In Lloyd's interference experiment, 10 fringes occupy a space of 1.5 mm. The distance between the source and the screen is 1.25 m. If light of wavelength 6000 `Å` is used, find the distance of the source from the plane minor.

Explanation

Here `beta = (1.5)/(10)mm=0.15xx10^(-3)m`
`D=1.25m`,
`lambda = 6000 Å = 6xx10^(-7)m`
As `beta = (D lambda)/(d)`
`therefore d=(D lambda)/(beta)=(1.25xx6xx10^(-7))/(0.15xx10^(-3))m`
`=50xx10^(-4)m=5.0 mm`
Hence distance of source from the plane mirror
`= (d)/(2)=2.5` min.