YDSE is carried out in a liquid of refractive index `mu=1.3` and a thin film of air is formed in front of the lower slit as shown in the figure. If a maxima of third order is formed at the origin O, find the thickness of the air film. Find the positions of the fourth maxima. The wavelength of light is air is `lambda_0 = 0.78 mum` and `D//d = 1000.` (##DCP_V05_C32_E01_086_Q01.png" width="80%">

Explanation

`beta= (lambda D)/( mu d)= (0.78)/(1.3)xx 10^(-6)xx 10^(3)= 0.6 mm`
Shift in the fringe pattern will be upwards
The refrective index of the air film is less than the medium Find the maxima `implies : 0.6mm" below "0`
`therefore -0.6mm`
`therefore 7xx 0.6mm= 4.2mm` above 0
The fourth maxima are `4.2mm` above 0 and `0.6mm` below 0.