Question:

Two light waves having the same wavelength \(\lambda\) in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1 while the second wave travels a path of length L2 through a medium of refractive index n2. After this the phase difference between the two waves is :

Show Hint

Phase difference = \(\frac{2\pi}{\lambda}\) times the optical path difference, and the optical path in a medium is \(nL\).
Updated On: Nov 5, 2026
  • \(\frac{2\pi}{\lambda}(n_1L_1 - n_2L_2)\)
  • \(\frac{2\pi}{\lambda}(n_2L_1 - n_1L_2)\)
  • \(\frac{2\pi}{\lambda}\left(\frac{L_1}{n_1} - \frac{L_2}{n_2}\right)\)
  • \(\frac{2\pi}{\lambda}\left(\frac{L_2}{n_1} - \frac{L_1}{n_2}\right)\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

The optical path of the first wave is \(n_1L_1\) and of the second is \(n_2L_2\).
Optical path difference \(\Delta = n_1L_1 - n_2L_2\).
Phase difference \(\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{\lambda}(n_1L_1 - n_2L_2)\), where \(\lambda\) is the wavelength in vacuum.
Answer: (1) \(\frac{2\pi}{\lambda}(n_1L_1 - n_2L_2)\)
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions

Top JEE Main Wave optics Questions

View More Questions