If fringe width is 0.4 mm, the distance between fifth bright and third dark band on same side is

Options

A. 1 mm
B. 2 mm
C. 3 mm
D. 4 mm

Answer

A

Explanation

Position of nth bright fringe from central maxima ` x_(n_(1))=(n_(1)lambdaD)/(d)` here `n_(1)=5`
`:. x_(n_(1))=(5 lambda D)/(d)`
Position of nth dark fringe from central maxima `x_(n_(1))=(n_(1) lambda D)/( d)` here `n_(1)=5 `
` :. x_(n_(1))=(5)/(2) (lambdaD)/(d)`
`x_(n_(1))-x_(n)=(2.5 lambda D)/( d)=2.5 beta `
Given ` beta =0.4 mm`
`implies x_(n_(1))-x_(n)=1 mm`