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In young's double slit experiment $\frac{d}{D} = 10^{- 4}$ (d = distance between slits, D = distance of screen from the slits). At a point P on the screen resulting intensity is equal to the intensity due to the individual slit I₀. Then the distance of point P from the central maximum is $(λ = 6000 Å)$

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0.5 mm

1 mm

2 mm

4 mm

answer is A.

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Detailed Solution

$I = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$

$I_{0} = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$

$\cos (\frac{ϕ}{2}) = \frac{1}{2}$

$or \frac{ϕ}{2} = \frac{π}{3}$

$or ϕ = \frac{2 π}{3} = (\frac{2 π}{λ}) \cdot Δx$

$or \frac{1}{3} = (\frac{1}{λ}) y \cdot \frac{d}{D} (Δx = \frac{yd}{D})$

$\Rightarrow y = \frac{λ}{3 \times \frac{d}{D}} = \frac{6 \times 10^{- 7}}{3 \times 10^{- 4}}$

$= 2 \times 10^{- 3} m = 2 mm$

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