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In young’s double slit experiment, the separation between two coherent sources $s_{1} and s_{2}$ is d and the distance of screen from the sources is D(D>>>d). In the interference pattern it is observed that a maximum is formed exactly infront of one slit. Then minimum possible wavelength used in the experiment is

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$\frac{d^{2}}{8 D}$

$\frac{d^{2}}{4 D}$

$\frac{d^{2}}{2 D}$

$\frac{2 d^{2}}{D}$

answer is C.

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Detailed Solution

Distance of nth maximum from the central maximum is given by $y_{n} = n . \frac{λD}{d}$
$Here y_{n} = \frac{d}{2} \Rightarrow \frac{d}{2} = n . \frac{λD}{d} [n = 0, 1, 2, 3, . ....]$
For minimum length, n=1
$∴ λ_{\min} = \frac{d^{2}}{2 D}$

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