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The distance between the double slits in the Young’s experiment is 1.0 mm. When the screen is 1 m from the slits, the fringe width is found to be 0.6 mm. The wavelength of the light used is

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$7800 Å$

$5896 Å$

$5890 Å$

$6000 Å$

answer is D.

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Detailed Solution

$β = \frac{λ D}{d}$
$0.6 \times 10^{- 3} = \frac{λ \times 1}{10^{- 3}}$
$λ = 6 \times 10^{- 7} m$
$λ = 6000 A °$

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