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In Young’s double slit experiment, Let $β$ be the fringe width and I₀ be the intensity at the central bright fringe. At a distance ‘x’ from the central bright fringe, the intensity will be

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$I_{0} \cos (\frac{x}{β})$

$I_{0} \cos^{2} (\frac{x}{β})$

$I_{0} \cos^{2} (\frac{π x}{β})$

$\frac{I_{0}}{4} \cos^{2} (\frac{π x}{β})$

answer is C.

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Detailed Solution

$I_{\max} = I_{0}; I = I_{0} \cos^{2} \frac{ϕ}{2}; β = \frac{λ D}{d}$

$Δ x = d \sin θ = d \tan θ$

$Δ x = \frac{d . x}{D}, ϕ = \frac{2 π}{λ} . \frac{d . x}{D}, ϕ = \frac{2 π x}{β}, I = I_{0} \cos^{2} (\frac{π x}{β})$

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