In Young's double slip experiment ,the width of fringes obtained from a source of light of wavelength 5000 `Å` is 3.6mm Calculate the frings width if the apparatus is immersed in a liquid of refractive index 1.2.

Explanation

Given, wavelength of source , `lambda = 5000 Å`
Fringe width , `beta = 3.6 mm`
Refractive index of liquid is
`mu = ("wavelength of light in vacuum")/("wavelength of light in liquid")`
`= (lambda)/(lambda.) = 1.2`
`implies lambda. = (lambda)/(1.2)`
Since, `beta prop lambda`
`:. (beta.)/(beta) = (lambda.)/(lambda) = (lambda//1.2)/(lambda)`
`implies beta. = (beta)/(1.2) = (3.6 mm)/(1.2) = 3mm`.