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MHT CET · Physics · Waves and Sound

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 \(\mu_1\), while the second ray travels a path of length \(L_2\) 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

  1. A \(\frac{2 \pi}{\lambda}\left[\mu_2 L_1-\mu_1 L_2\right]\)
  2. B \(\frac{2 \pi}{\lambda}\left[\frac{L_1}{\mu_1}-\frac{L_2}{\mu_2}\right]\)
  3. C \(\frac{2 \pi}{\lambda}\left[\mu_1 L_1-\mu_2 L_2\right]\)
  4. D \(\frac{2 \pi}{\lambda}\left[L_2-L_1\right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{2 \pi}{\lambda}\left[\mu_1 L_1-\mu_2 L_2\right]\)

Step-by-step Solution

Detailed explanation

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index \(\mu\) in time \(\mathrm{t}\) is given by

where \(v\) is the velocity of light in the medium. The distance traversed by light in a vacuum in this time,
\(\Delta=c t\)
\(=c \cdot \frac{d}{v}[\) from equation (1)]

This distance is the equivalent distance in vacuum and is called the optical path.
Here, the optical path for first ray \(=\mu_1 L_1\)
The optical path for second ray \(=\mu_2 L_2\)
Path difference \(=\mu_1 L_1-\mu_2 L_2\)
Now, phase difference \(=\frac{2 \pi}{\lambda} \times\) path difference \(=\frac{2 \pi}{\lambda} \times\left(\mu_1 L_1-\mu_2 L_2\right)\)