\left(n_{1} l_{2}-n_{2} l_{1}\right)$
  • C $\frac{2 \pi}{\lambda}\left(n_{2} l_{2}-n_{1} l_{1}\right)$
  • D $\frac{2 \pi}{\lambda}\left(\frac{l_{1}}{n_{1}}-\frac{l_{2}}{n_{2}}\right)$
Solution:
2512 Upvotes Verified Answer
The correct answer is: $\frac{2 \pi}{\lambda}\left(n_{1} l_{2}-n_{2} l_{1}\right)$
Optical path for ray $1=n_{1} l_{1}$ Optical path for ray $2=n_{2} l_{2}$ Phase difference,
$$
\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda}\left(n_{1} l_{1}-n_{2} l_{2}\right)
$$

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