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

Given `lamda = 6800xx10 ^(-10)m=6xx10 ^(-7) m`
` lamda . =?`
`beta = 6 xx10 ^(-3)m, beta ^(1) =4xx10 ^(-3)` m
`D^(1)=(D)/(2)`
wkt fringe width `beta = (lamda D)/(d)`
Hence `(beta .)/(beta) =( lamda^(1)D^(1))/( canceld) xx ( canceld)/( lamda D)`
`i.e., ( beta ^(1))/( beta ) = ( lamda ^(1) )/( lamda ) xx ( cancel (D) ^(1))/( 2 cancel D) therefore beta ^(1) = ( lamda ^(1) beta )/( 2 lamda )`
`or lamda ^(1) = ( 2 lamda beta ^(1))/( beta )`
`i.e., lamda^(1) = ( 2xxcancel(6) xx10 ^(-7) xx4xx10 ^(-3))/( cancel (6) xx10^(-3)) m`
i.e., `lamda ^(1) = 8 xx10 ^(-7) m or lamda^(1) = 8000 Å`
The wavelength required to produce fringes of width 4mm is 8000 `A^(@)`