In YDSE experiment, the separation between the slits is equal to $3 λ$ where $λ$ is the wavelength of the light incident on the plane of the slits. A thin film of thickness $2 λ$ and refractive index 2 has been placed in front of the upper slit. Location of central maxima on the screen is

$\frac{D}{3}$

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$\frac{D}{2}$

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$\frac{2 D}{3}$

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$D$

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Solution

The correct option is C
$\frac{2 D}{3}$

$\frac{y d}{D} = Δ x = (μ - 1) t$

$y = \frac{D}{d} (μ - 1) t$

$= \frac{D}{3 λ} (2 - 1) 2 λ = \frac{2}{3} D$

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