In a Young's expt., the width of the fringes obtained with the light of wavelength `6000 Å` is `2.00 mm`. What will be the fringe width if the entire apparatus is immersed in a liquid of `mu = 4//3` ?

Answer

`1.5 mm`

Explanation

Fringe width, `beta = (lambda D)/(d)`
When the entire apparatus is immersed in water,
`lambda' = lambda//mu`
`:. Beta' = (beta)/(mu) = (2.00)/(4//3) = 1.5 mm`