A beam of light consisting of two wavelengths 500 nm and 400 nm is used to obtain interference fringes in Young's double slit experiment. The distance between the slits is 0.3 mm and the distance between the slits and the screen is 1.5 m. Compute the least distance of the point from the central maximum, where the bright fringes due to both the wavelengths coincide.

Explanation

Given `lambda_1=500xx10^(-9)m, lambda_2=400xx10^(-9) m, d=0.3 xx10^(-3) m , D=1.5 m `
We know that `x=(n lambda D)/(d ) `
So that `x_1=x_2`
I,.e., `n_1 lambda_1=n_2 lambda_2`
`therefore (n_1)/(n_2)=(400xx10^(-9))/(500xx10^(-9))=4/5" ",n_1=4, n_2=5`
Here 5th bright fringe of `lambda_2` coincides with 4th bright fringe of `lambda_1`
Distance `x=(n_1 lambda_1 D)/(d)=(4xx500xx10^(-9)xx1.5)/(0.3xx10^(-3))=10^(4)xx10^(-9)xx10^3 m =10^(-2) m `
`x=0.01 m`