In a biprism experiment, light of wavelength 5200 Å is used to get an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards biprism by 50 cm. Find the distance between two virtual image of the slit.

Explanation

Given, `lambda=5200"Å"=5.2 xx 10^(-7)m`
`X_(1)-X_(2)=1.3 mm =1.3 xx 10^(-3)m`
`D_(1)-D_(2)=50cm=0.5m`
We know that, `X_(1)-X_(2)=(lambda D_(1))/(d)-(lambda D_(2))/(d)`
`X_(1)-X_(2)=(lambda)/(d)(D_(1)-D_(2))`
`therefore d=(lambda(D_(1)-D_(2)))/(X_(1)-X_(2))`
`=(5.2xx10^(-7)xx0.5)/(1.3xx10^(-3))`
`=0.002 m`
`=0.2 mm`