In Young's double slit experiment, the distance of the $n^{t h}$ dark fringe from the centre is-

n (\frac{λ D}{2 D})

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n (\frac{2 d}{λ D})

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(2 n - 1) (\frac{λ D}{2 d})

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(2 n - 1) (\frac{4 d}{λ D})

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Solution

The correct option is C $(2 n - 1) (\frac{λ D}{2 d})$
For $n^{t h}$ dark fringe path difference is
$Δ x = (2 n - 1) \frac{λ}{2} = \frac{d y}{D}$
$y = (2 n - 1) \frac{D λ}{2 d}$

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In Young's double slit experiment, the distance of the nth dark fringe from the centre is-

The correct option is C (2n−1)(λ D2d)For nth dark fringe path difference is Δ x=(2n−1)λ2=dyD y=(2n−1)Dλ2d

In Young's double slit experiment, the distance of the $n^{t h}$ dark fringe from the centre is-

The correct option is C $(2 n - 1) (\frac{λ D}{2 d})$
For $n^{t h}$ dark fringe path difference is
$Δ x = (2 n - 1) \frac{λ}{2} = \frac{d y}{D}$
$y = (2 n - 1) \frac{D λ}{2 d}$