In Young's double silt experiment, the fringe width obtained is 3 mm in air. If the apparatus is immersed in water `(mu=4//3)`,what will be the new fringe width?

Explanation

`beta = 3 mm = 3 xx 10^(-3) m`
Refractive index `= mu = 4//3`
Let `lambda =` Wavelength of light
d = slit width
D = Distance of screen from slit
`beta = (lambdaD)/d = 3 xx 10^(-3) m`
When the whole apparatus is immersed in water wavelength of light in water = `lambda.`
`=(lambda ("Wavelength in air"))/mu`
Fringe width in water `= beta. = (lambda.D)/d`
`beta. = 1/mu ((lambdaD)/d)`
`beta. = 9/4 = 2.25mm`