The slits in Young's double slit experiment, are $0.5 m m$ apart and interference pattern is observed on a screen distant $100 c m$ from the slits. It is found that the $9 t h$ bright fringes is at a distance of $8.835 m m$ from the second dark fringe. The wavelength of light will be

7529 A^{o}

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6253 A^{o}

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6779 A^{o}

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5890 A^{o}

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Solution

The correct option is C $5890 A^{o}$
Let $y_{1}, y_{2}$ be the positions of the 2nd dark fringe and 9th bright fringe respectively.
Therefore, $\frac{y_{1} d}{D} = \frac{3 λ}{2}$
And $\frac{y_{2} d}{D} = 9 λ$
Subtracting the above equations and rearranging gives,
$λ = \frac{2 (y_{2} - y_{1}) d}{15 D} = 5890 A^{\circ}$

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