In YDSE the distance between screen and slits is 1.5 m and light of wavelength 500 nm is used. If angular fringe width is `0.3^(@)` in normal conditions then what will be the angular width (in degree) of the fringe if the entire experimental apparatus is immersed in water? [Take `mu_("water") =(4)/(3)`]

Answer

`00.22`

Explanation

since a fringe width `beta` is formed on the screen at distance D from the slits, hence angular width,
`theta =(beta)/(D) =((Dlambda)/(d))/(D)=(lambda)/(d)`
`rArr d=(lambda)/(theta)`
In water, wavelength changes `(lambda.)` and hence angular fringe changes `(theta.)" " .....(ii)`
`:.d=(lambda.)/(theta.)`
From equations (i) and (ii),
`(lambda)/(theta)=(lambda.)/(theta.)`
`rArr theta.=(lambda.)/(lambda)xx theta=(lambda//mu)/(lambda)xx theta ...( :.lambda.= (theta)/(mu))`
`:.theta.=(0.3^(@))/(4//3)=0.225^(@)~~0.22^(@)`