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In double slit experiment, the distance between two slits is $7.7 μm$ and these are illuminated with light of wavelength $5000 A °$ . The angular position of the third dark fringe on the screen at a distance from slits will be

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${10.9}^{\circ}$

${0.15}^{\circ}$

${11.3}^{\circ}$

${9.4}^{\circ}$

answer is D.

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Detailed Solution

n = 3

$θ = \frac{(2 n - 1) λ}{2 d}$

$θ = \frac{5 λ}{2 d} = \frac{5 \times 5 \times 10^{- 7}}{2 \times 7.7 \times 10^{- 6}} \times \frac{180}{π} = 9 . 4^{0}$ =

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