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

Given, `beta=0.4mm`
Position of nth bright fringe from central maxima
`x_(n_(1))=(n_(1)lambdaD)/(d), " where " n_(1)=5`
`therefore x_(n_(1))=(5lambdaD)/(d)`
Position of nth fringe from central maxima
`x_(n)=((2n-1)lambdaD)/(2d), " where "n=3`
`x_(n)=(5)/(2)(lambdaD)/(d) rArr x_(n_(1))-x_(n)=(2.5lambdaD)/(d)`
`=2.5beta=2.5xx0.4`
`rArr x_(n_(1))-x_(n)=1mm`