A young’s double slit experiment is carried out in water having refractive index $μ_{1}$ as shown in the figure. A glass plate of thickness t and refractive index $μ_{2}$ is placed in the path of $S_{2} .$ The magnitude of the phase difference at O is (Assume that $λ$ is the wavelength of light in air, O is symmetrical with respect to $S_{1} and S_{2}) .$

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$\frac{2 π}{λ} |(\frac{μ_{1}}{μ_{2}} - 1) t|$

$\frac{2 π}{λ} |(μ_{2} - μ_{1}) t|$

$\frac{2 π}{λ} |(\frac{μ_{2}}{μ_{1}} - 1) t|$

$\frac{2 π}{λ} |(μ_{2} - 1) t|$

answer is C.

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Detailed Solution

Path difference $Δ x = (\frac{μ_{2}}{μ_{1}} - 1) t$

Phase difference $δ = \frac{2 π}{\frac{λ}{μ_{1}}} |(\frac{μ_{2}}{μ_{1}} - 1) t|$ = $\frac{2 π}{λ} |(μ_{2} - μ_{1}) t|$

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