Two light rays having the same wavelength ' λ ' in vacuum are in phase initially. Then, the first ray travels a path ' L 1 ' through a medium of refractive index ' μ 1 &#0

Question

Two light rays having the same wavelength ' λ ' in vacuum are in phase initially. Then, the first ray travels a path ' L 1 ​ ' through a medium of refractive index ' μ 1 ​ ' while the second ray travels a path of length ' L 2 ​ ' through a medium of refractive index ' μ 2 ​ '. The two waves are then combined to observe interference. The phase difference between the two waves is

TestHub · Accepted Answer

The optical path length of a light ray traveling a distance $L$ in a medium of refractive index $μ$ is given by $μL$ .

For the first ray, the optical path length is $x_{1} = μ_{1} L_{1}$ .

For the second ray, the optical path length is $x_{2} = μ_{2} L_{2}$ .

The optical path difference between the two rays is $Δ x = x_{1} - x_{2} = μ_{1} L_{1} - μ_{2} L_{2}$ .

The phase difference $Δ ϕ$ is related to the optical path difference by the relation $Δ ϕ = \frac{2 π}{λ} Δ x$ .

Substituting the value of $Δ x$ , we get $Δ ϕ = \frac{2 π}{λ} (μ_{1} L_{1} - μ_{2} L_{2})$ .

Answer: $\frac{2 π}{λ} (μ_{1} L_{1} - μ_{2} L_{2})$

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