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. It is partly reflected at the upper and the lower surfaces of the layer ant 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`.