In Young's double slit experiment while using a source of light of wavelength 4500 A, the fringe width is 5mm. If the distance between the screen and the plane of the slits is reduced to half, what should be the wavelength of light to get fringe width 4mm?

Explanation

Here :` lamda_1 = 4500 A&^@ , beta_1 = 5 xx 10^(-3) m , D_1 =D_1D_2 , beta = 4xx 10^(-3)m`
we have ` beta_1 = ( lamda_1 D_1 )/(d) , beta_2 = (lamda_2 D_2)/(d)`
`(beta_1)/(beta_2) = (( lamda_1 D_1)/(d))/((lamda_2 D_2 )/(d))`
`implies (beta_1)/(beta_2 )=( lamda_1 D)/( lamda_2 (D)/2)=? (beta_1)/(beta_2) = (2 lamda_1)/(lamda_2)`
` lamda_2 = (2 lamda_1 xx beta _2 )/( beta_1 ) = (2 xx 4500 xx 10^(-10) xx 4 xx 10^(-3))/(5 xx 10^(-3))`
` lamda _2 = 7200 xx 10^(-10) m`
` lamda_2 = 7200 A^@`