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In YDSE experiment shown in the figure, a parallel beam of light of wavelength $λ = 0.4 m m$ in air, is incident at an angle $θ = 30^{0}$ as shown. If the intensity due to each light source at point ‘O’ is $I_{0},$ then the resultant intensity at point ‘O’ will be (Refractive indices of media before and after the plane of slits are shown in the figure)

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$I_{0}$

$3 I_{0}$

$2 I_{0} [1 + \cos (\frac{40 π}{9})]$

Zero

answer is D.

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Detailed Solution

The extra path travelled by rays reaching $S_{2}$ is,

$Δ x = μ d \sin θ = (10^{- 3} \times \frac{1}{2}) \frac{4}{3} = 5 \times 10^{- 4} \times \frac{4}{3}$

$λ = 0.4 m m$

$\frac{Δ x}{λ} = λ + \frac{2}{3} λ \Rightarrow Δ ϕ = \frac{2 π}{λ} \frac{2}{3} λ = \frac{4 π}{3}$

$I = 4 I_{0} \cos^{2} (\frac{Δ ϕ}{2}) = 4 I_{0} \cos^{2} (\frac{2 π}{3}) = I_{0}$

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