A glass of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8 . Light of wavelength `lambda` travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interface . If `lambda = 648 nm`, obtain the least value of `t ("in" 10^(-8)m )` which the rays interface constructively.

Answer

`2mut=(n-(1)/(2))lamda` with `mu=1.8` and `n=1,2,3.. 90,t_(min)=90nm`

Explanation

AB denotes incident ray. It is partly reflected from the upper surface of layer as `R_(1).R_(1)` is reflected from a denser medioum. It undergoes a phase change of `pi`. Part of AB is relfected from surface of layer as `R_(2).R_(2)` is reflected from a rares medium
as `.^(a)mu_(m)=1.8` and `.^(a)mu_(g)=1.5`
`R_(1)` and `R_(2)` therefore passess an initial phase difference of `pi` before they undergo interference net phase ddifference should be `2npi` where n is an integer.
`Deltaphi=2npi-pi(2n-1)pithereforex=(2n-1)(lamda)/(2)`
since `Deltax=2(mu_(m))t=1.8xx2t=3.6t`
or `3.6t=(2n-1)(lamda)/(2)`
For least value of t is `n=1therefore3.6xxt_(min)(lamda)/(2)`
or `t_(min)=(648)/(3.6xx2)nm` or `t_(min)=90nm`.