In Young’s double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890 A^{0}$ is :-

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$5890 \times 10^{- 7} m$

$1178 \times 10^{- 9} m$

$1178 \times 10^{- 12} m$

$1178 \times 10^{- 6} m$

answer is B.

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Detailed Solution

$β = \frac{λ D}{d} = \frac{5890 \times 10^{- 10} \times 0.5}{0.5 \times 10^{- 3}}$

Distance between first and third bright fringe is $2 β = 2 \times 589 \times 10^{- 6} m = 1178 \times 10^{- 6} m$

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