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Two slits in Young's double slit experiment are placed 1mm from each other. Interference pattern is observed on a screen placed 1m from the plane of slits. What is the angular fringe width for a light of wavelength $400 n m ?$

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400 rad

$400 \times 10^{- 6} r a d$

$4 \times 10^{- 6} r a d$

$40 \times 10^{- 6} r a d$

answer is B.

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Detailed Solution

$d = 1 m m$ $= 10^{- 3} m$

$D = 1 m$

$λ = 400 n m = 400 \times 10^{- 9} m$

Angular width $θ = \frac{λ}{d} = \frac{400 \times 10^{- 9}}{10^{- 3}} r a d$

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