Young's double slit experiment is perfor-med inside water `(mu = 43)` with light of frequency `6 xx 10^(14)` Hz. If the slits are separated by 0.2 mm and the screen kept 1 m from the slits, find the fringe width. Using the same set-up, what will the fringe width be if the experiment is performed in air?

Explanation

Wavelength of given light in air can be obtained from the relation
` c = f lambda_("air") implies lambda_("air") = c/f = (3 xx 10^(8))/(6 xx 10^(14) )= 5 xx 10^(-7) m`
Wavelength of light of same frequency `(6 xx 10^(14) Hz)` in water will be different and ,
`lambda_("water") = (lambda_("air"))/(mu_("water")) = (5 xx 10^(-7))/(4//3)`
`= 3.75 xx 10^(7) m`
As the experiment is performed in water, so fringe width,
`beta = (lambda_("water") D)/(d)`
Here, `D = 1 m` and `d = 0.2 mm = 0.2 xx 10^(-3) m`
`:. beta = (3.75 xx 10^(-7) xx 1)/(0.2 xx 10^(3)) m`
or `beta = 1.87 mm`
If the experiment is performed in air, then fringe width,
`beta = (lambda_(air) D)/(d) = (5 xx 10^(-7) xx 1)/(0.2 xx 10^(-3)) m`
`= 2.5 mm`.