Calculate the fringe width of the interference pattern. Given wavelength of light used `=678` nm, distance between the slits `=0.35 `mm, distance between the screen and the double slit = 1 m.

Explanation

Given `lambda=678xx10^(-9) m , d=3.5xx10^(-4) m `
`D=1 m `
We know that the fringe width `beta=(lambda D)/(d)`
i.e., `beta=(678xx10^(-9)xx1)/(3.5xx10^(-4))=1.937xx10^(-3) m `
Fringe width obtained will be `1.937 xx10^(-3)` m.