Two light rays having the same wavelength ' λ ' in vacuum are in phase initially. Then, the first ray travels a path ' L₁ ' through a medium of refractive index ' μ₁ ' while the second ray travels a path of length ' L₂ ' through a medium of refractive index ' μ₂ '. The two waves are then combined to observe interference. The phase difference between the two waves is (a) (2π)/(λ)(μ₁ L₁ - μ₂ L₂) (b) (2π)/(λ)(L₂ - L₁) (c) (2π)/(λ)((L₁)/(μ₁) - (L₂)/(μ₂)) (d) (2π)/(λ)(μ₂ L₁ - μ₁ L₂)

Answer: (a). Optical paths: μ₁ L₁ and μ₂ L₂. Phase difference = (2π)/(λ)×(optical path difference) = (2π)/(λ)(μ₁L₁ - μ₂L₂) So the answer is (a).

Two light rays having the same wavelength ' λ ' in vacuum are in… | MHT CET 2024 (10 May, Shift 2)

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YDSE: Film, Medium & Two Wavelengths

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

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Answer: (a)

Optical paths: $\mu_1 L_1$ and $\mu_2 L_2$ .
Phase difference $= \frac{2\pi}{\lambda}\times(\text{optical path difference}) = \frac{2\pi}{\lambda}(\mu_1L_1 - \mu_2L_2)$
So the answer is (a).

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YDSE: Film, Medium & Two Wavelengths question from MHT CET 2024 (10 May, Shift 2)

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