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The intensity at the maximum in a Young’s double slit experiment is I₀. Distance between two slits is $d = 5 λ,$ where $λ$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?

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$I_{0}$

$\frac{3}{4} I {}_{0}$

$\frac{I {}_{0}}{2}$

$\frac{I {}_{0}}{4}$

answer is D.

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Detailed Solution

$Δ x = d \tan θ = d \times \frac{y}{D} = \frac{d \times \frac{d}{2}}{10 d} = \frac{d}{20} = \frac{5 λ}{20} = \frac{λ}{4}$

$\begin{array}{l} I_{max} = I_{1} + I_{2} + 2 I_{1} I_{2} \\ I_{0} = 4 I ∵ I_{1} = I_{2} = I \\ ∴ I = \frac{I_{0}}{4} \end{array}$

$Corresponding phase difference, ϕ = \frac{2 π}{λ} Δ x = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} .$

$Intensity at y is I_{y} = I_{1} + I_{2} + 2 I_{1} I_{2} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

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