In Young's double slit experiment the slits are 0.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slit. It is found that the 9th bright fringe is at a distance of 7.5 mm from the second dark fringe from the centre of the fringe pattern. The wavelength of the light used is

\frac{2500}{7} A^{\circ}

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2500 A^{\circ}

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5000 A^{\circ}

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\frac{5000}{7} A^{\circ}

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Solution

The correct option is C $5000 A^{\circ}$
Given,
$d = 0.5 m m$
$D = 100 c m$
$X_{9} - X_{2}^{'} = 7.5 m m$ . . . . . . . .(1)
Condition for bright fringe,
$X_{n} = \frac{n λ D}{d}$
for $9^{t h}$ bright fringe
$X_{9} = \frac{9 λ D}{d}$
Condition for dark fringe,
$X_{n} = \frac{(2 n - 1)}{2 d}$
$X_{2}^{'} = \frac{(2 \times 2 - 1) λ D}{2 d} = \frac{3 λ D}{2 d}$
From equation (1)
$\frac{9 λ D}{d} - \frac{3 λ D}{2 d} = 7.5 \times 10^{- 3} m$
$\frac{15 λ D}{2 d} = 7.5 m m$
$λ = \frac{2 d}{15 D} \times 7.5 m m$
$λ = \frac{2 \times 0.5 m m}{15 \times 100 c m} \times 7.5 m m$
$λ = 5000 \times 10^{- 10} m = 5000 A^{\circ}$
The correct option is C.

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