MHT CET · Physics · Wave Optics
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
- A \(\frac{2 \pi}{\lambda}\left(\mu_1 L_1-\mu_2 L_2\right)\)
- B \(\frac{2 \pi}{\lambda}\left(L_2-L_1\right)\)
- C \(\frac{2 \pi}{\lambda}\left(\frac{\mathrm{~L}_1}{\mu_1}-\frac{\mathrm{L}_2}{\mu_2}\right)\)
- D \(\frac{2 \pi}{\lambda}\left(\mu_2 L_1-\mu_1 L_2\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{2 \pi}{\lambda}\left(\mu_1 L_1-\mu_2 L_2\right)\)
Step-by-step Solution
Detailed explanation
Optical path difference \(=\mu_1 L_1-\mu_2 L_2\)
Phase difference \(=\frac{2 \pi}{\lambda} \times\) Path difference
Phase difference \(=\frac{2 \pi}{\lambda}\left[\mu_1 L_1-\mu_2 L_2\right]\)
Phase difference \(=\frac{2 \pi}{\lambda} \times\) Path difference
Phase difference \(=\frac{2 \pi}{\lambda}\left[\mu_1 L_1-\mu_2 L_2\right]\)
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