In a Young's double slit experiment , the slits are Kept `2mm` apart and the screen is positioned `140 cm` away from the plane of the slits. The slits are illuminatedd with light of wavelength `600 nm.` Find the distance of the third brite fringe. From the central maximum . in the interface pattern obtained on the screen frings from the central maximum. `3`

Explanation

Here `d=2 mm =2xx10^(-3) m, D=140 cm`
`=1.40 m`
`lambda=600 nm =600xx10^(-9)=6xx10^(-7)`
Position of bright fringes is given by `D`
`x_(n)=3lambda(D)/(d) =3xx6xx10^(-7)xx(1.40)/(2xx10^(-3))`
`10.08xx10^(-4)=1.008xx10^(-3)=1.01xx10^(-3)m=1.01 mm`
`therefore` Shift in the position of third bright fringe `=1.26-1.01=0.25 mm`.