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

Given : first case `lambda_(1)=4500 A`
`beta_(1)=5mm`
`=5xx10^(-3)m`
`beta_(1)=(lambda_(1)D)/(d)`
In the second case, the distance between the slit and .screen is — `(D)/(2)`
eqn (ii) and (i)
`(4xx10^(-3))/(5xx(10^(-3)))=(lambda_(2))/(2xx4500xx10^(-10))`
`4/5=(lambda_(2))/(9000xx10^(-10))`
`0.8xx9000xx10^(-10)=lambda_(2)`
`lambda_(2)=7200 xx10^(-10) m`
`lambda_(2) =7200 A`