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

Here, `lambda_(1) = 6000Å, beta_(1) = 2.0 mm`,
`mu = 1.33, beta_(2) = ?`
`lambda_(2) = (lambda_(1))/(mu) = (6000)/(1.33)`
`beta_(2) = lambda_(2)(D)/(d) and beta_(1) = lambda_(1)(D)/(d)`
`:. Beta_(2) = (lambda_(2))/(lambda_(1))beta_(1) = (6000 xx 2.0 mm)/(1.33 xx 6000) = 1.5 mm`