Two light rays having the same wavelength ' $\lambda$ ' in vacuum are in phase initially. Then, the first…

Two light rays having the same wavelength ' $\lambda$ ' in vacuum are in phase initially. Then, the first ray travels a path ' $\mathrm{L}_1$ ' through a medium of refractive index ' $\mu_1$ ' while the second ray travels a path of length ' $\mathrm{L}_2$ ' 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. $\frac{2 \pi}{\lambda}\left(\mu_1 L_1-\mu_2 L_2\right)$
  2. $\frac{2 \pi}{\lambda}\left(L_2-L_1\right)$
  3. $\frac{2 \pi}{\lambda}\left(\frac{\mathrm{~L}_1}{\mu_1}-\frac{\mathrm{L}_2}{\mu_2}\right)$
  4. $\frac{2 \pi}{\lambda}\left(\mu_2 L_1-\mu_1 L_2\right)$

Solution

Optical path difference $=\mu_1 L_1-\mu_2 L_2$ Phase difference $=\frac{2 \pi}{\lambda} \times$ Path difference Phase difference $=\frac{2 \pi}{\lambda}\left[\mu_1 L_1-\mu_2 L_2\right]$

Asked in: MHT CET 2024 (10 May Shift 2)

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