Question

Two light rays having 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 combined to observe interference. The phase difference between the two waves is

• A. \(\dfrac{2\pi}{\lambda}(l_2-l_1)\)
• B. \(\dfrac{2\pi}{\lambda}(n_1l_1-n_2l_2)\)
• C. \(\dfrac{2\pi}{\lambda}(n_2l_2-n_1l_1)\)
• D. \(\dfrac{2\pi}{\lambda}\left(\dfrac{l_1}{n_1}-\dfrac{l_2}{n_2}\right)\)

Hint:

Phase difference comes from optical path difference: \(\Delta = n_1l_1 - n_2l_2\). Then \(\Delta\phi=\dfrac{2\pi}{\lambda}\Delta\).

Answer 1

The Correct Option is B

Step 1: Optical path length.
Phase depends on optical path length (OPL):
\[ OPL = nl \] Step 2: Optical path difference.
\[ \Delta = n_1l_1 - n_2l_2 \] Step 3: Convert into phase difference.
Phase difference is:
\[ \Delta\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{\lambda}(n_1l_1-n_2l_2) \] Final Answer: \[ \boxed{\dfrac{2\pi}{\lambda}(n_1l_1-n_2l_2)} \]