Light of wavelength 6000 `overset(@)A` is used to obtain interference fringe of width 6 mm in a young's double slit experiment. Calculate the wavelength of light required to obtain fringe of width 4 mm if the distance between the screen and slits is reduced to half of its initial value.

Explanation

`beta=(lambdaD)/d`
`beta_1=(lambda_1D_1)/d beta_2=(lambda_2D_2)/d`
`beta_1/beta_2=(lambda_1D_1)/(lambda_2D_2)`
Substitution and simplification : Arriving upto `lambda^2` = 8000 Å
Given, `lambda=6800xx10^(-10)m = 6xx10^(-7)` m
`lambda.`=?
`beta=6xx10^(-3)m , beta.=4xx10^(-3)m`
D.=D/2
`rArr` fringe width , `beta=(lambdaD)/d`
Hence, `(beta.)/beta=(lambda.D.)/d xx d/(lambdaD)`
i.e., `(beta.)/beta=(lambda.D.)/d xx d/(lambdaD)`
`therefore beta.=(lambda.beta)/(2lambda)`
or `lambda.=(2lambdabeta.)/beta`
i.e., `lambda.=(2xx6xx10^(-7)xx4xx10^(-3))/(6xx10^(-3))`
i.e., `lambda=8xx10^(-7)` m
or `lambda.`=800 Å
The wavelength required to produce fringes of width 4 mm is 8000 Å.